for-the-matrices-in-exercises-15-17-list-the-eigenvalues-repeated-according-to-their-multiplicitie-3

Question

For the matrices in Exercises $15-17$ , list the eigenvalues, repeated according to their multiplicities.
$$
\left[\begin{array}{rrrrr}{3} & {0} & {0} & {0} & {0} \ {-5} & {1} & {0} & {0} & {0} \ {3} & {8} & {0} & {0} & {0} \ {0} & {-7} & {2} & {1} & {0} \ {-4} & {1} & {9} & {-2} & {3}\end{array}ight]
$$

Sam Sohn · Accepted Answer

So we have our matrix of three 0000 and then negative. Five 1000 three 8000 zero Negative. Seven to 10 Negative for one. Knowing negative too. Three. What&#x27;s great about this is that this is once again a triangular matrix. Because every entry above the diagonal line is zero living. All these entries in the bottom left half. So that means all we need to do to find the Eigen values is pick up all the Dagnall entries like this three. This one, this zero, this one and this three. So the Eigen values are going to be zero with multiplicity. One one with multiplicity, too. In three. What? Multiplicity, too? Yeah, we&#x27;re done.

Amy Jiang · Answer

Okay, So based on our damn, you know, there are Indian values. Argues the values on our that, you know. So that&#x27;s five Maigret 411

Sam Sohn · Answer

were given this matrix of four negative seven zero to zero three negative four six zero zero Hillary negative eight zero zero zero and one, and this is asking us to list the Eigen values now. Fortunately for us, this is what&#x27;s called a triangular matrix, where every non zero entry and where, where every entry is going to be above this triangular line and everything below is just zero. So that means, in order, find the Eigen valleys always do is look at the diagonal entries, and we can see very easily that they are 433 and one. In other words, the values the Eigen values are one with multiplicity one three with multiplicity, too, and four with Multiplicity one and we&#x27;re done.

Gennady Notowidigdo · Answer

Okay, so our pretty quick question. So we have a three by three matrix, and we want to find the Eigen values. Hey, so to do this, we&#x27;re going to refer to theorem number one from chapter five. This one here, which states that the Eigen values off a triangular matrix, are the entries off the diagonals. So we look att, this matrix here, Demetri, start written at the Star Old. These entries are zero. So we can therefore conclude that this matrix is a lower triangle of matrix which implies than that the entries off the diagonals, these ones here, Arthuis, Aiken values off the triangle image off this matrix. So the argon values are 40 and minus three. That&#x27;s a pretty quick so

Chris Trentman · Answer

were given a matrix A and were asked toward fog only diagonal eyes. This matrix giving an orthogonal matrix diagram matrix were also given that the Eigen values for this Matrix or Lambda equals three and five. The Matrix A is 401004101104001 04 Noticed that a transposed says the Matrix has the same diagonal for four. For four. We exchanged zero with zero one with one zero with zero zero with zero one with one and zero with zero and recognize that this is the same as the matrix. A. So it follows. That&#x27;s since a transfers equals a is a symmetric matrix. Therefore, it follows that a is north Ogden Aly Diagonal Izabal Your thought only diagonal eyes A when he defined complete set of orthe normal vectors. Eigen vectors for a So take lambda One to be the first Eigen value three. Then we have that. If the one is an Eigen vector then satisfies the equation. A minus three i v one equals zero vector The matrix a minus three I is the same as matrix except for before has become ones So we obtained the matrix. 10100101 10100101 in this simply reduces to the matrix. 10100101 0000 0000 from which we obtained a system of equations V 11 plus V one of three equals zero and V 12 plus view and four equals zero. This tells us that the 13 equals negative view on one and the 14 equals negative b 12 And so if we take V 11 to be the parameter s and he one to be the parameter t, we get that The general form for the I conductor V one is the vector s T negative s negative T which would be written as a linear combination s times 010 negative one or zero 01 My mistake s times 10 negative 10 plus t times +010 Negative one. So we see that this wagon spaces two dimensional. And if we take s to be one and t v zero, we obtained one Aiken Victor called V one is the Agon vector 10 negative one zero it would take Has to be zero. On TV one, we obtain a second Eigen Vector V two, which is linearly independent from V one, which is the connector 01 zero negative one. And so we have that set V one be to is a basis for this Eigen space. They also have that V one dotted with feet too is equal to zero plus zero plus zero plus zero, which is zero. And so we have that V one is or thought Donald to V two. And therefore we have this set view and be two of the north Ogle set norm of V one is the square root of one plus one, which is Route two in the normal. V two is the square root of one plus one, which is Route two. And so we have the unit vectors. You one, which is V one over. Norm of the one is equal to 1/2 zero negative. One of a route to zero. And the unit after you too, is the vector V two over the normal vector he to which is 0 1/2 zero negative 1/2. And we have that both. You want you to our Eigen vectors and are linearly independent in North Ogle. So that set you on you too. Isn&#x27;t Ortho normal basis for this Eigen space? Next, Let claimed to be the second and last Eigen value five. Then if the three is an Eigen vector associated with this Eigen value, we have that A minus five I times feet three is equal to the zero vector matrix. A minus five. I is the sings matrix A Except for the diagonals now negative one. Instead of positive one, we obtain the matrix. Negative. 1010 ones. Zero negative. One 01 10 Negative. 10 and 010 Negative one. This could be reduced to the matrix with reduction 10 Negative. 10 010 Negative. One and 00000000 from which we can obtain these system of equations. The 31 minus V 33 equals zero and be three to minus fee. 34 equals zero. So we have that. These 31 is equal to be 33 and V 32 is equal 33 4 So if we take B 31 to be the parameter s and V 32 to be the parameter t yet the general form for the I conductor V three is S T yes, T is equal to the linear combination s times 1010 plus T times +01 zero one And if we take us to be won t v zero weekends the Eigen Vector V three, which is 10 10 And if we take s to be zero and t to be one, get second Eigen Victor, before which is 0101 and by construction we have that these two wagon vectors are linearly independent and thus form a basis for this wagon space. We also have that V three died with before is equal to zero, which implies that V three is orthogonal to before. So it&#x27;s actually in orthogonal basis for this vector space. The norm of V three is equal to the square root of one plus one which is to in the normal before is the square root of one plus one which is equal to. And so we have the unit Vector you three, which is equal to V three over the norm of you three, Our normal V three. My mistake is a vector one over route to zero 1/2 zero. And the vector You four is a unit vector, which is V four over the norm of the four. This is the vector zero. Whatever too zero one of the route to and we have that both u three and U four our Eigen vectors for this wagon space and are valuable and are unit vectors. So it follows that set you three. You four is an Ortho Normal aces for this Eigen space. Since our Matrix A is symmetric, it follows that you won and you two are Ortho or thought Donald U three and U four. So it follows that to set u one u two you three you for is a complete Ortho normal set of Eigen vectors for this matrix. And we have that. If we take P to be the Matrix who&#x27;s calling Victor&#x27;s Air U one U two, you three and you four. This is the Matrix one of a right to zero negative one of a route to 01 of a route to zero negative 1/2 and whatever to zero Whatever. Route to zero and zero one of a route to 0 1/2 when we take matrix D to be a matrix whose entries our gagan values on the diagonal. So you have each Agnelli was multiplicity too. And so this is the matrix 3355 and then zeros at the worlds. And using these two matrices p and G, we can diet.

Bobby Barnes · Answer

so for us to die analyze this matrix. Since they don&#x27;t give us the Eigen values, we need to find those first. So remember to find that we look at the determinant of a minus. Lambda I and so a minus lambda I is going to be so let&#x27;s just write that So it would be determinant off eso for my slammed for my slammed to mind Slammed to my slammed We have a one in the corner here and then everything else is going to these zeros. And luckily for us, since we have it in this upper Earl Starr, this lower triangular form, we know that the determinant of this is just going to be equal to the diagonal multiplied together. What? I don&#x27;t know why I have a zero down here, but that zero should be, uh so this ends up becoming for or minus slammed squared times to mine. Islam two squared. And remember, this should be equal to zero since if we were actually thinking about what this is. We have a minus slammed I, Times X and this being able to the zero vector. So this thing here should be zero. Because, remember again we&#x27;re assuming that X is not zero. Come on now, since the determinant of that is going to be zero that just tells Islam dais for land is too. So we&#x27;ll use these values here to help us create our factors. So let&#x27;s go ahead. Pull this down. Let&#x27;s first check out when Lambda is equal to, um to actual. Let&#x27;s do four first. So a minus slammed. I, in this case, is going to be so 0000 very sparse matrix and sparse just means a bunch of zeros. Uh, that one stays there and then negative two. Okay, so we have this. And now, um, we&#x27;ll go ahead and really reduce this, which I actually went ahead and already did before and doing this e mean, we actually essentially just get what we have here. So I actually don&#x27;t even need to really do anything, because if we look at it from here, this is just telling us X one whiteness to X or is equal to zero, and then we have X or negative two x three is equal to zero. So first this implies X three is just equal zero. And then this top one here implies X one is equal to two x four and then notice we don&#x27;t have anything about X two. So x two is just some free variable. So now let&#x27;s go ahead and set up our actual factor, our Eigen vector. So we&#x27;re going to say X is equal to so it&#x27;s x one x two x three x four So x one over here we said was going to be two x four x two was just free. Nothing really There x three we said was zero and then x four is also free. So it&#x27;s just going to be itself. So maybe here I should say X two x four, both free and now we can break this up. So we have X two here. Eso we only have a next two in row two. It&#x27;ll be 0100 plus x four times. So we have to and row 11 in row to none in a row. Three. I&#x27;m sorry. Uh, there&#x27;s none in row two. That was to their I mean, that was executed on down here. We just have one. So remember these are going to be our to Eigen vectors. or Lambda equal to five right now, Not be able to five family able to four lambda you before. Now we&#x27;re going to repeat the same process, but with two this time. So let&#x27;s sweep this down. So now when Lambda is equal to two, we have a minus two I, which is going to give us 20000200 And then everything else becomes zero. Except for this one in the corner. Um, and then this here. Well, the one in the corner down there, just as Excellency zero. So we could just kind of ignore that. Um So what this says, though, is two x one is equal to zero. The second one is two. X two is equal to zero. And then the last one just says X one is equal to zero. So not just implies X 100 x two is equal to zero. And then we have x three on explore being are free variables. So once again, we go and set up that specter So x one x two x three x four. So we had x one x two book being zero and then x three x four just themselves. So if we split this up just like we did before, we only have an X three in a row. 30010 and then x four we only have in row four. So 0001 and then both of these go with Lambda is equal to two. So let&#x27;s go ahead and set up our diagonal eyes and matrix now So we have d is equal to um So I&#x27;ll do 2 to 44 and then everything else is going to be zeros. And now it doesn&#x27;t get better if you do 2 to 44 or 4422 a. Za, Long as you have matched up the Eigen value with the Eigen vector which we do in a second. But this is just, I would say the most standard way to do it from smallest to largest now to get P. So we need to place a Eigen vector in the first two columns here and actually let me see how I set this up earlier. Um, I don&#x27;t need to necessarily do it the same way, but I just want to do it so we get the same answers that I wrote down. So we don&#x27;t have to do this all by hand again. Um, since everything already plugged into my calculator Um yeah, so you could see here. Eigen, Vector one Eigen Vector to Eigen. Vector one, Eigen Vector to And you could also flip these It doesn&#x27;t really matter. You just get a slightly different in verse now for four s. Let&#x27;s go back up here and look at what those were. So that was 0100 and 2001 Um, so let me see which order I had placed those in. So I did 0100 And then I did the other one 2001 Yeah, so there&#x27;s gonna B P now we need Thio. Get P in verse. You can do this pretty much any way you want, but my favorite way is to just set up the augmented matrix and reproduce it into the are set up in augmented matrix, where this is equal to the identity and then just roll reduce it. So I&#x27;m gonna put P over here. So, 00020010 10000101 And then over here, I put the identity matrix, and then we&#x27;re going to ro reduce this. And I went ahead into this beforehand, and that gives us the matrix. Zero 010 negative. One half 001 01 00 and then one half 000 So, uh, sorry. This is not exactly what it was because this should be the right side of the matrix scandal. Head of myself. So the left side, Woodrow reduce down to the identity. Yeah, and then the right side will roll reduce down to our inverse. So this is P universe. So let&#x27;s go ahead and write out our diagonal ization of you So there&#x27;s gonna be a is equal to so p d. P in verse. And so p we had was 0002 00101000 0101 and then d we had so 2 to 44 along the diagonals and then everything else zeros and then are inverse that we got was 0010 negative one half 001 0100 one Half 000 and so this is going to be our diagonal ization. And like I was saying earlier, you might have got something slightly different from what I got here. But it may be still a valid way to write the diagonal ization. As long as you multiply everything together and it gives you the original matrix say it doesn&#x27;t really matter, because again, this is not a unique thing. We will have multiple different ways that we can write any diagonal ization.

Problem

It can be shown that the algebraic multiplicity o…

Problem 17 Medium Difficulty

Answer

The eigenvalues are $0,1$ and $3,$ with multiplicity $1,2,2$ respectively.

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

Linear Algebra and Its Applications

Heather Z.

Caleb E.

Kristen K.

Samuel H.

Vectors Intro

Vector Basics - Example 1

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The eigenvalues are $0,1$ and $3,$ with multiplicity $1,2,2$ respectively.

Linear Algebra and Its Applications

Heather Z.

Caleb E.

Kristen K.

Samuel H.

Vectors Intro

Vector Basics - Example 1

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

Heather Z.

Caleb E.

Kristen K.

Samuel H.

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

Linear Algebra and Its Applications