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## Homework Statement

Let T [tex]\in[/tex] L(V, W), where dim(V) = m and dim(W) = n. Let {v

_{1}, ..., v

_{m}} be a basis of V and {w

_{1}, ..., w

_{n}} a basis for W. Define the matrix

**A**of T with respect to the pair of bases {v

_{i}} and {w

_{j}} to be the

*n*-by-

*m*matrix

**A**= (a

_{ij}), where

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

The vector spaces V and W are isomorphic via the bases {v

_{i}} and {w

_{j}} to the spaces F

_{m}and F

_{n}, respectively. Show that if

**x**[tex]\in[/tex] F

_{m}is the column vector corresponding to the vector x [tex]\in[/tex] V via the isomorphism, then

**Ax**is the column vector in F

_{n}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**.

## Homework Equations

## 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:

[tex]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},[/tex]

and for Ax:

[tex]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}).[/tex]

I don't know how to make the jump from v

_{i}to w

_{j}. Or have I gone completely in the wrong direction? Any help would be greatly appreciated.

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