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Question

Suppose $A$ is a symmetric $n 	imes n$ matrix and $B$ is any $n 	imes m$ matrix. Show that $B^{T} A B, B^{T} B,$ and $B B^{T}$ are symmetric matrices.

Pagadala Kishore Reddy · Accepted 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 want to prove that if A and B are symmetric matrices and it&#x27;s probably important they have to be the same dimension than the product, A B will be symmetric as well. So the fact that A and B are symmetric tells us that Element A. I J he&#x27;s going to equal elements A. J I and similarly element be i j is going to equal Element B J I. It&#x27;s now if we look at the index definition of the Matrix product A B elements i j, they&#x27;re going to equal the sum over. Let&#x27;s I assume they are end by or m by N matrices. So we have some from k equals one to n of elements. I&#x27;ll use lower case is toe specify the elements it of element ai j Times Element B I. J. Well, if we consider those two elements because we know that the major seas are symmetric or ah symmetric, then a I J is the same thing as Element A. J. I and B I. J is the sell the same thing as Element B. J I, which is the same thing as the J life element of the product. A B so therefore a B has to be symmetric as well

Fan Yang · Answer

Yes, Magics A is, uh, and times and my chicks Foon A The transfer magics off a ico is, uh, in times magics so we can know a transfer. The transfer metrics off a tons magics A yes, on my chicks off and tough and times no tricks with, um, Tom&#x27;s and Matt Chicks. The paradox off thes two magics is who will be, uh, in times and my chicks so, so chance for all full. It is also well, be the transfer off our magics, which is incomes. And it&#x27;s also well, pee in terms and magics, no less Kong. Third, we also can know the A tire&#x27;s a transfers. Transfer metrics will be Ah, um tongues Magics. So so do you. Uh, so the a chance for oh, so a times a transfer magics the products off thes and the transfers and magics Well, ico chu a transfer from four times a transfer, which you called Chu about chicks a times a transfer soon. So a times a trance first transfer magics equals a times a transfer so we can see it is, uh, my and minus se mi chicks mad tricks

Angelo Rendina · Answer

we want to show their A plus a transpose it symmetric matrix, which means that a plus a transpose old transposed is equal to itself. So I processed transport to do that, we look at its genetic i j element. It is by the finish off transposed a plus a transpose imposition J I. Now, by definition, off some. This is the J Iife entryway, plus the J. I&#x27;ve entry off a transpose. But now we use again that the division of Transposed so the first some money is just a transport in position, i j. And the second some Monday&#x27;s a transpose transposed in position I day. So these a transposing position i j plus while the second someone we have the transport of our transposed. And that&#x27;s just the matrix itself. So that&#x27;s a position I j. And now, by definition, off some disease, a transpose blase in position I j. And finally using the fact that the sum is communicative. This is just a placid, transposed in position, i j. But now the stinking green and this thing in brew of the same. And so we checked that both mattresses have the exact same entries and therefore that the two matrices are equal

Lucas Finney · Answer

for the first part of the problem we want to answer If a is symmetric What r B and C So what are its symmetric and sq symmetric parts? So yeah, intuitively we can think about Okay, well, be by definition, it&#x27;s the symmetric part in C is the skew symmetric part. If a is symmetric, then be should just be the same thing as a But we can prove this formally And we can also show that C is going to be zero. So see or sorry, B is equal to 1/2 of a plus a transpose. But by definition, if a is symmetric than a transpose is equal to a that&#x27;s actually gonna be 1/2 of a plus a That&#x27;s just going to be a and similarly see is 1/2 of a minus a transpose. But in a world where a is equal to a transpose, then that means that that&#x27;s just going to 1/2 of zero is gonna be zero then for part B, we want to answer if a is skew symmetric Excuse my hand writing here What are B and C Well, in this case again, I&#x27;m just looking at it going to say B has to equal zero and C has to equal a But we can prove this formally so we&#x27;ll have. B is equal to 1/2 a plus a transpose. But if a skew symmetric, then that means that a is equal to negative a transpose that&#x27;s equal to 1/2 a minus a, which is equal to zero and see what&#x27;s going to equal 1/2 a minus a transpose, Yeah, minus a transpose. It&#x27;s going to equal 1/2 of a minus negative A which is going to be obviously inside a plus a So we&#x27;re gonna we&#x27;re gonna have 1/2 a plus a so that&#x27;s just gonna equal a there we have it.

Matthew Allcock · Answer

in this video, we&#x27;re gonna go through the answer to question number 41 from chapter 9.3. So we&#x27;re given a definition off what it means for matrix and in my own matrix to be symmetric eso the matrix A asymmetric if the transpose of a is equal to itself. So we&#x27;re asked to show that if a is in my own matrix, then a plus a transpose is a symmetric matrix so less let be beat matrix A plus a transpose Then that&#x27;s considered what be transpose is okay, that&#x27;s gonna be a plus a transpose transpose. We know that Trump&#x27;s Bo, the operator of transposing is distributive across edition. So this is there gonna be a transpose plus a transpose transfers. So this is gonna be a a transpose close, and then transposed transpose is just three identity operator, So that&#x27;s just gonna be a and then we know that when we add major cities together, it doesn&#x27;t matter which order we after him. So not a transpose a or we can try a transpose s. So this is gonna pay just a plus. Hey, transfers, which is itself be so therefore be transfers is equal to bay, so be

Problem

Suppose $A$ is invertible and orthogonally diagon…

Problem 28 Hard Difficulty

Suppose $A$ is a symmetric $n \times n$ matrix and $B$ is any $n \times m$ matrix. Show that $B^{T} A B, B^{T} B,$ and $B B^{T}$ are symmetric matrices.

Answer

since $A$ is symmetric, $\left(B^{T} A B\right)^{T}=B^{T} A^{T} B^{T T}=B^{T} A B,$ and $B^{T} A B$ is
symmetric. Applying this result with $A=I$ gives $B^{T} B$ is symmetric. Finally,
$\left(B B^{T}\right)^{T}=B^{T T} B^{T}=B B^{T},$ so $B B^{T}$ is symmetric.

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

Linear Algebra and Its Applications

Catherine R.

Heather Z.

Caleb E.

Maria G.

Absolute Value - Example 1

Absolute Value - Example 2

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since $A$ is symmetric, $\left(B^{T} A B\right)^{T}=B^{T} A^{T} B^{T T}=B^{T} A B,$ and $B^{T} A B$ issymmetric. Applying this result with $A=I$ gives $B^{T} B$ is symmetric. Finally,$\left(B B^{T}\right)^{T}=B^{T T} B^{T}=B B^{T},$ so $B B^{T}$ is symmetric.

Linear Algebra and Its Applications

Catherine R.

Heather Z.

Caleb E.

Maria G.

Absolute Value - Example 1

Absolute Value - Example 2

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

Catherine R.

Heather Z.

Caleb E.

Maria G.

since $A$ is symmetric, $\left(B^{T} A B\right)^{T}=B^{T} A^{T} B^{T T}=B^{T} A B,$ and $B^{T} A B$ is
symmetric. Applying this result with $A=I$ gives $B^{T} B$ is symmetric. Finally,
$\left(B B^{T}\right)^{T}=B^{T T} B^{T}=B B^{T},$ so $B B^{T}$ is symmetric.

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

Linear Algebra and Its Applications