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Let A aud B be n following identity:matrices ad B; be the ith column 0l B for = [We want t0 prove thedet([ABB,] +det([Bi ABzBa]~det([BABa] = tr(A) det (BIF You S pr...

Question

Let A aud B be n following identity:matrices ad B; be the ith column 0l B for = [We want t0 prove thedet([ABB,] +det([Bi ABzBa]~det([BABa] = tr(A) det (BIF You S prore the above dlirectly; do it . Otherwise follow the following steps;Show that for any fixed B the mapping T : R"x" _ R defined byT(A) = det([ABBal det([B AB,B,]det( [B1AB,]linear transformation; Let {E;j:i.j=1,2 "} be the standard basis of R"xn where E,j is the n X n matrix with in the (i. j)-entry and with 0 in

Let A aud B be n following identity: matrices ad B; be the ith column 0l B for = [ We want t0 prove the det([AB B,] +det([Bi ABz Ba] ~det([B ABa] = tr(A) det (B IF You S prore the above dlirectly; do it . Otherwise follow the following steps; Show that for any fixed B the mapping T : R"x" _ R defined by T(A) = det([AB Bal det([B AB, B,] det( [B1 AB,] linear transformation; Let {E;j:i.j=1,2 "} be the standard basis of R"xn where E,j is the n X n matrix with in the (i. j)-entry and with 0 in other entries_ Show that T(E;;) tr(E;j) det(B). Show that T(A) tr( A) dlet ( B).



Answers

Verify the uniqueness of $A$ in Theorem $10 .$ Let $T : \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$
be a linear transformation such that $T(\mathbf{x})=B \mathbf{x}$ for some $m \times n$ 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.

Hello there. Okay, so for this exercise we have a transformation from the space of matrices. Uh square matrices two by two Into the two dimensional cleaning space. The transformation is the finest photos here. We pick as an input A matrix A, B, C and D. And we return a picture. Not corresponds to the sum of a plus B plus C. And the second term in the vector is just 1/4 term in the matrix. We're going to consider the following basis. B will be the standard basis for the space of matrices. The prime will be the standard basis for the two dimensional Cleveland space, and B Double prime will be defined. The basis defined by these two vectors 1, 1 and -1. Or task is to find the the representation of this transformation relative to the basis B and B prime and BMB double prime. So you can observe that here. It is easiest to work with the standard basis. So we're going to start with the easiest matrix representation uh relative to the basis. B and B prime. So in order to construct this, we need to consider the basis. B. That is just a standard basis for Mitrice's. Just to remind you how is defined this this basis. Well corresponds to the following 1000 0100. The Matrix 0010. And the matrix here 0001. Okay, so this is a standard basis for this base of matrices. And then we need to apply the transformation that we have to find it here t to each of these basic elements in order to construct this. This transformation this basis here. Okay, so first we need to apply to the first matrix here. So that means to the matrix here. 1000. And the result of this will be the victor 10. Fortunately here, we're working with the standard basis for the garden space, two dimensional clean space. So this is already written in the basis of the prime for the for the rest will be I'm going to put some labels to make this simpler. So I'm going to pull this and one, two, 23 and four. So if we apply t to the matrix to the second matrix we will obtain 10 as well. Because how is defined this transformation two and 3 We obtain the same. Here's the vector one, zero. And finally to M four. So T applied to the 4th matrix will return as the director 01. All these factors are already in the basis be prime. Right? So yeah, the the the the last thing that we need to do in order to obtain our matrix is just to put each vector as a column in the matrix. So this is different. B and B. Prime. And the matrix will be one want 1110 and 000, one. Great. Now we need to define this transformation with respect to the other basis. But the only thing that we're going to change is how we're right. And these vectors here. So technically we're going to obtain the same. So let me copy this to make things faster. So now we're going to right. The representation for this transformation with respect to the basis, B and B. Double price be double prime Is defined by two factors 1, one And the Vector -1, 0. This is a basis for me to and the basis. B. We know how it is defined well, and two and 1 and two And three and 4. And you already know what is the structure of these matrices? So when we apply the transformation we're going to obtain the same victors. But here we have a change. Let me read this. These vectors are in the standard notation or in the standard basis for the right. This here. Okay, okay, so all these vectors are in the basis of in the standard basis for us to and we need to write in these basics here. Okay, so, well, technically, well, I'm going to make more space here from Italy there, is it? Okay, so this is the first transformation to the first element in the basis of matrices. So we need to find how to grant this vector In the basis of B two. The way to do this is we'll find The linear combination. That results into this vector 1.0 means all for one, 11 Plus, out for two minus 10. You can observe that here is really easy to find. This. Alpha one should be equal to zero and alpha two will be equal to minus one. This implies that this vector written in the basis be double prime will be the vector zero minus one in the basis. Be the prime. So that means that all the factors that is 10 corresponds to the vector 0 -1. In the basis be double price. The same for 23. Don't here take this here. So there's also 10. So this corresponds to the vector zero minus one B double play. and four days we need to repeat. Right? So we put here the vectors 01 and we need to find the linear combination Alpha one Times A Victory 1, 1 plus over two and here minus 10. No, you can observe that here. Alpha one should be close to one and alpha two should be close to one as well. So the victor In the basis be double prime will be the Victor 11. And we have both the correspondent vectors to write the transformation the representation of this transformation in the basis with respect to the bases, B and B. Double price. So the matrix will be the following zero 001 -1 -1 minus 11. We put each victor as a column for these matrix. Great. So that's the first part. Now we have 22 matrices with respect to different bases. And now we need to apply this so we need to calculate the transformation of the following matrix 1, 2, 3, 4. But we need to use the matrices that we have to find. So that means that we need to follow the three step procedure. So this is the first matrix do we obtain? It is defined by one, 1, 1, 0. And here 0001. All right. With respect to the to the basis, B and B double prime is a matrix is just above this bit of prime equals to the matrix 000, one -1 -1 -1 and one. Okay, so we have here these two matrices define way to follow the three step procedure in order to obtain the transformation of these matrix. So the first thing is to write these matrix. So that's the first step. I'm going to start with this matrix first. So let's start with this matrix B. And this is the prime. So, as I mentioned, the first step is great. These matrix into the basis. B B is the standard basis of matrices. That means that this Matrix can be represented by the Vector 1, 2, 3 and four in the basis be. That's the first step. In the second step we need to apply the transformation but in order to apply the transformation, we need to multiply these matrix. The matrix of the transformation times deep vector in this case I'm going to call it X B times the vector X B. And we're going to obtain as a result the application of this transformation to the victor in this space be prime. Okay. So that's what the this equation it's saying. So we just need to multiply the matrix 1110, 0001 by the vector that we how to find it here. So by the Vector one 2, 3 and four. Their assault is the vector six and four. And this of course is written in the basis be prime, which corresponds to the basis, the standard basis for our two. So this is already correct. Okay, we we don't need to transform anything. This is already in the the result that we want. So we escaped the third step. The third step consists in taking the vector that we have in the basis that we have obtained it and then eliminate the put in the standard basis of this space. But in this case the prime is already the standard basis for our two. So we don't need to do anything here is that is a result of this transformation. Now let's do this with the second transformation matrix that is considering here relative to the basis B and B double prime. So again here we need to use we have the following transformation that we want to use 123 and four. And the transformation I think that they've written here and here we have the transformation. Okay so the first step we know we need to pass these matrix into the basis beat. But that means considering the vector 123 and four, that is just as a previous case. I'm going to go this X. B. Of course In the 2nd step, what we need to do is just multiply the transformation matrix. But by the picture 0001. Maine is 1 -1 -1 applies to the vector one 234. And the resulting vector in this case is the vector four minus T. But this of course isn't the basis be double pride. Let's remember the double prime Is defined by the Raptors. 1, 1 And -10 here we need to apply the third step. So for the third step we need to pick this vector here and transform it into the standard basis for our two. So this is indeed double prime but this represents just four times The victory. 1, 1 -2 times the vector -10. And you can observe that this result into the vector six four. That is the same as the previous results we obtain in this part 64. Okay. And the last thing is to what happened if we apply directly in the calculation. So that means considering just the rules of this transformation. 2, 1, 2, 3, 4. Now let's remember that if we apply directly this this transformation means taking one plus two plus three, and here we put the number four And you can observe that. Indeed, the result is six four, just like in the previous cases.

Thank you. Cause to with Reese aim bean. See anti B equals toe threes. Yes, Jeanne. Which Okay, close. Be transfer schools in course. Gasohol aid Must be. He's a last you c plus G b plus the plus inch transpose. All thes is born a plus in see close G B plus deep last inch. No, the calculation of differently transpose all a It calls to a see be de I'm transport will be No, because to e dreaming Yeah, ich so a trance pools plus be transposed equals a plus a c plus she if plus you deep loss Which, which means that a plus be Thrones pools equals two a trance pools last be transposed lorded be a transport trans schools because to a see be deep trance pools Commission wants to Okay, being thievy the we should wants to Okay, for the sea A that three people's too a e lost clear BG b itch most a the G lost the the H see flow A be transpose It wants to b g plus a the G plus see b itch last a If the h yes see also we have them be transposed a trance pools meted waas full e g age wants to build boy a C be the which waas toe exactly a B transpose.

Hello there. Okay. So in this actually says we're going to work with to transformation. The first response to in the transformation and the second one to a linear operator. We're going to work in the spaces of poor animals Of most degree one and most agree to. So p. one and p. two. The first transformation is the finest Philip we pick the polynomial and then we multiply X. Instead this second transformation, it's going to be a linear operator actually. And what's going to modify is the argument of the point of view For both spaces. B one and B. two. We're going to work with the standard basis for each space. So the first thing that we need to do is try to do is to calculate or obtain the transformation matrices of these transformations we have here to want to of the composition. So let's start with the simplest one happened with T. one. The transformation from the to be prime. So there's the first that we're going to work with. Okay, so this transformation technically pick P. One and transform it into P. Two. So we need to consider the basic elements of the one that is one X. And we need to apply the transformation rule in this case T. One. Just to remind you you just multiply X in front of the polynomial. So we need to construct this these matrix by computing by applying this transformation to each basis element. So T. Of one work responds to X. And instead T of X. Is X square. Now we can represent these vectors in the basis B. This point was in the basis be so this X. is 010. Mhm in the basis B. And similarly X squared is just the element 001 in the base beef. This will help us to construct the matrix. And from this we obtained that the matrix associated to the first transformation within relative to the basis. B. And B. Prime. Mm. This is yeah, this is the prime B. Prime. Yeah, he's just produced these vectors as columns of the matrix 010 and 001. This is in the basis b. one. Great. So that's 41. Now we need to do the same 42. So for the two we are going to remain in the space. In the coordinate be prime in this case this transformation just change the argument of the polling models. Couldn't be we put Be of two x plus one. Okay, so in this case the basis, what is the prime is the standard basis of the two? So one X and X square. And again we need to repeat the procedure as before we apply T. Two to the basis, element one. And that means just what T applied to X is two explosive 1. And finally did two of x squared is just four. X square plus four plus four plus four. X plus one. Again this point I must have a A better representation in this basis. So there is 100. There's Polynomial is the factory 1- zero. And finally this is the vector 144. And from this we can construct the matrix that we need. So the matrix relative to the basis be prime will be defined by 100, 1-0 and one 44. Right now what happened with the composition? So let's see what it's going to be the composition of this. Well basically T 2 0. T. one will be a transformation relative to the basis. B. And B. Pride. How it's going to be defined this composition? Well, Taiwan composed with two of a polynomial text will be first taking X times P. X. And then we need to apply the second transformation to this point. And this will be just two x plus one times well with a different color. There will be two X plus one times the polynomial. And which is the argument to ex post war. Okay, so let's do this for the basis elements in B. That means just want an X. So T. 2 0. T. one of one Is like T. 2 2 x. But this is just we just have to calculate there's an ease to explosive one. Now we need to do the same for tea for X. This is the second element in the basis B. So this will be applied the second transformation to the polynomial X. Square at this is just four X. Square plus four X plus of one. Again, we need to represent this as a basis in the basis of the prime. So this is the factory 1-0 in the basis. Big prime. And similarly this will be one for for the basis B. Price from these two pictures. We can construct the transformation the representation of this composition of transformations and it will be the matrix 1- zero and 144. Yeah. Okay, great. Now in the part B we need to state. Okay, uh they're related a formula relating the matrices in part A. That means and There is a relation. Okay, we know that he composed with T one in the basis. Be prime, be can be written as a multiplication of matrices and that is the multiplication of the matrices of the two in the prime on T one relative to the basis. B and B. Prime. Okay, that's the relation that we have. And of course we need to verify that this is true. So that's further part C. So we knew already what's going to be these matrix here but we don't know what's going to be the result of this matrix multiplication. So that's what we need to do here. Did to be prime. is defined by the vector by the Matrix 1, 1, 1, 02, four and 004. The Matrix T one Is defined by 01. zero, Okay. And when we multiply these two matrices together, we obtain the matrix 12144, zero. And you can observe that this expression with the previous one here, Well, this is the mineral multiplication. And here I'm going to copy 2, 2, 21. These matrix was 1-0 and 144. So, well, I think that at this point is clear. This, too, are the same. So this relation that were great in be history.

Were given a vector space V with a basis be with basis vectors, be one through bien and were asked to find the bee matrix for the identity transformation I from B to itself. So we have the B coordinate vector of B J. Well, this is simply going to be e j, which is the standard basis vector for RN. This is because, for example, we have that P J is going to be zero times be one all the way up to zero times B J minus one plus one times B J plus zero times B J plus one all the way up to zero times bien. So it follows that thebe coordinate vector of the image of B. J. This is the same as Thebe. Coordinate Fechter of B J Since I is just the identity which we are too determined is E J and therefore the identity matrix relative to be This is going to the matrix with columns standard basis vectors e one e two through eat n. We see that this is the same as the identity matrix dimensions and by n


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