exercises-22-26-provide-a-glimpse-of-some-widely-used-matrix-factorizations-some-of-which-are-disc-3

Question

Exercises $22-26$ provide a glimpse of some widely used matrix factorizations, some of which are discussed later in the text.
(QR Factorization) Suppose $A=Q R,$ where $Q$ and $R$ are $n 	imes n, R$ is invertible and upper triangular, and $Q$ has the property that $Q^{T} Q=I .$ Show that for each $\mathbf{b}$ in $\mathbb{R}^{n},$ the equation $A \mathbf{x}=\mathbf{b}$ has a unique solution. What computations with $Q$ and $R$ will produce the solution?

Chris Trentman · Accepted Answer

in this problem were introduced to the method of Q R factory ization. So I suppose that A is equal to Q times are where Q and R R n by n this means that, of course, is an n by N Matrix. Our is in vertebral and upper triangular, and Q has the property at its transposed because it&#x27;s in verse so cute transpose times Q. Is the identity matrix, whereas to show that French column Vector B The equation X equals B has a unique solution. So because Q is square and transpose times, Q. Is equal toe. I. We have that Q is in vertebral by inverse matrix. The&#x27;re, um, and we have the queue in verse. As I mentioned before, this is going to be cute transpose so it follows that A is the product of convertible matrices Q and R, and therefore it follows that a itself is in vertebral. So since a is in vertebral, it follows that the equation X equals B has a unique solution for X for all column vectors. Be now from our equation. X equals B. This implies that Q r X equals B, and so it follows that huge transpose Q r X equals Q transpose be so we have since Q. Transposed Q. Is the identity matrix I. R X equals Q transpose be. Or, in other words, our X equals q transposed be. And then we have that X is equal to are in verse. Que transpose be so That was part A or Sorry, not party. Now a good algorithm for finding X is to compute que transposed be then Reverend juice, the augmented matrix or huge transposed be. What will end up with is the identity matrix on the left and are inverse que transpose b which is X on the right. And we had that The reduction is fast since matrix are is upper triangular. This is why we&#x27;re choosing this method of computation. All we have to do is apply row replacement operations to the rose above each entry on the diagonal

Chris Trentman · Answer

were introduced to the singular value decomposition method of matrix factory ization. So suppose that a is equal to u times d tend to be transposed where u and v r n by n matrices with the property that you transpose u equals I and that v transpose v equals I. And where d is a diagonal matrix with positive entries. Sigma one through sigma n on the diagonal were asked to show that the matrix is in vertebral and to find a formula for a inverse. Now we have that both u and V transpose are square. So the conditions on U and V imply that you and the transpose are both in vertical matrices by the in vertebral matrix. The&#x27;re, um so it follows that you inverse equals u transpose and we have that be transposed in verse is simply ve Now The diagonal entries of D are non zero. Since none of these entries or zero it follows that D is an in vertebral matrix, and we have that the inverse of D. It&#x27;s a diagnose metrics as well civically with entries Sigma one in verse through sigma and in verse on the diagonal. So it follows that a is a product of convertible matrices. So it follows that a itself is in vertebral. Not only is a convertible we have the a inverse is the inverse of you d be transposed and this is going to be be transposed in verse The inverse universe which is equal to the de in verse um, verse and these air Easy to our Sorry you transpose since you universe is u transpose and these air easy to compute

Chris Trentman · Answer

in this problem, you&#x27;re introduced to reduced l U factory ization. So given a as in practice problem, whereas to find a five by three matrix B and a three by four matrix C such that a is equal to B. C. And then to generalize this idea to the case where A and M by N Matrix a is factory Izabal and has l u factory ization l u and you has only three non zero robes. So first of all, from a practice problem, we have the A on the practice problem Matrix A is the four by four matrix to negative four negative 23 six naked nine negative 58 to negative seven negative 39 four negative to negative 21 or negative one. In fact, this is actually four by five matrix in the last room is negative. 6334 So to solve this problem first you want to find l u factory ization for a and then we want to roll reduce eight echelon form. Okay, so and roll reduced by eliminating entries under the first column. In doing so, I obtained the matrix with first row to negative four negative 23 031 negative. One zero Negative. Three negative. One six 062 Negative. Seven zero Negative. Nine Negative 3. 13. And in this problem, I can eliminate entries under the three in column two using only raw replaced in operations. In doing so, I obtained the Matrix to negative four. Negative. 23 Uh huh. Excuse me. Uh huh. Zero 31 negative one. Part 0005 000 Negative five became 000 10. And finally, I&#x27;ll ro reduce a using only replacement applications on the roads below road three. And so I obtained the Matrix to negative four. Negative. 23 0311 Negative one. I mean, 0005 0000 and 0000 So we found a matrix. You This is a you from the l u factory ization. And then we&#x27;re going to find El using the algorithm from the book. So the first pivot column This was 2624 negative six in dividing Pivot column by the pivot Which is to we get that The first column l is 1312 negative three. And next The second pivot column was three negative 36 negative nine and dividing by the Pivot, which is three. We obtain zero one negative one to give three In her third pivot column is five Negative 5 10 divided by the pivot, which is five. We obtain one negative 12 00 one negative 12 and will use the last two columns from the identity matrix to fill in l 00010 and 00 001 So we have our value factory ization for a now notice that the bottom two rows of you containing the zeros now for using the Row column method defined L. U. The entries in the final two columns of L won&#x27;t be used because these entries will be multiplied by zeros from the bottom two rows of you. So we&#x27;ll pick be to be the first three columns of l and see will be the top three rows of you. So in other words, we have that be will be matrix. 100 310 one negative 11 to to negative one and negative three negative 32 And likewise, see, this is going to be to negative four Negative. 23 031 negative one and 000 five So we obtained the desired matrices B and C, and we have that b c is equal to l times you by the comment from above, which we know is equal to a So it also has the desired property. And of course, we can generalize when a is m times end. So suppose is an M by N matrix. I am going to suppose that a has an L U factories ation that you has only 390 rose. Well, we know that these three none zero Rose They&#x27;re going to be on the bottom. So we&#x27;ll take be to be the first three columns of l and will take c Yeah, to be the first three rose of you, then we have that&#x27;s and

Chris Trentman · Answer

were introduced to the spectral factory ization method for matrices. So suppose that three by three matrix a has a factory ization a was PDP inverse Repeat. It is an in vertebral three by three matrix. So the PM versus defined in de is the diagonal matrix with entries 11 half and one third on the diagonal. Whereas to show this factory ization is helpful in computing high powers of A by finding simple formulas for a squared cubed and a d. K where K is positive injures using P in the entries. Indeed, first of all, we have a squared is equal to by definition p d. P in verse times PDP inverse and using associative it you see it the p and P inverse multiply to get I And so this reduces to pee d squared p in verse and we have that d squared could be obtained. They simply multiplying the entries of be together. This is 11 half one third times the matrix with entries 11 half and one third. This is simply the matrix with entries 100 0 1/4 0 0019 And so it follows that A squared is equal to P. Times 1000 1/4 000 1/9 times p in verse. That&#x27;s a fairly simple formula for a squared in terms of entries me and P. Likewise, we have that a cube is going to be, well, a squared times a which above we have this is tee times d squared times p in verse, times p tends detained p in verse, which reduces to P D Cube P in verse. And so it follows that a cubed is equal to p times, and then we&#x27;re going to cube the entries of each on the diagonal. So now we have one cube 0001 half cubed, which is 1/8 0001 3rd Cube, which is one 27th p in verse. So this formula for a cube Yeah, this formula for a squared. And so we might guess at the formula for a que well, this is going to be P D to the K P in verse, which, of course, is equal to P. One of the K is still one 00 zero one over to to the K zero 00 1/3 to the K P in verse. So we have a simple formula for eight, Okay, in terms of P and the entries of D.

Chris Trentman · Answer

this problem were introduced to the method of rank factory ization. So we&#x27;re going to suppose that M by N Matrix a has a factory ization. A close CD or C is an M by four and D isn&#x27;t a four by N In part a were asked to show that is the sum of four outer products. This is actually quite simple. Well, just express each row of D as the transpose of a column vector. So we have, first of all, that a equals C D equals. We have column vectors for C of C, C one, C two, C three and C four, since he is an M by four matrix and then he can be expressed as the one to do you three do you four. And because we&#x27;re multiplying, this is going to be the transpose of the column Vector. Well, not quite is the transpose of each column, Victor, so we have d one transposed due to transpose d three transposed for transport. In this way, we obtain a four by N Matrix. Now, with this representation, it&#x27;s easy to see that using matrix multiplication, we get see one the one transpose plus C two de to transpose plus C three d three transpose plus C four Do You four transpose, which is the sum of four outer products now in Part B, were given M and N and were asked to explain why a computer programmer might before to store the data from Matrix A in the form of the two matrices C and D. Well, we have the M is 400 and and he&#x27;s 100. So it follows that the dimensions of a R 400 by 100 the A has 400 times 100 which is or 40,000 entries. We see that see, which is an end by four. Matrix has 400 times four, which is 1600 and trees and the which is a for biometrics has four times 100 which is 400 entries. So storing C and D together requires a total of 2000 entries. And of course, we see that 2000 is only 5% of the entries for storing a so this factory ization or the advantage that takes much less memory to store the matrices C and D, and using our outer products above, we can easily compute a using matrices C and D

Subham Jyoti Mishra · Answer

for set off. The question is, is you could do be who have so listen or oh, bad dreams or B in our, um Yeah, the diamond Sinar column Space off. He must be goto m know when this is true. In the previous question, we have calculator. We have that here on their times, you know, Hold on space off air paths, diamonds and up No space off a transpose is equal m no sense. This is m Therefore, this must be zero. So this is true. I&#x27;m leaving the diamonds and off another face off you Transport is a zero on this is zero The solution metrics. Uh, a transpose is trivial. That is mhm when another hour diamonds and honor off your transfer zero Then you get to do the here Transfers into X has trivial Feddersen. So therefore yeah, this is possible only and only when a transpose x trivial solution Mhm

Problem

Exercises $22-26$ provide a glimpse of some widel…

Problem 24 Easy Difficulty

Answer

$\mathbf{x}=\mathbf{R}^{-1} \mathbf{Q}^{\mathrm{T}} \mathbf{b}$

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

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Kayleah T.

Caleb E.

Absolute Value - Example 1

Absolute Value - Example 2

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$\mathbf{x}=\mathbf{R}^{-1} \mathbf{Q}^{\mathrm{T}} \mathbf{b}$

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Kayleah T.

Caleb E.

Absolute Value - Example 1

Absolute Value - Example 2

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

Catherine R.

Anna Marie V.

Kayleah T.

Caleb E.

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

Linear Algebra and Its Applications