How to calculate the Matrix of a given Linear Transformation?

Question

Let $V = F^3$ and $W = F^4$ and we define the following functions:

$p\in {\cal L}(V,F)$ given by $p((x,y,z)) = 3x + 4y + 2z$

$q\in {\cal L}(W,F)$ given by $q((w,x,y,z)) = 2w + 5x + 7y + 11z$;

$T\in {\cal L}(V,W)$ given by $T((x,y,z)) = (x,x+y,x+y+z,y+z)$; ${\cal B}_V = ((1,1,1),(1,-1,0),(1,1,-2))$ and ${\cal B}_W = ((1,1,1,1),(1,-1,0,0),(0,0,1,-1),(-1,-1,1,1))$ are ordered bases for $V$ and $W$ respectively.

Now I would like to find the matrix of $T$ relative to ${\cal B}_V$ and ${\cal B}_W$ and the matrix of $T^t$ relative to the dual bases. I am in search of some hints for solving this problem.

Just to add that this is not a homework problem which I must submit. I have found this in a problem sheet given out by some professor. Any help would be appreciated greatly.

Answer 1

Putting $\mathcal B_V=[b_1,b_2,b_3]$ (so $b_1=(1,1,1)$ etc.) compute $T(b_1)$, $T(b_2)$ and $T(b_3)$ using the definition of$~T$ (all three are elements of $W$), and express each of those vectors in coordinates with respect to $\mathcal B_W$; put those coordinates in three successive columns, and you've got the matrix of$~T$ with respect to the bases $\mathcal B_V$ of $V$ and $\mathcal B_W$ of $W$. The matrix has size $4\times 3$.

The dual bases are composed of the coordinate functions with respect to $\mathcal B_W$ (which you already used in the first part to find the entries of each of those columns) and with respect to $\mathcal B_V$. To find the matrix of $T^t$, play the same game as before: apply $T^t$ to each of the coordinate functions associated to $\mathcal B_W$, and write the resulting functions in terms of the coordinate functions associated to $\mathcal B_V$; for each, the coefficients give the entries of one column of the matrix of $T^t$. The matrix will have size $3\times 4$, and if doen correctly will be the transpose of the matrix found in the first part.