Matricial Representation and change of basis

Question

I was doing a exercise for Tom. Apostol Volume II about linear transformations, it´s this:

"A linear transformation $T: V_2 \rightarrow V_3 \text{ maps the basis vectors as follows: } T(i) = (1, 0, 1), T(j) = (-1,0, 1).$ Find bases $(e_1, e_2)$ for $V_2$, and $(w_1, w_2, w_3)$ for $V_3$, relative to which the matrix T will be in diagonal form."

So I found $(e_1, e_2) = (i, j)$ and $w_1 = (1, 0, 1), w_2 = (-1, 0, 1), w_3 = (0, 1, 0)$, but the book's answer is $(e_1, e_2) = (i, i+j)$, $w_1 = (1, 0, 1), w_2 = (0, 0, 2), w_3 = (0, 1, 0)$.

In fact, I noticed the book's answer is a linear combination of my answer, but this wouldn't have for contradict the fact of each matrice is associated with only one basis? Or have I made some mistake in my approach?

I would aprecciate any help. Sorry for my bad english.

Answer 1

Your answer is correct!. A linear map with respect to two different bases can have the same matrix.

Let $V$ and $W$ be finite dimensional vector spaces. Let $L(V, W)$ denote the set of all linear maps $T : V\to W$. Let $F^{m,n}$ denote the set of all $m\times n$ matrices with entries from $F$. Let $M : L(V, W)\to F^{m,n}$ be the linear isomorphism that takes a linear map $T\in L(V, W)$ to a matrix in $F^{m,n}$ with respect to some choice of basis $v_1,\dots,v_n$ of $V$ and $w_1,\dots,w_m$ of $W$. Let $M(T, (v_1,\dots,v_n), (w_1,\dots,w_m))$ denote the matrix of a linear map $T$ with respect to the basis $v_1,\dots,v_n$ of $V$ and $w_1,\dots,w_m$ of $W$.

The way the isomorphism $M$ should be interpreted is that: For every matrix $A\in F^{m,n}$, there exists a choice of basis $v_1,\dots,v_n$ of $V$ and $w_1,\dots,w_m$ of $W$ such that there exists only one linear map $T\in L(V, W)$ such that $M(T, (v_1,\dots,v_n), (w_1,\dots,w_m))=A$.

Since we already know that $M$ is an isomorphism, we know that there does not exist a choice of basis $v_1,\dots,v_n$ of $V$ and $w_1,\dots,w_m$ of $W$ and a linear map $S\ne T$ such that $$M(T, (v_1,\dots,v_n), (w_1,\dots,w_m))=M(S, (v_1,\dots,v_n), (w_1,\dots, w_m))$$

Here the bases picked for $T$ can also be different to the bases picked for $S$. Since it is an isomorphism, a matrix in $F^{m,n}$ can only be associated with one linear map. Although the choice of basis with respect to which we take the matrix can be different.

Thus your answer is completely correct. The authors might have had a different way for you to approach the problem, that is why there is a different answer. But once you compute with your answer and their answer, the matrices are completely the same. Namely,

$$M(T, (i,j), (w_1,w_2,w_3))=\begin{bmatrix}1 & 0 \\ 0 & 1\\ 0 & 0 \end{bmatrix}$$

and with respect to the bases in the answer in the book we have

$$M(T, (i,i+j), (w_1,w_2,w_3))=\begin{bmatrix}1 & 0 \\ 0 & 1\\ 0 & 0 \end{bmatrix}$$