verify-the-uniqueness-of-a-in-theorem-10-let-t-mathbbrn-rightarrow-mathbbrm-be-a-linear-transformati

Question

Verify the uniqueness of $A$ in Theorem $10 .$ Let $T : \mathbb{R}^{n} ightarrow \mathbb{R}^{m}$
be a linear transformation such that $T(\mathbf{x})=B \mathbf{x}$ for some $m 	imes n$ matrix $B .$ Show that if $A$ is the standard matrix for $T,$ then $A=B .$ IHint: Show that $A$ and $B$ have the same columns.

Michael Jacobsen · Accepted Answer

in this video we&#x27;re considering of transformation T, which has standard matrix A and we also have an interesting situation where Thi Vex is also equal to be X, and this is going to hold for all X and RN. Well, let&#x27;s start with first that the standard matrix of the transformation T is a what this means is that the Matrix A can be written as TV one for the first column. TV two for the second column all the way through t of e n For the inthe column. Recall that the Matrix is going to be of size M by n so we have and columns altogether. The Matrix B is going to be of the same size, but let&#x27;s just label the columns as generically b one b two all the way through B N. So now we have two major cities, and since a is the standard matrix, we know that T of X is equal to eight times X. But at the same time, T of X is also equal to B times X. We wonder next is a equal to be, or is it possible for leaner transformations that these two matrices can be different. Well, one way to tell is to go. And first look at this entry right here. Recall that T of X is equal to be X for all X and e one is one such X. So we can say that TV one is going to be equal to B times. He won, but this is equal to a matrix B times This column vector will be a one in the first century. Then the remaining entries are all zeros. If we multiply B by. Such a are such a vector, e one. We in fact pick up the first calm of B B one so so far a is equal to TV one, but the first calm is B one. Likewise, if we look at say, TV too, this is b times E to which gives us be too. So we have a B two here that came from this position. Then going down the line. TVN would also be equal to be in, and that tells us that the matrix A is in fact equal to be so Our work is complete. We know now that a must be equal to B, but this work tells us even more than that. It tells us when we find a linear transformation and we form it&#x27;s standard matrix. This standard matrix A is unique. If we try to have a second matrix B, satisfy the same information that a must satisfy, we just come up with a conclusion that a is in fact be

Jimmy Yao · Answer

way need to apply. Just urine. Four Queen 9.1. So it is the ring, um, to rink nal ity serum. So it stays for any A is m by and the rank of A Because the minority of a it&#x27;s the culture in into this province It gave up its condition. That&#x27;s the common space. So that there Colin Space, it&#x27;s you co two than now space So they aren&#x27;t you go. So if there you go one thing is the dimension of the Colin space this Nico to the dimension of the now space. Right? So the dimension of the constipated should go to the rank of A into by definition, the dimension of the now space. It&#x27;s really equal to the reality that a so here this end of this are the same number. Say, if this is X and this will be X, then end will be two x considered the Dimension X. Here is there is the interchange. So again, it&#x27;s two times I muted. So Ben is even en If even the any it even means there Colin Number is even so a have even college

Jimmy Yao · Answer

don&#x27;t just the problem here we want to argue are a Is that many independent? So So there The corner of a linear independents is equivalent to say a s incredible. So it is actually a is inevitable if, in the only if agents pose a it is also you, Bertel. So they said speakers If we see this terrorist in so the so so here for this demolition If a the colon is they&#x27;re Carlin&#x27;s of a it&#x27;s linear independence. Then we know a is you&#x27;re irritable. So if a is the convertible and so we also know a transpose is inevitable So this is because the rink of a it&#x27;s ricotta rink of transposed So yes, so they have the humors so safe. But he knows the U S of a transport be ancient spell beavers and the humorous of a will be a uber&#x27;s. So we say a transpose Huber&#x27;s and a universe we&#x27;ll be the humors of a transpose eight. This is because it&#x27;s their time. Together, it&#x27;s echo to a uber&#x27;s in a transpose universe, a transposed in a and become to a transpose i A. If you go to a inverse, a issue could you buy. So this means it is the inverse matrix of a chance was eight. So agents pose a is convertible, so for two this direction so we know a chance. Close A. It&#x27;s the Invincible due to their rink knowledge t serum. We know that the rink of agents pose a trust that not a team of aid you This chance was a It&#x27;s Nico to in the hair agents will agency in veritable So we know this rink. It&#x27;s because you in so the knowledge he should be sure that you come to narrow So we know if agents pose a eggs. If you co 20 you have to have X You cut your hero, right? They were going to consider the now space of eighth. So for the now space of a so it looks like it&#x27;s satisfies a X equal to zero. So we choose any experience now space okay thing than now of age. So this aches had a property. A ex coaches there. So also, if we transposed, there&#x27;s we&#x27;re gets shows X transferal agents bill should also be there. Oh, if we time that to them together we will get X transport a gentle a X if you cut through there. Mm, No, I don&#x27;t want to do that into this is enough. So if X it&#x27;s lifting this now space a X is equal to zero and we times a transpose to two this vector. So we know agents will A X is equal to zero since we already know if there now space for a transpose a is contents in the their oh, so we can derive that extra butte there. So if exit zero then we means this means this now space for any the X in the snow space We can deduce that X is zero. So this means there&#x27;s now space only contend content zero So according to their well so well, I don&#x27;t want to do that. So for any So we know there no space here, so only condemned zero. So the knowledge t for a it&#x27;s their so called into the rank nal ity. Mr Room. We know the rank of a it&#x27;s in. So here we know the the colon, the colon that&#x27;s a his denia independence

Lucas Finney · Answer

for this problem. We wants to prove to statements. When is that given a matrix? A. That eight times its transposed is a symmetric matrix. The second is that given Major sees a, B and C that the transpose of the product of those three matrices is equal to see transpose sense be transposed onto a transpose. So first looking, answering part A. We know that using the index definition of the Matrix product that a times a transposed the I Jeff element of that is going to be the some from K equals one upto end, assuming it&#x27;s a m by n matrix of elements ai que times the k j element of a transpose. But we know from definition that for a transposed, it&#x27;s i J element is equal to the J. I&#x27;ve element of a so setting that as an aside, that means that we can rewrite this sum up above some from K equals one up to end ai que as so we need to reverse the indices that we have here. So it&#x27;ll be a i k Times element a J K. Now, the next thing that we can dio, you can recognize that these two elements are just scale er&#x27;s unlike matrices scale er&#x27;s do commute so we can reverse their order there. Hagel&#x27;s one up to end of a J K A. I k. And this here is the same thing as of the J IFE element of a times a transpose that&#x27;s proving the first statement Now from be here we have a times, b, times C and we want to find the i J element of its transpose. So the idea of element of the transpose of ABC first of all, going to be same thing as the J IFE element of a times B times C the J I have elements of a times B times C so we&#x27;re going to have to do a do two consecutive sums here. So, first of all, we want to have some from K equals one up to end of a I K. Sorry. In this case, since we have the indices reversed, that should be a um yeah, sorry. That should be a J k times the element A B c element k i BC element K I we can write as second here that element there. It&#x27;s going to be the some from l equals one up to n both these air up to end Still of be elements K l times see Element Uh, l I now we can bring the sums to be next to each other equals one upto end times These some from l equals one up to end So you have a double summation there a j k b k l c l i. So now what? We can dio way we recognize him. I know I didn&#x27;t got a little bit lazy with using the square braces, but these air referring to the elements of the matrices we can recognize is that each of these elements they&#x27;re being multiplied together. Those are all scale ear&#x27;s so we can rearrange the order of that multiplication cake was one up to end times some from l equals one up to end of C L I B k l A J K. The next thing is that these elements here no element cli. Actually, I&#x27;ll write my summation Summations here with the limits of the summation unchanged CLI is the same thing has element I l of C transpose and B k l is the same thing as element. Okay. Oops. That should be for be transposed there. Moment l k times a transpose element k j which is equal to I will say, that was K. That was l some over k of C transpose I l Actually, I&#x27;ll use l as the outer summation here some over l of see transposed element I l times the some over k of be transposed element L k times a transpose element KJ, which is equal to the sum over l of C transpose element I l times second here be transposed a transpose element l j which is equal to element i j of c transposed be transposed a transpose, thus proving the second statement.

Bryan Lynn · Answer

prints. So prom 22 here deals with symmetry, and so was one quickly de fine what it means for a matrix to be symmetric. It&#x27;s that we take the transport of the Matrix. The result is theory journal matrix. Okay, so we&#x27;re given they&#x27;re going to show that a matrix multiplied by the transpose is symmetric. So to do that, I mean to show that the transpose of that is going to be equal to what we started with, right? The original matrix. Okay, we want to start with our matrix multiplied bye, transposed and take the transports of that. We can use an identity this at, uh, two matrices multiply by each other that transpose it with a transpose it a times B is going to be the transpose it be times the transposed a case. We pry that here. And we&#x27;ll get then that the transpose of the transposed times just a transposed is equivalent. Well, we also have this identity that d transposed of the transpose is equal to our original matrix. Right? So if we apply that, then what we have here we then get just a times the transpose right. But that is indeed what we started with case we took the transpose and got back out our original matrix and so therefore it is symmetric.

Ernest Castorena · Answer

Manny once approved that the universe of a matrix is unique. Well, let&#x27;s say were given to me. Thank. Let let&#x27;s say it had to and versus being see then. Since B is an inverse, we know this is true. They&#x27;re both the same no matter where you multiply it because you just kept the identity. And now the same is true for us. He you end up that being the same no matter where you multiply it cause you gotta be ability Think it so whenever we want to show that this inverse is unique calling it be here we&#x27;re going to see here you want to end up with the equation. B is equal to see the one. We actually just multiply both sides of this equation. A times B is equal time by the matrix Eat one We multiply like that. We just have that scene time saying time speak is equal to see According to that correct foreigner matrix, multiplication is associate. So we could just as easily put the parentheses or on sea time Thank and then multiply be after. But we know that see times a is also the a day. So this is equal to the identity. Sometimes a ranger moving in with the identity times. Anything is just equal. So that matrix, So we&#x27;ll be. But there is a chain of equalities here, and we&#x27;ve seen that we do have B is equal to see.

Problem

Why is the question "Is the linear transformation…

Problem 33 Hard Difficulty

Answer

$A=B$

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

Linear Algebra and Its Applications

Grace H.

Anna Marie V.

Caleb E.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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$A=B$

Linear Algebra and Its Applications

Grace H.

Anna Marie V.

Caleb E.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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

Grace H.

Anna Marie V.

Caleb E.

Michael J.

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

Linear Algebra and Its Applications