show-that-if-an-n-times-n-matrix-a-is-positive-definite-then-there-exists-a-positive-definite-matrix

Question

Show that if an $n 	imes n$ matrix $A$ is positive definite, then there exists a positive definite matrix $B$ such that $A=B^{T} B .[	ext { Hint: }$ Write $A=P D P^{T},$ with $P^{T}=P^{-1} .$ Produce a diagonal matrix $C$ such that $D=C^{T} C,$ and let $B=P C P^{T} .$ Show that $B$ works. $]$

Runpeng Li · Accepted Answer

and and by a matrix A is positive. Definite. Then there exists a positive definite matrix Be such that a equals B transposed times beat Okay, to solve these to solve this problem on DDE the first thing we need to notice that we can write a as the product off product off p times t times he transposed. So this is just our more textbook that we can write down symmetric right down the the Matrix as Thea p. Times the diagonal times a diagonal make matrix and where p is ah, eyes orthogonal matrix and times p transpose with such p transport ese Exactly the reverse. All right. And the next thing we can observe that if we consider the matrix a diagonal matric see such dead the square off see is D that is just like taking the a square root Tell off Thea diagonal matrix D because scuse me because we for the diagonal matrix deep we&#x27;re just taking the square would help each entry on the diagonal though you be and that is, see and we find the most matrix multiplication of such a C that is the same as the Dagenham. A 60 so that? It seemed to say, since C is also a diagonal matrix hens, that implies the transpose us. I&#x27;ll see his the same matter. See, we have c transposed time. See Equals D. All right. And here we need to make a note. That is a trick. See, see, is also diagonal. Okay, The next thing is, consider as first let&#x27;s define the matrix B to be product off P see transpose off p. And now we re calculate product product out Pete transpose and be transposed and b so that gives can you see transpose I suppose times p see he transports. So the first, the first records things in the inside the breakfast deep sighs take the transpose We have P c transposed and he transposed kinds. You see, Petey and scenes The numbers off p is equal to the transpose Uppity can suppose off p So that means to transpose lp times p will be the identity matrix as we have p times. See transposed times. See? Yeah, this part this part will be cancelled. Okay. I now recall that the product embassy transports and see is exactly deep. So his p times three times he transposed and this is exactly the matrix A as we assumed here. So we&#x27;re that

Joe Lesueur · Answer

the following problem tells us that matrix A is m by n matrix Matrix B is a p by n matrix and we must prove that a times the transpose of be all transposed equals matrix b times the transpose of matrix A. To start this problem, we will use index notation. So a times that trans, whose of matrix b all transposed is I&#x27;m Jay Using that notation, we can take away the final transpose of the entire major cities, have a times transpose a matrix B and we can just foot those and put J i since we&#x27;re transposing the entire operation. And this is also equal to the summation cake. Want em of a J K be transpose Okay, I right now these case are just placeholders were mainly focusing on the the I N J notation. You can see a transfers to just the JK notation all the transpose of B to the K nine notation. I&#x27;ll find them together. The case with cancelling you&#x27;d have a B j I. So the next step of this problem would be to take away the transposed and move it to a We can set this equal to summation K equals 1 10 of be I today. We&#x27;re just taking that transpose here and applying it by switching these around and this is multiplied by a okay, J transposed. We&#x27;re doing the opposite here, or that gene K now flips because we&#x27;re adding the transpose operation. And now, since this notation can also be written back in the form of matrix be multiplied by the transpose of matrix

Pagadala Kishore Reddy · Answer

so you can see the the transport equals you and be any M volume on mathematics. No be clowns fools You be pulled down, schools equal. You can&#x27;t force food we can&#x27;t pull That becomes meet on sports the transport Because you it will be so we can&#x27;t be Hold on We can&#x27;t force you so you can see the You see No by a playing hey equals toe We can see that beat on sports We is also you might think and no, I mean in tow the transport cool middle school The not equal I mean to So you can say that be I mean to be transport. You hear? Medic! Thank you for watching.

Lucas Finney · Answer

for this problem, we have to matrix functions A and T where a is m by N and B is end by p and we want to prove that d a b by d t is equal to a and e b by d t plus d a by d t times be so that the matrix equivalent of the product rule. So what we can do is we know that d by d t a b the i j element of that guy is going to be the same thing as the derivative of the I J element of a times B The i j element of a times B. Using the index definition of the of the Matrix product is going to be the some from K equals one up to N of a element I k Times be Element K J. And we can then bring that derivative inside of the summation there. I&#x27;m just going to write some over K at the bottom. So that&#x27;s the same thing as the sum over the derivative of AI. Important detail these two elements because we are told that these are matrix functions. Those elements are functions of tea so it be of T k j there. So these are scaler functions. Each element is going to be a scaler function. So when we look at the individual elements like that, we can now apply the product rule, as we have been doing since just ordinary calculus. So d by d t of a i k t times be of t k j is going to be d by d t a of t element i k times be of tea element KJ plus in here D by d t be of tea element kj times a of t element i k So that gives us the i J elements there. But if we look at this, you know, we have I wrote this is one Some can rewrite this as being two different sums. Here we have some from k equals one up to end of I&#x27;m just gonna copy and paste that down there for expedience here. Plus some from K equals one up to n. So the next step along here if we look at this first term here, by definition, this is the same thing as the I J element of the I J elements. of D by D T a Times be home and I j there. But if we look over here, you need to be careful. So we have the d by D T k J A f T i k. So when we wants to express something in the index form of the Matrix product, we have the the sum or the index of summation. That has to be the second element of the thing that we have on the left. And it has to be the first element of the thing that we have on the right. So this guy on the right hand side there that guy is the same thing as the i J element of a times D B by D. T. And so that then implies that D by d t of a b is equal to a times db by d t plus d A by D. T times be

Chris Trentman · Answer

you&#x27;re giving the A&#x27;s. They blocked me tricks, eh? Not 00 be not where? Not beyond our lot Mitrice&#x27;s and here asked to prove ET 80 as a certain form, since a zey block matrix we have this is going to be Call it in and by n Matrix and since May not be diagonal lock Mitrice&#x27;s with yeah, that a mouse won&#x27;t be so and by him said Meter, Seas and you know of some and by L or lbl juices. But a not be not Earl suppose diagnosed so that say, you know, metrics Ah e e. And what this terrible Suze it&#x27;s that if a Z two, he said by J, then piece of I J is equal to we have you&#x27;re different cases lies in the day, said Isaac J. And we had at eyes less than or equal to. And and we have that a. I j. This once you diagonal entries that not just won&#x27;t do a not and I working on Jake, is that the same? And if we&#x27;re still in the diagnosis that I was J now we have it, and plus one is less than ableto. It&#x27;s less than two to And yet that Jay Z to you know, I finally I was not a J. This is simply zero. So he said that we do have a diagonal matrix here. So a is one b diet on patriots with entries you want one you got and not one possession I length. And so not a one off. Be not, and minus. So, since a diagonal matrix from a previous sexual side thanks each of the 18 As for me, not one. You know, l t We recognize that this is a diagnosed hatreds and this is actually going to be two different block diagonal matrices. Dagnall blocking associate. There&#x27;s two different Meet your sees Diagonal block Mitrice&#x27;s It&#x27;s actually going to fall more angels with and tree. This is not just an entry that this is a some you treats the nuts eaten. Not so before we actually do this firm. That&#x27;s again as we&#x27;ve shown before since May not be not both, uh oh major cities, same as diver music&#x27;s trees did it? Not one team, and that needs to be not too is the bank of Beatrice entries. And so we see from this that are diagonal matrix 82 diagonal matrix with Look, the angle Neutra sees or diagonal block matrices to be a not too needs of the team is equal to lock diagonal matrix. It&#x27;s not zero zero, did you not? And this is what we want proof from your gun.

Lucas Finney · Answer

So for this problem were given a Matrix A, which is an end by P Matrix matrix d just a diagonal matrix with the one up to D n along its diagonal. And we are asked to show that d a the product of the two it&#x27;s found by multiplying the I throw vector of a by D I where one is less than or equal to I is less than or equal to end. I note here, that&#x27;s for the diagonal matrix the i J. The element is just going to be d I times the chronicler Delta So that is going to the same thing as D. J here. Sorry, I should put that outside of the square braces. That&#x27;s the same thing as D I J. So now we can look at this from the point of view of looking at the index form for the matrix product. So d a element i j is equal to the sum from K equals one up to n uh de element I times. Hey, Element K. J. We know what d element I K is going to be. That is going to be one second. Here was one of two end this D I k is going to be de i Delta I k times the k j eighth element of a But now, because we have this chronicler Delta here, this is going to make this some b are each term in the summer B zero whenever I isn&#x27;t the same thing is K or K isn&#x27;t the same things I so because that kills off all the terms, were just going to have the results. It&#x27;s going Teoh d i times Delta II which is just going to you one times a element i j or that&#x27;s going to be d I names element A Hi, Jay. So now if we consider what the, um I throw of this d a matrix is going to look like So we have d a is equal to, you know, have one dot, dot, dot, dot dot i dot, dot, dot going along s so we will have de i a. I won. Then we&#x27;ll have d I a I to and so on up to d I a i p. Now, since we have these d eyes along each element of this road factor, this is the same thing as doing the row operation of just having d I times that row of ai one a i to up to a i p But this thing here, this row vector that we&#x27;re considering that is the same thing as the I throw of our starting matrix. Hey! And so that proves the that proves the statement.

Problem

Let $A$ and $B$ be symmetric $n \times n$ matrice…

Problem 26 Hard Difficulty

Show that if an $n \times n$ matrix $A$ is positive definite, then there exists a positive definite matrix $B$ such that $A=B^{T} B .[\text { Hint: }$ Write $A=P D P^{T},$ with $P^{T}=P^{-1} .$ Produce a diagonal matrix $C$ such that $D=C^{T} C,$ and let $B=P C P^{T} .$ Show that $B$ works. $]$

Answer

Therefore, if the matrix $A$ is invertible $n \times n$ matrix and if $A$ is positive definite then there exists
positive definite matrix $B$ such that $B^{T} B=A$

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

Linear Algebra and Its Applications

Heather Z.

Maria G.

Kristen K.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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Therefore, if the matrix $A$ is invertible $n \times n$ matrix and if $A$ is positive definite then there existspositive definite matrix $B$ such that $B^{T} B=A$

Linear Algebra and Its Applications

Heather Z.

Maria G.

Kristen K.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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

Heather Z.

Maria G.

Kristen K.

Michael J.

Therefore, if the matrix $A$ is invertible $n \times n$ matrix and if $A$ is positive definite then there exists
positive definite matrix $B$ such that $B^{T} B=A$

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

Linear Algebra and Its Applications