I want to know if I solve this exercise in the right way. The exercise reads: consider the matrix $$M = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}$$
and a basis of $\mathbb{R}^2$: $A = \{(1, 1),\ (1, 2)\}$
Consider two applications: $F: \mathbb{R}^2 \to \mathbb{R}^2$ and $G: \mathbb{R}^2 \to \mathbb{R}^2$.
Be $F$ such that $F_e = M$ (the notation means that $M$ is the matrix associated to $F$ wrt the canonical basis); calculate $F_A$ and perform a verification of the resul.
Conversly, be $G$ such that $G_A = M$ (notation as before), calculate $G_e$ and perform a verification.
Solution
About the first part, I used the fact that if one of the two basis is the canonical one, then the matrix representing the passage from the basis $A$ to the canonical one, $F_{eA}$ is simply the matrix whose column are the vectors of the basis $A$, hence
$$F_{eA} = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$$
From this, we have $F_{Ae} = F_{eA}^{-1}$ so then
$$F_{Ae} = F_A = \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}$$
At this point I don't really know what he meant with "verify the result".
I thought I had to calculate $F_A$ in the classic way, so then I wrote
$$(1, 0) = \alpha(1, 1) + \beta(1, 2)$$ $$(0, 1) = \gamma(1, 1) + \delta(1, 2)$$
Finding $\alpha, \beta, \gamma, \delta$ and I found them correctly obtaining $F_A$ again.
About the second one, I have to find $G_e$ and I can use the fact, with my professor notation:
$$G_{ee} = G_{eA} G_{AA} G_{Ae}$$
Where $G_{AA} = G_A = M$.
Again, the matrix with repsect to the canonical basis $e$ from $A$ is the matrix whose columns are the basis $A$:
$$G_{eA} = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$$
again
$$G_{Ae} = G_{eA}^{-1}$$
as before.
Eventually
$$G_e \equiv G_{ee} = G_{eA} G_{AA} G_{Ae} = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}$$
$$G_e = \begin{pmatrix} -2 & 3 \\ -3 & 4 \end{pmatrix}$$
Thence
$$G_e(x, y) = (-2x + 3y,\ 3x + 4y)$$
- As I said: I don't really know what he wants when he asks for the verification. "Verify the result", what does it mean?
Also, what is the meaning of $G_e(x, y) = (-2x + 3y,\ 3x + 4y)$? It doesn't seem to me that $G(1, 1)$ returns $(1, 0)$ and $G(1, 2)$ does not return $(3, 1)$...
Thank you!
You could guess that your interpretation/approach are not quite right because your solution never references the matrix $M$.
In part (a), the intended meaning of the exercise is: Let $F$ be a linear transformation such that its matrix with respect to $e$ (the standard basis) is $M$. Then you need to find the matrix of $F$ with respect to $A$.
Let's think about part (a). When we say "$M$ is the matrix of $F$ with respect to $e$, it means that if $F(a e_1 + b e_2) = c e_1 + d e_2$ then $M \pmatrix{a \\ b} = \pmatrix{c \\ d}$. To put it in words: If you write your input vector and output vector in terms of the basis $e$, then applying transformation $F$ is the same as multiplying by matrix $M$.
Now we want a new matrix $F_A$ such that if we write inputs and outputs in terms of $A$, then applying transformation $F$ is the same as multiplying by matrix $F_A$. The standard way to do this is to find a "change of basis" matrix $P_{Ae}$ that helps us take a vector written in terms of $A$ and write it in terms of $e$ instead. In other words, if $x = r a_1 + s a_2$ and $P_{Ae} \pmatrix{r\\s} = \pmatrix{c\\d}$ then we should also have $x = c e_1 + d e_2$.
If we need a change of basis matrix where the output basis is the standard basis $e$, then we just put our starting basis vectors in the columns of the matrix, getting $P_{Ae} = \pmatrix{1&1\\1&2}$. (This is the same matrix you called $F_{eA}$.) Now think about starting with some vector written in terms of $A$, like $r a_1 + s a_2$. We want to find the matrix $F_A$ such that $F_A \pmatrix{r\\s}$ gives the $A$-coefficients of $F(r a_1 + s a_2)$. To do this, we can use multiple steps.
Putting it all together, we find $F_A = P_{Ae}^{-1} M P_{Ae}$. Once you get used to it, this should feel intuitive: $M$ tells you how to apply $F$ if you have a column vector in terms of $e$, and you start on the right with a column vector written in terms of $A$. So you should translate your vector to be in terms of $e$ instead, then apply $F$ (multiply by $M$), then translate back in terms of $A$. Remember that the first step is written on the right because of how composition works.
Let's compute the final matrix explicitly: $P_{Ae}^{-1} = \pmatrix{2 & -1 \\ -1 & 1}$, and multiplying all the matrices gives $F_A = P_{Ae}^{-1} M P_{Ae} = \pmatrix{7 & 12 \\ -3 & -5}$.
Finally, we can verify our result. If our starting vector is $a_1 = \pmatrix{1\\1}$, then applying $F$ using $F_e$ gives the result vector $F_e \pmatrix{1\\1} = \pmatrix{4\\1}$. Alternatively, we could apply $F$ to $a_1$ using $F_A$. This time we need to write the input vector as a linear combination of $A$, so we write $\pmatrix{1\\0}$ aka $1 a_1 + 0 a_2$. Now we get $\pmatrix{7 & 12 \\ -3 & -5} \pmatrix{1 0} = \pmatrix{7 \\ -3}$, so the result is $7 a_1 - 3 a_2$. Simplifying, we find $7 \pmatrix{1\\1} - 3 \pmatrix{1\\2} = \pmatrix{4\\1}$, which matches the other result we computed. We could run through the same check starting with the vector $a_2$ instead.
I'll leave you to practice doing the similar computation for finding $G_e$, but I'll give an outline. This time you are given the matrix $G_A = M$, so if you write your input vector as a linear combination of $A$ then you can apply $G$ by multiplying by $M$. Your input vector is given to you as a linear combination of $e$. This is very similar to the above example, but this time the steps are: