# Prove that all entries in $M(T, (v_1, … ,v_n), (w_1, … , w_m))$ are zero except for the entries in row j, column j

"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.

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.