in-exercises-37-40-use-a-matrix-program-to-find-the-eigenvalues-of-the-matrix-then-use-the-method-2

Question

In Exercises $37-40$ , use a matrix program to find the eigenvalues of the matrix. Then use the method of Example 4 with a row reduction routine to produce a basis for each eigenspace.
$$
\left[\begin{array}{rrrr}{9} & {-4} & {-2} & {-4} \ {-56} & {32} & {-28} & {44} \ {-14} & {-14} & {6} & {-14} \ {42} & {-33} & {21} & {-45}\end{array}ight]
$$

Gennady Notowidigdo · Accepted Answer

Okay, so we are given This are four by four matrix, which I will hard light in green. So what, we&#x27;re gonna do it be going to use a computer package? So I used Mabel. If you want to use Matt, Matt or Mathematica, I feel free to do so or something else off that, like again, feel free to use whatever computer package you desire. So once I have this four by formation, so I want to find the Eiken values and the corresponding Aiken vectors, or sort of a basis for the Eiken space, so to speak. All right, so two. So, once I used the computer package, so let me just sort of figure out what this character is. It polynomial is. So if I obtained the characteristic polynomial, it&#x27;s going to be our to the four miles to are cute, minus 311 are square plus 312 aw, plus, 24,000 336. And I&#x27;m sure that this fact is nicely, um, so factors into, uh, plus 12 times our minds 13. A love that square good. And when we solved the zeros off the characteristic polynomial which is in this case minus 12 and 13 are then what we called the Aiken Valleys. So for this particular matrix, we have to Aiken values minus 12 and 13 case are so that our oppressed it yet. Okay, So for the Eiken battle, it&#x27;s for the Eiken value. Say Lambda equals two. I&#x27;m with these arms. That lab, just to make it easier, are equal to 13. Then if we compute, uh, our time, the identity matrix minus the matrix that we have, let&#x27;s call it a Okay, so let&#x27;s call it a Are I minus A. This ends up being for 4 to 4. 56 minus 19 28 minus 44. And then we just raised this just to make make it need us to be minus 19. Um, just dish it black. It&#x27;s minus 19. Then you got 14. 14 seven on 14. That should be 14 uh, minus 42. 33 Mona&#x27;s 21 and 58. Okay, now the Eiken space for our the Eigen value, our eagles 13 is pretty much just the null space off our times. I minus a or in other words, the Colonel. Okay, if you want you can. You can interject two things. Although I reserve the word colonel for lenient maps and no spaces for matrices. Okay, Now this matrix reduces too to to want. So I&#x27;ve done it in its simplest form. So it ends up reducing to something like this. Okay, so at this point, you can use back substitution and you&#x27;ll end up for the basis for the null space ends up being It will be minus 10 to 0 and one minus 40 and three. So obviously, if you did back substitution, you would set these two quantities to be one, and you work backwards, but I&#x27;ve just sort of normalized. It&#x27;s another normal, but I&#x27;ve made it fraction free just for our simplicity purposes. Okay, so we&#x27;ve got a basis for the eye Gance base for articles 13 and similarly, I will do in the next page, sir. I&#x27;ll keep it in black financing for the Eigen value. Let our used ob for I mean, it doesn&#x27;t really matter. I just like using are because we&#x27;re sticking to English letters rather than moving to Greek letters. But in any case, so for the Eiken value are equal to minus 12. Are I modest, eh? You know, when you write out what its people to sort of figure that out on your own. But it reduces to Mona&#x27;s 21 for 24 Sara Wan. Modest one of one. And in two bars, off zeros on a basis for the null space. So for the Eigen space for article 2 12 minus 12. Sorry is 2770 and zero. Modest one, sirrah. One. Okay, so I&#x27;ll leave it there. Uh, please verify older cat places that I&#x27;ve made Adlai Vilo on the computer or by hand, But in any case, thank you very much.

Chris Trentman · Answer

were given a matrix and were asked to or thought generally. Diagonal eyes. This matrix the Matrix is 0.63 negative 0.18 negative 0.6 points or 4.4 negative 0.4 0.12 negative point of six negative point. 04 0.72 negative 0.12 negative point of four 0.12 negative 0.12 166 I noticed that matrix a transpose is equal to the diagonal stays the same. So we have 0.63 point 4.72 point 66 no, and we exchange negative point went for negative 0.0.18 negative point of six For negative point of six. They get appointed four for negative point before negative point of four for negative point of four 0.12 for point went to and the negative 0.12 for negative 0.12 We see that a transpose is the same as a and so it follows that a is a symmetric matrix and therefore, since a symmetric it follows that a is orthogonal e diagonal izabal to or thought Nelly diagonal eyes A. They want to find its Eigen values. You can be found by solving the terrorist equation. So we have the determinant of a minus Lambda I is equal to now. I could write this out, but the thing is, this is a four before matrix, so calculating the determinant is going to be rather difficult. This is where you want to use computer October system or maybe a program like maple find the Eigen values so you never know. Pause here. And so, using Wolfram Alfa, I found that this characteristic polynomial could be written in factored form as Lambda minus one lambda minus one lambda minus 10.8 lambda minus 0.55 Lambda minus 0.5. You want to find the roots of the characteristic polynomial, so we&#x27;ll set this equal to zero. And so we obtain the Eigen. Values are Lambda one equals one, Amanda two equals 20.8 and the three equals 30.55 And Linda four equals 40.5. Yes, the is an ID injector associated with Lambda one, and it satisfies a minus. I the equals zero vector The matrix a minus eyes the same as the matrix. A. Except for now, the entries in the diagonal or one less this is going to be and I could write this out, but I&#x27;ll go straight from here to reduce row echelon form of this matrix A minus. I so have the A minus. I can be reduced to menopause. Here we have this reduces to matrix 1001 010 negative too 001 0.5 In a row of zeros which we obtain the system of equations V one is equal to negative before the two is equal to positive. To be four in V three is equal to 0.5 times before or negative 0.5 before. So if we take the four to be the parameter negative to T, then the general form for the vector V is going to be the one is negative. Negative two tier two t He too was two times negative to tea or native 40 b three is negative 30.5 times negative to t or simply T. And what V four is negative two t which is the same as the vector tee Times K to negative four one negative too. And if we take TV one, we obtain the Eigen vector V one which is Tiu Negative for one negative too. The norm of this I in Vector V one is the square root of four plus 16 plus one plus four. This is 25. So this is this. Where? 25 5 And so we have that The unit vector you one which is V one over the normal view. One is the vector 2/5 Negative. 4/5. 1/5 Negative. 2/5. Likewise. If the is an Eigen vector associated with the Eigen, value landed too. It satisfies a minus 0.8 I the equals zero. And again pause here. I&#x27;m going to use a reduction to reduce a minus 0.8. I after wrote a reduction. I obtained the matrix 100 negative. One 10.5 and 0012 In the rest of zeros, which I obtain a system of equations V one is equal to positive before V two is equal to 0.5 before and the three is equal to negative too. This is native 0.5 before my mistake V three is negative two before. So if I take the four to be the parameter t t let&#x27;s choose to t the general form for the Eigen Vector V is now going to be to t negative 0.5 times to tea is negative. T negative two times two t is negative 40 and to t This is the same as the fact as t times the vector to negative one negative four to And if we take t to be one, we get the I injector be to which is equal to to negative one negative four to the norm of V two is the square root of four plus one plus 16 plus four where 25 which is five. And so we have the unit vector you two is equal to 2/5 Native, 1/5 native 4/5 2/5. If v isn&#x27;t Eigen vector associated with William to three then it satisfies a minus 0.55 i the equals is your vector The matrix a minus 0.55 I rib uses to. But when the pause here after a reduction introduces to the matrix 100.25 10.5 and 001 negative 10.5 which we obtained A system of equations. V one is equal to negative 0.5 point 25 before for written other way, this is four V one equals negative before, Maybe I shouldn&#x27;t do that. We have. The V two is equal to negative 20.5 before and V 30 with a positive 300.5 before. If we take the four to be the parameter negative four t in the general form for the Eiken Victor V is negative 0.25 times negative four, which is simply won t negative 0.5 times negative four, which is positive to t 0.5 times negative four, which is negative two t and 40 just the same as t times. The vector one to negative too four. And so if we take tea to be one, we obtain the Eigen Vector B three, which is one to negative 24 The norm of E three is the square root of one plus four plus four plus 16 which is equal to the square 25 or five. And so the unit vector you three is the vector V three over the norm of V three, which is 1/5 2/5 negative. 2/5 4/5. Finally, if he isn&#x27;t I? Connector associated with? I can value Lambda Four and it satisfies the equation. A minus 0.5 I the equals zero The matrix A minus 0.5 I reduces to meant one. A pause here produces to the matrix 100 negative. 4010 Negative. 2001 negative too. From which we obtained system of equations the one is equal to for before V two is equal to two he four he three is equal to before. If we take before to be the parameter t, we have the general form of the Eigen Vector V is 40 to t to t t or t times the vector for two to one from which we obtain if t equals one the Eigen vector for 2 to 1. The norm of the four is the square root of 16 plus for plus for plus one which is the square to 25 or five in the unit Vector you four is the vector before divided by the norm of before, which is the vector 4/5 2/5 2/5 1/5 And because of symmetric matrix, we have that the set of unit vectors. U one You too. You three and you four are a North. A normal set of vectors, each of which is the basis for its Eigen space. It is an idea nectar and so two diagonal wise a Take the matrix p to be the matrix. Whose columns are you went through you for? Or two fists? Negative 4/5. 1/5 mega tooth It&#x27;s you two is 2/5 negative. 1/5 Negative. 4/5. 2/5 u three is 1/5 2 5th Negative. 2/5 4 5th Can you? Four is forfeits 2/5 to Fitz 1/5 and take the matrix D to be the diagram matrix. Whose entries are the Eigen values? Corresponding. Do you went through you four. So we start with Lambda one, which is one then point eights than points 5.5 when you could write these infraction for him if he wanted. But this is also acceptable. And the rest of the entries are zeros

Chris Trentman · Answer

were given a matrix, you were asked two orthogonal E di act allies. This matrix Matrix&#x27;s 8 to 2 negative. 69 282 Negative 69228 Negative 69 Negative six. Negative six. Negative six 24 nine 99 nine nine Native 21 and I could read out the transpose and show that it&#x27;s equal to a But I think, just been looking at this matrix. It&#x27;s pretty clear that the entries opposite each other on the diagonal across the diagonal are equal. So it&#x27;s clear that a transpose is equal to a which means that A is a symmetric matrix and therefore follows that a is fourth organ elite diagonal Izabal. So this problem makes sense to our Thorgan only diagonal eyes. A first we&#x27;ll find the icon dies of a characteristic polynomial of a is the determined of a minus. I am toe I, and this is a five by five metrics, so calculating the determinant could take a very long time. Instead, I&#x27;ll use a computer program to do this, and that went to do the same and pause here. So, using a computer algebra system, I calculated that the characteristic polynomial is in factored form Lambda plus 30 times Lambda minus 30 times Linda minus 15 times Linda minus six squared. You want to find the solutions to the characteristic equation? Just once equals zero. So every to the Paul, no meal. And so we have their Eigen values are Lambda one equals negative. 30 I&#x27;m 22 equals positive 30 and the three equals 15. And Linda four, is equal toe. Linda five is equal to six. So it has a multiplicity of to If he is an Eigen vector associated with Lambda one, it satisfies the equation. A plus three i times v equals zero vector to find B. Let&#x27;s reduce the matrix a plus 30. I pause so you can do this yourself. After reduction, we obtain the matrix 10004 or when fourth I mean so you&#x27;re a 100 1/4 zero zero&#x27;s 10 1/4 000114 This red last year was actually just zero room and so we obtained the equations. The one is equal to negative 1/4 before, which is also equal to the two, which is equal to be three And so if we take the four to be the parameter negative 40 and we obtain general form of the Eigen vector, the is negative. 14 attempts in 84 has won t t t native 40 core tee times the vector 111 negative 14 Therefore I mean to take TV one. We obtained the Eigen vector one, which is 111 negative four the normal V one. Well, we have that The unit vector you one is the one over the norm of the one which is not everyone is one plus one plus one plus 16 square rooted, which is Route 19. So that&#x27;s going to be one of a Route 19 one of the Route 19 one of her 19 and negatives 4/19. I actually didn&#x27;t make the steak. This should be 1 1/4 here as well and then another row of zeros. And so they should actually be that this is all equal to negative. 1/4 the five. And so if we take the five to be the parameter negative 40 he obtained T T t t native 40 which is 111 one negative four now the norm is 16 plus one. 17 18 19 squared 20 Which is going to be to rid five. This is one over to route five. One of her to rid five one over to room five and negative 4/2 5 which is negative to over Route five. Likewise, if he isn&#x27;t Aiken Victor associated with Linda to it satisfies a minus 30 I the equals zero vector. Find a V will have to reproduce a minus 30. I might wanna pause here this time we obtain Matrix 100 1/3 0 010 1/3 0 in 001 When 3rd 0 0000 one in a row of zeros. So we obtain the system of equations. V one is equal to negative 1/3 before V two is also equal to negative on third before, as is the three we have that fee five is equal to zero. So if we take the four to be the parameter negative three t, then we have the General I conductor V. It&#x27;s going to be t t t negative three t zero or tee times a vector 111 negative 30 If we take tea to be one we obtained the Eigen vector be to which is 111 negative three zero unit Vector U two is V two over the normal V two and the normal V two is square root of 12 which is 233 And so this is going to be one over to three. 1/2 or three. 1/2 or three. Negative 3/2 3 and zero. Next. If V is an Eigen vector associated with Lambda three, it satisfies a plus 15. I, our a minus 15. I mean, I times v equals a zero vector. So we want to reduce matrix a minus 15 i And when a pause and do this yourself after row reduction, we obtain the matrix. 1000 negative. 10100 neck of 10010 Negative warning 0001 negative one And the rest of zeros. From this we obtain the system of equations. The one equals B five, as does B two b three and before. So if we take the five to be the parameter t and we have General form V is simply t t t t t or tee times, all ones. So if you take TV one we obtained you I connector V three, which is all ones, and we have that you threw. The unit vector is B three over the normal V three, which is normal. V three is going to be the square root of five. So this is one of unverified. Whatever Route five. Whatever. Route five. One of every five. Whatever. Five. And next we have that. If he isn&#x27;t Eigen vector associated with Lambda Four or lamed of five that falls that it satisfies the equation. A minus six i times V equals zero. So we want to reproduce the matrix a minus six. I I want to pause while you do this on your own. After a reduction, we obtain the matrix. 11100 0001000001 And the rest of the entries are zero. So we obtained the system of equations. Excuse me. V one equals negative two minus 83 and V four and the five are equal to zero. And so if we take V two to be parameter negative s in the three to be the parameter negative t we obtain That V is the vector s plus t negative s native T 00 is equal to the linear combination s times one negative. One 000 plus two times one zero negative one 00 And if we take s to be one and t v zero, we obtained the Eigen vector before, which is one negative 1000 And if we take s to be zero in t to be one, we obtain the second Eigen vector the five She&#x27;s 10 negative warm 00 by construction. We have that before and the fiber linearly independent. However, we have that since their dot product is non zero even though they&#x27;re linearly independent and said they&#x27;re a basis of I didn vectors you have that before out of it be five is equal to one which is not zero and therefore we have it before is not orthogonal to be five Tina North Ogle basis of Eigen vectors will use the Gram Schmidt or thought generalization procedure. So let&#x27;s w four equal to be four ticked up You five to be be five minus the projection of the five under the four his before That would be five over the normal before times before this is equal to the five. This one&#x27;s your negative. 1000 Here is your I guess, minus. We calculated this to be one over the normal before, which is route two times before, which is one negative. 1000 This is equal to one minus one of the route to zero plus one of route to negative one minus zero and 00 They should actually be the norm of the four squared here. So instead of being divided by route to this is divided by two. So this is actually not one minus one of her two, but one minus 1/2 which is 1/2 of illnesses. Also, 1/2 this is negative ornament. 00 And so we have that. You four is the unit vector, which is W four over the normal w four just going to be one of the route to negative one of two 000 And do you five this unit vector W five over the norm of w fine, which is equal to instead doing this. I&#x27;ll do to W five over normal to W five sense or thought banality, and I can tell you stole preserved and this is going to be. While the norm of two w five is one plus one plus four square rooted just squared. Six. This is going to be one of a Route six 1/6 negative two of the route six 00 and we have that u four and u five former North Normal basis where this wagon space. And so we have that the set of vectors u one you too. U three u four and you five are a complete set of work, the normal Eigen vectors of a and so we can diagnose eyes A by taking p to be matrix. Whose column vectors are you went there, You five War one of her to rid 51 of her to rid 51 of her two word five negative two of every five. You, too, which is one of her two or 3 1/2 or 31 of her two or three negative 3/2 or 30 one of a Route 51 of the routes life one of a roof life in every fight whenever. Five one of her to negative one of 2000 And whatever Route 61 of a Route six Negative two of her Route six since year zero. This is one option for P and for B take you to be the Matrix just diagonal into his entries of the Eigen values in order. So we start with negative 30 then 30 then 15 and six and six. The rest of the entries are zero. This is our answer.

Amy Jiang · Answer

Okay, so we use it on like Catholic calculators to solve this. Three chicks we could 70 So, So, So, So. Three year old girls out there are always little dreams. Oh, it&#x27;s out todo 050000 fight. So this is our diagonal matrix. What&#x27;s five inches? And there&#x27;s three with multiples is two and five with multiplicity.

Chris Trentman · Answer

were given a matrix, you were asked two orthogonal lee diagonal eyes. This matrix matrix is 6 to 9 Negative 626 Negus 69 nine Negative 662 Negative 69 to 6. You have that The matrix a transpose is still six is on the diagonal and then the rest We have it to exchange for two nine exchanged for a nine negative six exchanged for a negative six. A negative six exchange for negative six A nine exchanged for a nine and two exchanged for two and we see that a transposed is the same as a And so it follows that it is a symmetric matrix and therefore a is or fog Nelly diagonal Izabal My theory to tore thought grimly diagonal eyes This matrix first need to find the item values This is going to be the lamb. Does it satisfy the characteristic equation? The determinant of a minus Lambda I equals zero. We have that The determinant of a minus claimed the I is the determinant of the matrix any Except for now there are Lando&#x27;s So six minus slammed down the diagonal instead of six. Then 29 Negative six Negative 692 and fill in the rest. Using symmetry. This is a four by four matrix, so calculating a determinant might take a while. In this case, I&#x27;m going his computer algebra system. Or if you have a program set up through something like Maple, you can use that to calculate the Eigen values. And so, using Wolfram ALF, I obtained that the characteristic polynomial is landed in the fourth, minus 24 landed cubed minus 26 minutes, 26 Lambda squared plus 2904 Lambda minus 4 11,095 So this is equal to zero. Another form of this characteristic polynomial could be written in, and you can do this just by plugging in and finding the factors. But does the same Islam to minus 19 times Linda minus 11 times lambda, minus five times Linda plus 11. And so we have the Eigen values or now go from least a greatest lambda. One is negative. 11. We&#x27;ll take landed to to be 5 23 is 11 and Lambda four is 19 so we have her four agon values, each with a multiplicity of one, so each Eigen space has a dimension of one by a serum from the book. And so in the case of Lambda One, we have that if the is an Eigen vector associate with limbed a one, and it satisfies a minus negative 11 or plus 11 I times V equals zero Vector The matrix A plus 11 eyes, the same as the Matrix. A. Except for now, instead of sixes in the diagonal. We have seven teens in the diagonal, so this is going to be 17 to nine. Negative. Six, 17 17 17 negative 69 to and you symmetry to fill in the rest. My one to pause here and use it. Row reduction. I&#x27;m going to use produce reduction. Materazzi&#x27;s echelon form After run reduction. We obtain the matrix 100 negative. One 01010011 than a row of zeros. This totals. We have one degree of freedom and we obtain the system of equations. V one minus V four equals zero v two plus before equals zero. And these three plus before equals zero, which tells us that the one is equal to be four and the two is equal to the three which is equal to negative before. So if we take the one to be the parameter t in the general form for the Eiken Victor V, his V equals t negative t negative T t which is the same as t times the vector one negative one negative 11 And so if we set t equal the one we obtain the Eigen vector the one which is one negative one negative 11 We want to find an Ortho normal basis for this IED in space. So you already have. This is a basis for the second spaces orthogonal to normalize it, we have that you one is the unit vector which is V one divided by the normal view one which is equal to no. Everyone is going to be the squares of four, which is too. So this is 1/2 native, 1/2 negative, 1/2 1/2 or land it too. Likewise, if he is an Eigen vector associate with limbed to than it satisfies a minus five i the is equal to zero vector the matrix a minus five I is the same as matrix A except for now diagonal is all ones instead of all six is so we have one 111 29 Negative six Negative. 692 And you symmetry to fill in the rest pause. Here. I&#x27;m going to apply for a reduction to get produced. Rocha one form upon row reduction. I obtained a matrix. 10010101 001 Negative. One 0000 So I have one degree of freedom and I obtain system of equations. B one plus V four equals zero b two plus before equals zero and the three minus before equals zero. So I have that V one is equal to negative before the two is equal to negative before and V three is equal to before. So if I take the four to be parameter negative t in the general form For Gagan Victor V is T Teoh negative t negative team which is the same as tee times a vector 11 negative one negative one If I take tea to be one. I obtained Eigen Vector V two, which is 11 negative one negative one. This is already a basis for this wagon space and therefore it is an orthogonal set of basis vectors for this Aiken space. Normalize this. We have that The unit vector You too. Which is V two over the normal B two. Well, the normal V two is going to be the square root of four, which is one. So this is it&#x27;s very forwards to So this is 1/2 1/2 negative. 1/2 negative. 1/2 this isn&#x27;t worth normal. Set for this wagon space. Or if V is an itin vector associated with gland of three, we have that. It satisfies a minus. Lander three, which is a minus 11 I times V equals zero vector the matrix a minus 11 eyes. The same is a except for your place. Dagnall, with instead of sixes will be six minus 11 which is negative fives. So we have now negative five negative five negative five negative. Five, 29 Negative six negative. Six, nine two And then you symmetry to fill in the rest. This is 2986 negative. Six nine to pause here, I&#x27;m going to apply reduction upon row reduction. This reduces to the matrix. 100 Negative. 1010 Negative 1001 negative one The number of zeros. You see that we have one degree of freedom and we obtain the system of equations V one minus before equals zero be to minus before equal zero and the three minus before equals zero. So that V one is equal to V two is equal to B three is equal to before and if we take this to be the parameter t them to get that The general form for Gagan Victor V is the vector t t t t sanest. He times the vector 1111 And if we take TV one, we obtain the I connector b three, which is 1111 We have that v three is and instead of orthogonal trivially basis vectors for this, like in space Me one north A normal set. So we have that you three is the Yoon effect er b three over the norm of the three, the normal V three is square root of four which is to So this is going to be a vector 1/2 1/2 1/2 1/2 We have that you three isn&#x27;t orphan normal basis for this Eigen space Finally, if these satisfies was associated with Lambda four, which is 19 it satisfies a minus 19 I the equals zero matrix a minus 19. I is the same as Matrix A except for the Diagne or in trees are now six minus 19 which is negative. 13 instead of six. Yes, we have Tina Matrix. Negative six. Negative 13. Pause here. I&#x27;m going to apply real reduction. Conroe reduction. I have Teen the Matrix. 1001010 Negative. 10011 from which I obtained the system of equations B one plus before equals zero. We will be two minus before equals zero n v three plus he for equal zero which I obtain That V one is equal to negative before V two is equal to V four and V three is equal of you. One is equal to negative before and if I take before to be the parameter negative t I have seen the general form for the vector V is going to be t negative. T t negative T or t times the vector one negative one one negative one If I take TV one. I obtained the Eigen vector before, which is one negative 11 negative one. And I have that before is already and or thought Gunnell set of basis vectors for this Agon space. The vector you for is the unit vector, which is before, over the normal before the normal before is going to be square to four, which is to So this is the vector. 1/2 negative, 1/2 1/2 negative, 1/2. And so I have that you four is a Ortho normal basis for this, like in space. And because the Matrix A is symmetric, it follows that u one u two u three u four north ogle in fact, there, Ortho Normal since there are unit vectors And so diagonal eyes The matrix a take p to be the matrix. Who&#x27;s calling vectors? Are you one through you four? We&#x27;re the vectors. 1/2 negative, 1/2 negative. 1/2. 1/2 You too. Which is 1/2 1 half negative. 1/2 negative one Have you three, which is 1/2? 1/2 1 half, 1/2 And you four, which is 1/2 negative. 1/2 1 half negative, 1/2. So this is the Matrix being and pick Matrix D to be the diagonal matrix whose entries are the Eigen values. I want to go in the same order as the Eigen vectors. So he is equal to the Matrix, starting with Lambda one all the way to lend before so negative 11 5 11 19 and the rest the entries are zero entries.

Samuel Goyette · Answer

Okay, So once again, we want to diagnose lies a four by four matrix and not again. That&#x27;s not super difficult. He&#x27;s just very long in terms of calculations, and you you want to make sure you end up making any calculate calculation mistake. So we&#x27;re going to get started right away. And the question states that we can use our own program to find he again values off of a so to find the agent values. I won&#x27;t suggest any particular tool you can use. Some calculator can compute a gin values. We have a lot of APS or both on phone, mobile phone or on browser, where you can simply, uh, enter how to come play online computation of alien values that will and this will be done for you. So if you have any way of finding the Hagen values, you should find one minus two and five as the egg and values of, uh, personally. The tool that I use doesn&#x27;t tell me the multiplicity multiplicity of beach again values. I only know there&#x27;s three so foredeck matrix to be diagonal sizable. One of these and haggard value must lead to an hag in space of dimension to, um for their their This matrix is diagonal sizeable and warning. There&#x27;s a lot of calculation ahead, so let&#x27;s get started. We gotta find the Eigen vectors or the basis of the hag in space of a eso. The first taken value is one. So we saw discipline their system. So a minus identity times X people zero. We solve that system for X, which looks like this. So minus 7409 minus three minus 116 minus one minus 200 It&#x27;s pretty nice to zero minus 44 06 Yeah, the more yourself in your system. Eventually it gets a bit boring and tiring. So when you see a bunch of zero, you&#x27;re always happy because you know that a line full of zero means a quicker solution, quicker way to a solution. So here we have, the air system will want to solve. Let&#x27;s rewrite it as equations, which I prefer to do, because I think it&#x27;s easier to visualize every component on its own. Um, if you prefer to solve it using matrix operations, that&#x27;s fine, too. As always, we&#x27;ll get to the same answer. So minus seven plus four x two plus nine x four equals zero minus three x one plus not plus so you gotta be careful minus x two. Thus x three plus six x four equals zero a minus one x one minus two x two equals zero and minus four x one minus, ah minus or plus plus four x two plus six exports or have you haven&#x27;t noticed by now, It&#x27;s very easy to make tiny mistakes when diagonal rising a matrix or when solving linear system by hands. So if it&#x27;s possible for you, always solar in your system, using a computer to avoid computation. Mistake. Let&#x27;s look at this system very easy. Third line. What does it tell us? Tell us that X one is equal to minus two x two. If we inject that information into every other line, will get the following 18 x two x two First nine x four equals zero five x two plus x three First six x four equals zero and 12 x two across six x four equals zero. So, already, just by injecting X one equals minus extra well, where you have a much simpler system and we can rewrite it basically like this. So we have X one equals minus X minus two x two. That doesn&#x27;t change. We&#x27;re gonna have the same thing for export if we divide everything lying four by six and we put the X two over on the right hand side. But we have x four. There&#x27;s no dot just a tiny writing mistake. X four is also equal to minus two x two and let&#x27;s take a look at x three. So knowing that X four is equal toe minus two x two, we can inject that in the second in the second line right here if I can scroll up. So if we all right X for is equal minus two extra, we can have x three equals toe seven x two. So there you have it. Your solution to this in their system will look something like this So x times X two times minus 21 seven and minus two. So here&#x27;s the basis for the first Haijun space it&#x27;s of them mentioned. Want So we have a in value having victor. Let&#x27;s move on to second, uh, hag in space. I want to write in black. So for lender to being rules. Five. Uh, we want to solve this linear system. So a minus five times identity, times X equals zero district. Your system well, look like this minus 11. Four 09 minus three minus 516 minus one minus two minus for zero and minus for 402 x one x two x three eggs. It&#x27;s four sore Director X and every day 000 So, at first glance, this system is a little less friendly than the previous one. There&#x27;s not a single roll with toward three zeros. So we have a it more work to do to find a solution for this system. So ah, I prefer to write it in simply equation and to work with maitresse. Ease So minus haven&#x27;t x one plus four x two plus nine x four equals zero once again, if you can solve that system anywhere you like, but you should get an answer that&#x27;s similar to mine. Um, not plus, but my goodness five x two and every operation you do by hand with matrix. Once again, I&#x27;m repeating myself. It&#x27;s always a game of making sure that you&#x27;re not making animistic and you writing everything correctly. So sometimes you can have the right calculation. But you write minus is a plus or a plus as a minus and that will derail everything. So always better to go slow and steady than to try to go too fast and have to come back many times to fix your previous mistake and minus four x one because four X two plus two X four equals zero. Okay. Ah, what can we do here? Well, let me check my notes here to make sure. So the first thing I decided to do was to divide Ah, the third, the fourth line by two and to and basically that&#x27;s it, I think. Oh, yes. So you can divide it signed by two and have x four on the left and side, so x four will be equal to two x one minus two eggs, too. Uh, knowing that we can, uh, we can right exporters to x one minus two x usual. The first line will be minus 11 x one plus four x two Uh, yes. 18 x one minus 18 x two equals zero. So there&#x27;s zero right there, and the 2nd and 3rd line are left unchanged for this step. So minus three x one minus five x two press x three six Export If you wanted to inject the new value of X four, you could do it. I as I said, I prefer toe work. Ah, a bit slower and and make it everything one step at the time. So that way, if you make a mistake, it&#x27;s a good habit for me. Because if I make a mistake, it&#x27;s easy to go back and, ah, figure out how to fix minus day if I you everything in one step. Sometimes it&#x27;s a bit harder to retrace your step and you end up just starting over, which is very time consuming in frustration. So now that we only have X one and X tools on the first line Ah, we have, uh we can just simplify everything on the first line. So we have a seven x one minus 14 x two equals zero and the 2nd, 3rd and 4th lines are unchanged. So ah, just to save made this video slightly shorter. I won&#x27;t rewrite all those lines. I will just right sane. Think. Of course, if you&#x27;re writing this question for a homework. You should write everything. Ah, but for time&#x27;s sake, I won&#x27;t. And that e that leads us to a kind of simpler system. So we know that X one is equal to two x two, uh, knowing that X four becomes equal to two x two also a to X one is four x two. So for extra minus 22 is to is to ah and then we can simply inject the fact that x one is two x two and exports two extorting this third and the 2nd and 3rd line here and we get something like this. So x three plus x two equals zero and minus four x two is for X two minus four. X three equals zero and we end up everything in relation to extra. So x one is two x two x three is minus x two and x four Uh is two x two All right, so we have another i injector. X equals x two times to another basis for against basic for a bit more accurate. So to one minus one and two, That will be another column of our matrix p when we are ready to die. Analyze A. The last thing we need to do is to find another basis for the third hike in space again. Nothing very complicated. Ah, nothing very hard is just a bit long and boring. So we have land a three, which is equal to minus two. We solve this in their system. So careful here we have a plus to a types identity. So that&#x27;s a very common mistake. Forgetting that when I am, does negative disciplinary system is different, which will look like this now for a to be diagonal visible. We know that this a hike in space must be of dimension to minus four for 09 minus 3216 minus one minus two 30 minus 4409 So already we have a good indication that the system will be of dimension. To decide in space will build dimension to since the first and the fourth line off this matrix of this linear system or equal, they&#x27;re the same thing so we can just focus on We don&#x27;t have to consider the fourth thing. We can just focus on the first, 2nd and 3rd line since the 1st and 4th line or the same Once again. Uh, I will write my assistant in Be right to property. Mm. Okay. Minus four. Mine. I don&#x27;t want to raise it. Minus four x one plus four x two purse nine x four equals hero minus three x one plus two x two plus x three for six x four equals zero and minus x one minus two x two plus three x three is equal to zero. Ah, so since we already it&#x27;s very easy. We can simply put x one is equal to minus two x two to us x at three x three sometimes book plus three x three. Now that&#x27;s extra one. We gonna Sorry there&#x27;s this little typo here. This is X two. So, as I said, you always have to be careful even when you already did it and you have notes to follow. You always have to be careful. We can inject that information in the 2nd and 3rd line that will give us 12 x two plus minus 12 x three plus nine x four equals year old eight x two minus eight x three plus six x four equals zero And if we were closely. The 1st and 2nd lines are the same because if we divide this blind by three will get four x two minus four x three plus three x four And if we divine that line by two, we&#x27;re gonna have four x two minus four x three plus three Explore. So we&#x27;re left with only two lines that don&#x27;t have redundant information, which are four x two minus four x three plus three x four equals zero and uh x one equals minus two x two plus three x three. So since x one is already dependent on x two and x three, we&#x27;re gonna write X for the same way. X four is equal Tu minus four third so time stakes to cross four older, three times x three and x one is equal to minus two x two plus three x three. Therefore, the solution to this final linear system looks a bit like this. Uh so minus two x two plus it&#x27;s very x three x two x three and minus 4/3 x two plus four times over three x three We can distill it to its most simplest form minus 210 minus 4/3. Thus X three times three c 01 for over three. So here you have it. We have our final two basis for I in space. The third Dragon Spaces of damage to a stagnant. I said both. And just to I&#x27;m gonna write a in the form of a times P. I did a tiny bit bigger if I can scroll down is perfect a the Matrix P, we have minus 217 minus two to one minus one to minus 21 zero minus for over three and 301 for over three. So we have our four. This is a fight in space for literally independent vectors. We gonna place the corresponding hang in various or 15 minus two and minus two with zeros outside off type, you know, and we can rewrite P while I write p. I would like to, um notice. I would like for you to notice that if you look at the text answer of this problem on the website, it&#x27;s not exactly the same. Ah, which I did on purpose. You can columns of p. You can change him as long as you changed the you can, and you can do permutation on the columns of P as long as you do according permutation on the columns of tea minus 210 minus 4/3 and 301 for over three minus. Once a very long solution to a very easy problem. Don&#x27;t say anything. The only difficulty in those questions is Do not make any mistake. Win too computer. You do any calculations, but basically you find the age and values Using a computer. Erin, you find basis for hag in space by hand. And if you don&#x27;t make any calculation mistakes, you should have something like this eight times p a equals P Times D times PNG&#x27;s.

Problem

In Exercises $37-40$ , use a matrix program to fi…

Problem 38 Hard Difficulty

Answer

$\left\{\left[\begin{array}{l}{2} \\ {7} \\ {7} \\ {0}\end{array}\right],\left[\begin{array}{c}{0} \\ {-1} \\ {0} \\ {1}\end{array}\right]\right\}$

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

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$\left\{\left[\begin{array}{l}{2} \\ {7} \\ {7} \\ {0}\end{array}\right],\left[\begin{array}{c}{0} \\ {-1} \\ {0} \\ {1}\end{array}\right]\right\}$

Linear Algebra and Its Applications

Kayleah T.

Samuel H.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Kayleah T.

Samuel H.

Michael J.

Joseph L.

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications