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In each item write down the representing matrix of the given linear transformation T : R: + R: with respect to the standard basis of RS, {(1,0,0), (0,1,0) , (0,0,1)...

Question

In each item write down the representing matrix of the given linear transformation T : R: + R: with respect to the standard basis of RS, {(1,0,0), (0,1,0) , (0,0,1)}. (a) T(a,b,c) = (a _ b+c,-2a _ b+ 3c,a + 4c).T is such that (1,0,1) and (0,1,1) are eigenvectors of eigenvalue 2 and T(0,0,1) (1,1,1).T(1,0,1) = (,2,0) and a basis of ker(T) is {(0,2,1), (-1,1,0)}.

In each item write down the representing matrix of the given linear transformation T : R: + R: with respect to the standard basis of RS, {(1,0,0), (0,1,0) , (0,0,1)}. (a) T(a,b,c) = (a _ b+c,-2a _ b+ 3c,a + 4c). T is such that (1,0,1) and (0,1,1) are eigenvectors of eigenvalue 2 and T(0,0,1) (1,1,1). T(1,0,1) = (,2,0) and a basis of ker(T) is {(0,2,1), (-1,1,0)}.



Answers

Determine the matrix representation $[T]_{B}^{C}$ for the given linear transformation $T$ and ordered bases $\bar{B}$ and $C$. $T: M_{2}(\mathbb{R}) \rightarrow \mathbb{R}^{2}$ given by $$T(A)=(\operatorname{tr}(A), \operatorname{tr}(A))$$ (a) $B=\left\{E_{11}, E_{12}, E_{21}, E_{22}\right\} ; C=\{(1,0),(0,1)\}$ (b) $B=\left\{\left[\begin{array}{cc}-1 & -2 \\ -2 & -3\end{array}\right],\left[\begin{array}{ll}1 & 1 \\ 2 & 2\end{array}\right],\left[\begin{array}{cc}0 & -3 \\ 2 & -2\end{array}\right],\left[\begin{array}{ll}0 & 4 \\ 1 & 0\end{array}\right]\right\}$ $C=\{(1,0),(0,1)\}$

So in order to solve this problem, we first need to do one find t of 0100 t of 0010 and he 00 01 So the first thing to notice is that tee off 1100 minus t of 1000 is gonna equal 51 minus three negative to which, of course, on the right hand side will be equal to 23 But if you look on the left hand side, you'll note that T is only in your transformation, which means I can factor t out of my expression. In this case, what it will look like is this t 1100 minus 1000 which, of course, will simplify too t of 0100 equal to 2/3 again, if we also look at t of 1110 minus t of one 100 that's gonna equal. Ah, negative. 10 minus 51 Which, of course, on the right hand side will equal neg of sex. Negative one. But looking at the left hand side, the transformation is a linear. So again we can factor out out of the terms. In this case, we're going to get tee off 1110 minus 1100 And this simplifies to t of 0010 which equals your six negative one. So finally, define our last term. Uh, look, at t of 11111 minus t of 11 one zero. And that's gonna equal negative one. Actually, that's gonna equal to to minus negative 10 Which, of course, the right hand side is gonna equal 30 and on the left hand side again for the same reason. Because to use linear weaken, factor it out of our expression. In this case, we get tee off. 111 one minus 11 one zero. It was simplifies to t of 0001 is equal to three zero. So putting off together the linear transformation or the lin the matrix representation of our linear transformation is going to be three minus two 2/3 minus six minus one 30

Problem Number two, which is a weavers, say that T one is equal to one I love plus Dean Wax. Last year, experts so secret one mind the three times zero and two times one plus zero mind it's true times here which you will be equal to 1 to 40 x we have here at X quarter fission which is corresponding to zero minus three times zero and two time zero plus one minus two times here was will be equipped to bear in mind for the X Square. So here we have only expert proficient and this will give and get it to be ending it too. Ah to show that ah, he won and the ex ante expert are at Landry combinations. So we can in a t 11 times one and deal plus two times you're at one anti NCC 20 time is one and zero plus one times you don't want and t X squared is equal to negative one and zero plants negative to zero in one we give negative three and bigots too as shown here So we can say that t one. Oh, see what I wanted to feel the access that you don't want. It's a square in dentistry and negative tour, so that tee off CMB is one of 201 and negative three and against similarly, for a coaching be we say that you want as we to hear it wanted to the same this year. 41 plus X Here we have Moscow efficient on eggs and it gives one and three and one exports X square. So here we have three coefficients here is it's one. So this gives begins and what we have to throw that t one and t one plus X and 31 plus X perfects. It is where it is on any recombination or better see So here we can see that t one which is equal to one to acquit toe a a month the blind by one and negative one plus B month buying by toe and one by solving these equations which gets from here we say that a is equal to negative one and music which one? Eso Similarly using this approach, we can get a bad T one plus acts. They would donate 5/3, uh, one and negative one plus 4/3 21 C plus one x x squared to go negative for entry one the negative one minus one on +13 21. So TCF be one on one with the fiber. Three over. 4/3 but the forward three over negative one over three.

In this example, we have a transformation t that's going to be linear, and what it does is it takes the first column of the two by two identity matrix, and it maps it to this element and are too. Likewise, the second element of the identity matrix is mapped to negative 5 to 00 of our two. Our goal, then, is to find a matrix. A such that t of X could be expressed as a Times X If we confined such a matrix A. Then we can use The Matrix to evaluate T at any Vector X, not just these two vectors. So the first step is to note that the standard Matrix A if we have these two elements, is given as follows. The Matrix say will be t evaluated one for the first column and t evaluate at E two for the second call. So for this situation, we know that t one is 3131 TV two is negative. Five to 00 And so this matrix A is the standard matrix for the transformation. T


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