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"Suppose $V$ and $W$ are finite-dimensional and $T$ is a linear map from $V$ to $W$. Prove that there exist a basis of $V$ and a basis of $W$ such that with respect to these bases, all entries of the matrix of the linear map $T$, $M(T)$, are $0$ except that the entries in row $j$, column $j$, equal $1$ for $1 \le j \le \dim(rangeT)$."

My solution:

Let $V,W$ be finite-dimensional and let $T:V \to W$ be a linear map such that all the entries of $M(T)$ are $0$ except for the entries in row $j$, column $j$ , where they are $1$.

Let $w_1,...,w_m$ be a basis of $W$. From the definition of a matrix of a linear map, there exists a basis of $V$ that satisfies the condition of $M(T)$.

Conversely let $v_1,...,v_n$ be a basis of $V$. From the definition of a matrix of a linear map, there exists a basis of $W$ that satisfies the conditions of $M(T)$

Is the solution correct or have I misunderstood the definition?

Edit

Let $v_1,...,v_n$ be a basis of $V$. Let $T:V \to W$ be a linear map and let $w_1, ... , w_n \in W$ $$Tv_k = w_k$$ Since $v_1,...,v_n$ is a basis of $V$, any vector $v \in V$ can be written as a linear combination of $v_1,...,v_n$. Therefore $$ Tv = a_1w_1+...+a_nw_n $$ Therefore $w_1,...,w_n$ spans the range of T. Let $u_1,...,u_m$ be a basis of $W$. Any $w_j$ in the equation above can be written as a linear combination of $u_1,...,u_m$. $$w_j = A_{1,j}u_1 + ... + A_{m,j}u_m $$ where $A_{k,j} \in F$ Therefore we can rewrite $Tv_k$ to $$Tv_k = A_{1,k}u_1 + ... + A_{m,k}u_m$$

Since all the entries of $M(T)$ are $0$ except that the entries in row $j$ column $j$ equal 1, we can simply put $A_{k,k} = 1$ and $A_{1,k}=...=A_{m,k}=0$ for $Tv_k$. We get $$Tv_1 = u_1 + 0u_2 + ... + 0u_m$$ $$.$$

$$.$$

$$.$$ $$Tv_n=0u_1 + ... + 0u_{n-1} + u_n + ... + 0u_m$$

Therefore there exists a basis in $V$ and a basis in $W$ that satisfies the conditions.

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The problem is that your argument is circular. $M(T)$ is defined with respect to a given basis. That is $M(T)$ is really just a stand-in for $M(T;v_1,v_2,\dots,v_n;w_1,w_2,\dots,w_m)$ where $v_1,\dots,v_n$ and $w_1,\dots,w_m$ are bases of $V$ and $W$ respectively. As such, by asserting that there's a matrix where $M(T)$ satisfies the conditions, what you're really doing is asserting there's a basis such that $M(T)$ satisfies the conditions. Instead, what you need to do is focus on finding two bases such that $M(T)$ happens to satisfy the conditions.

Edit: The problem with your post stems from the statement "let $T:V\to W$ be a linear map such that all the entries of $M(T)$ are 0 except for the entries in row $j$, column $j$, where they are 1." This you can't make statements about what $M(T)$ is without saying what basis you're using. For example, Let's pick $T:R^2\to R^2$ is the identity. In addition, let $v_1=(1,0)$ and $v_2=(0,1)$ and also let $u_1 = (1,1)$ and $u_2=(-1,1)$. Then $M(T;v_1,v_2;v_1,v_2)$ is simply the identity while $M(T;u_1,u_2;v_1,v_2)=\begin{bmatrix}1 & -1 \\ 1 & 1\end{bmatrix}$. Simply speaking, it generally doesn't make sense to talk about the matrix for a linear map without knowing or at least implying what that is. As such, you should begin by trying to thing about how to pick linearly independent $v_1,\dots,v_n\in V$ and $w_1,\dots,w_n\in W$ such that $Tv_i=w_i$ as those can be extended to bases which will have a matrix that satisfies the conditions.

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    $\begingroup$ Thank you for the reply. I have edited the post with a new answer. Can you check it? Thanks once again. $\endgroup$ – E.Bob Oct 11 '18 at 17:35
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    $\begingroup$ That still doesn't work as I explained above $\endgroup$ – memerson Oct 11 '18 at 19:51
  • $\begingroup$ Thank you! Got it now. $\endgroup$ – E.Bob Oct 11 '18 at 20:00

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