# Linear algebra proof (matrices and linear transformations)

## Homework Statement

Let T $$\in$$ L(V, W), where dim(V) = m and dim(W) = n. Let {v1, ..., vm} be a basis of V and {w1, ..., wn} a basis for W. Define the matrix A of T with respect to the pair of bases {vi} and {wj} to be the n-by-m matrix A = (aij), where

$$T(v_{i}) = \displaystyle\sum_{j=1}^{n}a_{ji}w_{j}, 1 \le i \le m, 1 \le j \le n.$$

The vector spaces V and W are isomorphic via the bases {vi} and {wj} to the spaces Fm and Fn, respectively. Show that ifx $$\in$$ Fm is the column vector corresponding to the vector x $$\in$$ V via the isomorphism, then Ax is the column vector in Fn corresponding to T(x). In other words, the correspondence between linear transformations and matrices is such that the action of T on a vectorx is realized by the matrix multiplication Ax.

## The Attempt at a Solution

I can see this is true, and when we learned this in class, it was pretty clear to me. But now I'm going through some linear algebra book and the stuff is introduced in a different way that confuses the hell out of me.

What I've tried to do here is just show what T(x) and Ax get you, and that they are equal. But I just can't get to that, it seems.

For T(x), I get the following:

$$T(x) = T(\displaystyle\sum_{i=1}^{m}b_{i}v_{i}) = \displaystyle\sum_{i=1}^{m}b_{i} \displaystyle\sum_{j=1}^{n}a_{ji}w_{i},$$

and for Ax:

$$Ax = \displaystyle\sum_{j=1}^{n} (\displaystyle\sum_{i=1}^{m}a_{ji}x_{i}) = \displaystyle\sum_{j=1}^{n} (\displaystyle\sum_{i=1}^{m}a_{ji}b_{i}v_{i}).$$

I don't know how to make the jump from vi to wj. Or have I gone completely in the wrong direction? Any help would be greatly appreciated.

Last edited:

I believe there is a problem with matrix multiplication in your solution. The first warning: how did you get $$v_i$$'s into the result? It clearly shouldn't have happened, since the dimensions don't match. I'd say
$$Ax=\sum_{j=1}^n\left(\sum_{i=1}^m a_{ji}b_i w_j\right)$$