Question on basis change and endomorphism This is a question given in the book Mathematics in Machine Learning.

I know how to solve the first part a. of the question, finding the kernel and image of the transformation matrix.
But I don't understand the second part of the question.

*

*I'm not sure what the fact that it's an endomorphism tells me. I understand that endomorphism means that it is mapping to the same vector space and that it is linear. Does that mean it's mapping to a vector subspace of this vector space? (since it's not bijective) But the transformation matrix seems to say otherwise.


*I don't understand what the "(with respect to standard basis in R^3)" part means. Does it mean, the 2 basis sets we are mapping are wrt. to the standard basis, or that we are mapping the standard basis to a new basis? Or something else?


*Finally I don't understand what the b. part is asking me to do. I don't understand the "with respect to the basis B" part either.
I've written the formation of a transformation matrix step by step, but I don't really understand what the question is even saying to infer anything.
I do know there is a formula for finding the transformation matrix when a basis is changed,
A' = (T^−1)(A)(S) (T and S being the transformation matrices to map old basis with new).
But I don't know how I would apply it for this question.
 A: 

*

*I'm not sure what the fact that it's an endomorphism tells me. I understand that endomorphism means that it is mapping to the same vector space and that it is linear. Does that mean it's mapping to a vector subspace of this vector space? (since it's not bijective) But the transformation matrix seems to say otherwise.


But in this example, $\Phi$ is bijective (since the rank is 3, and the kernel is $0$). It's exactly as you said: "endomorphism" just means a linear map from a vector space to itself (there is no assumption on whether or not it is bijective...it might be, but it might not).



*I don't understand what the "(with respect to standard basis in R^3)" part means. Does it mean, the 2 basis sets we are mapping are wrt. to the standard basis, or that we are mapping the standard basis to a new basis? Or something else?


That's right: it means both the domain and range are written in the standard basis. If we call them
$$e_1 = \begin{pmatrix}1\\0\\0\end{pmatrix}, \quad e_2 = \begin{pmatrix} 0\\1\\0 \end{pmatrix}, \quad e_3 = \begin{pmatrix}0\\0\\1\end{pmatrix} $$
then saying $A_\Phi$ represents $\Phi$ in the standard basis means
$$ \begin {align*} \Phi(e_1) &= e_1+e_2+e_3 \\ \Phi(e_2) &= e_1-e_2+e_3 \\ \Phi(e_3) &= e_3 \end{align*} $$
This is just obtained from looking at the columns of $A_\Phi$ (the coefficients of $e_1,e_2,e_3$ on the right-hand sides are the columns of the matrix).



*Finally I don't understand what the b. part is asking me to do. I don't understand the "with respect to the basis B" part either.


Now, if we write the new basis as
$$ v_1 = \begin{pmatrix}1\\1\\1\end{pmatrix}, \quad v_2 = \begin{pmatrix}1\\2\\1 \end{pmatrix}, \quad v_3 = \begin{pmatrix} 1\\0\\0\end{pmatrix} $$
Then the new matrix $\widetilde{A}_\Phi$ for $\Phi$ in the new basis is
$$ \widetilde{A}_\Phi = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} $$
where these coefficients satisfy
$$ \begin{align*}
\Phi(v_1) &= a_{11}v_1 + a_{21}v_2 + a_{31}v_3 \\
\Phi(v_2) &= a_{12}v_1 + a_{22}v_2 + a_{32} v_3 \\
\Phi(v_3) &= a_{13}v_1 + a_{23}v_2 + a_{33}v_3
\end {align*}
$$
So part $(b)$ is asking you to find these coefficients $a_{ij}$ of this matrix $\widetilde{A}_\Phi$.
The formula you gave is the correct one: $\widetilde{A}_\Phi = T^{-1} A_\Phi T$, where $T$ is the matrix whose columns are $v_1,v_2,v_3$:
$$ T = \begin{pmatrix} 1&1&1\\1&2&0\\1&1&0 \end{pmatrix} $$
