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

Question

For the matrices in Exercises $15-17$ , list the eigenvalues, repeated according to their multiplicities.
$$
\left[\begin{array}{rrrr}{4} & {-7} & {0} & {2} \ {0} & {3} & {-4} & {6} \ {0} & {0} & {3} & {-8} \ {0} & {0} & {0} & {1}\end{array}ight]
$$

Sam Sohn · Accepted 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.

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

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.

Pagadala Kishore Reddy · Answer

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Bobby Barnes · Answer

so they want us to orthogonal eyes this matrix. So first thing we need to do is actually find are arguing vectors so we could go ahead and try to normalize thumb. Eso will need to I&#x27;ll call this a So we need to do a Maya slammed I, which is going to be one minus Lambda Negative five negative 51 minus lambda. And then we take the determinant of this. And remember, we&#x27;re assuming that this is equal to zero. Yeah, so do the determined we do. The downs binds the ups. So that would be one minus lambda, uh, squared and then negative five times negative. Five is 25. That b minus 25 is equal to zero. So, yeah, we can actually go ahead and add that over. So one minus lambda square is you go to 25 take the square, root each side, and that&#x27;s going to have one minus. Lambda is able to plus minus five and then we I&#x27;ll add land over and I&#x27;ll subtract The plus surmised fives that just as lambda isn&#x27;t one plus minus five or, in other words, lambda is equal to six or land us all Lambda one is six and then lamp too is negative or so. Let&#x27;s go out and find what are Eigen Vectors are going to be now and let me move this down here. So when Lambda is equal to negative for, I&#x27;ll do negative for first, Um, a minus minus four I or, in other words, we&#x27;re adding for should give us so would be five negative five negative 55 And then you can see that we could clear this to reduce it down to just one negative one 00 So then this tells us that X one minus x 20 or X one is equal to x two. So when we&#x27;re creating our eye in vector over here, If we start with X one x two, then this is just going to be x two x two or x 211 Now we need to normalize this. So let&#x27;s go ahead and, uh, find the length of it. So remember the length is just going to be both of those components squared square rooted. So the one squared plus one squared square rooted. So that&#x27;s gonna be route to. So we multiply this by problems are. Divide this by route to and then that&#x27;s going to give one over route to one over, route to And then this is going to be our Eigen vector. So this will be be too. And so this was associate with Lambda to equal to negative for all right. And now let&#x27;s go ahead and do this with our other one. So this would be a plus. Um, What was it? Six were actually minus six, minus six I. So again, if we look at what a waas on you to skip this down, they want to get all that. So just yes, explain me this town and I&#x27;ll speed this up. Yeah, eso we would do one minus six of being negative five negative five. And then here we have negative five negative five and then notice that this would reduce down to just 1100 So it&#x27;s x one plus x two is equal to zero. Or, in other words, X one is even too negative X two and now over here we have x one next to and this should be equally so negative. X two x two Factor that out So the x to negative 11 and now we need to normalize this one as well. So same thing is before. So we find the length of this negative one squared plus one squared square. Root it, and that&#x27;s going to give us a route to again. So we divide this by route to, and that&#x27;s going to give us negative one over route to one over route to. And then this is going to be our first one when Lambda One was equal to six, right? So now to create are normalized vector, so we&#x27;re going to have So let&#x27;s first create the diagonal. So it&#x27;s D. So we had negative 46 and then we have 00 and then for P. This is going to be wealthy one associated with negative for the Eigen vector. Let&#x27;s come back appearance, see where it waas. There&#x27;s one over route to one of her route to so we have one over route to one over route to, and then the one with six. We just found to be negative one over route to 1/2, which, if you want, you could pull out the one of the route to and just write this as one over route to and then one negative. One 11 Um, so that&#x27;s one way you could do it. And do we need to find P embers? Yeah, we did. We just need to find this. So this is pretty much where we can leave it like this. So that&#x27;s our diagonal ization. And then that would be our orthogonal matrix.

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.

Problem

For the matrices in Exercises $15-17$ , list the …

Problem 15 Medium Difficulty

For the matrices in Exercises $15-17$ , list the eigenvalues, repeated according to their multiplicities.
$$
\left[\begin{array}{rrrr}{4} & {-7} & {0} & {2} \\ {0} & {3} & {-4} & {6} \\ {0} & {0} & {3} & {-8} \\ {0} & {0} & {0} & {1}\end{array}\right]
$$

Answer

4, 3, 3, 1

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

Linear Algebra and Its Applications

Kayleah T.

Caleb E.

Samuel H.

Michael J.

Vectors Intro

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4, 3, 3, 1

Linear Algebra and Its Applications

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

Kayleah T.

Caleb E.

Samuel H.

Michael J.

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

Linear Algebra and Its Applications