Problem 1, Exercise 3.4, Linear Algebra, Hoffman and Kunze The fundamental question is this: Let $V$ be a n-dimensional F-vector space. Is the matrix $A \in F^{n \times n}$ of $T \in L(V,V)$ relative to $\mathcal{B}$ is unique upto row-equivalence or not. With this out, let me discuss what is my issue with the problem.
The problem says: Let $T \in L(\mathbb{C}^2,\mathbb{C}^2)$ be defined as $T(x_1,x_2)=(x_1,0)$. Let $\mathcal{B}=\{e_1, e_2\}$ with $e_1 = \begin{pmatrix}1 \\0 \end{pmatrix}$ and $e_2 = \begin{pmatrix} 0 \\1 \end{pmatrix}$.Let $\mathcal{B}' = \{\alpha_1, \alpha_2\}$ with $\alpha_1 = \begin{pmatrix}1 \\i \end{pmatrix}$ and $\alpha_2 = \begin{pmatrix} -i \\2 \end{pmatrix}$. Then:
a) What is the matrix T relative to the pair $\mathcal{B}, \mathcal{B}'$.
My approach: We know $Te_j = A_{ij} \alpha_i$. So, columns of A are $A_1, ... , A_n$ such that $A_i = [Te_i]_{\mathcal{B}'}$. Now, we can find a $2 \times 2$ matrix Q such that $\begin{pmatrix} \alpha_1 \\ \alpha_2 \end{pmatrix} = Q \begin{pmatrix} e_1 \\ e_2 \end{pmatrix}$ and then $P := Q^{-1}$ gives $X = P X'$, where X is the coordinates of a vector in $\mathbb{C}^2$ wrt the basis $\begin{pmatrix} e_1 \\ e_2 \end{pmatrix}$ and $X'$ is the coordinates of the same vector in $\mathbb{C}^2$ wrt the basis $\begin{pmatrix} \alpha_1 \\ \alpha_2 \end{pmatrix}$ (based on section 2.5-6 of H&K).
Then, $A_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}_{\mathcal{B}'}=P\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ -i \end{pmatrix}$. Similarly, $A_2 = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$. So the answer is \begin{pmatrix} 2 & 0 \\ -i & 0 \end{pmatrix}
This is all right, as it matches with [1,2]. The problem is with b, which says:
b) What is the matrix T relative to the pair $\mathcal{B}', \mathcal{B}$.
My approach: $A_i = [T \alpha_i]_{\mathcal{B}}$, then $A_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}_{\mathcal{B}}=Q\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ i \end{pmatrix}$. Similarly, $A_2 = \begin{pmatrix} -i \\ 1 \end{pmatrix}$. So the answer is $Matrix_1 = \begin{pmatrix} 1 & -i \\ i & 1 \end{pmatrix}$.
Here is the confusion: the answer I got from online solutions (See [1], [2]) is $Matrix_2 = \begin{pmatrix} 1 & -i \\ 0 & 0 \end{pmatrix}$.
$Matrix_1$ and $Matrix_2$ are row equivalent but not the same. So where have I gone wrong?
I got similar results for c) and d), which asks: c) T relative to $\mathcal{B}'$ and d) T relative to $\{ \alpha_2, \alpha_1 \}$
My approach: c) $A_1 = [T \alpha_1]_{\mathcal{B}'} =  \begin{pmatrix} 1 \\ 0 \end{pmatrix}_{\mathcal{B}'}= I \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$. Similarly, $A_2 = \begin{pmatrix} -i \\ 0 \end{pmatrix}$
Thus my answer is: $Matrix_3 = \begin{pmatrix} 1 & -i \\ 0 & 0 \end{pmatrix} $. reference gives $Matrix_4 = \begin{pmatrix} 2 & -2i \\ -i & -1 \end{pmatrix} $ which is row-equivalent to $Matrix_3$. Again, the same question?
d) Similar approach gives me: $Matrix_5 = \begin{pmatrix} -i & 1 \\ 0 & 0 \end{pmatrix} $. reference gives $Matrix_6 = \begin{pmatrix} -1 & -i \\ -2i & 2 \end{pmatrix} $ which is row-equivalent to $Matrix_5$. Again, the same question?
References:
[1] G. Grant, Solutions to Linear Algebra H&K, p64, 
http://greggrant.org/hoffman_and_kunze.pdf
[2] https://linearalgebras.com/solution-hoffman-kunze.html
 A: I have figured it out. I was wrong and the right answer is:
$T = [T]_{\mathcal{B}}=\begin{bmatrix}  1&0\\0&0 \end{bmatrix}$
Let $\alpha = \begin{bmatrix} \alpha_1 \\ \alpha_2 \end{bmatrix} $ and $e = \begin{bmatrix} e_1 \\ e_2 \end{bmatrix} $. Then $\alpha = P e$ where, $P =   \begin{bmatrix}1 & -i\\ i&2 \end{bmatrix} $ and $P^{-1} =   \begin{bmatrix}2 & i\\ -i&1 \end{bmatrix} $.
a) $[Te_1]_{\mathcal{B}} = \begin{bmatrix} 1\\0 \end{bmatrix}$ and $[Te_2]_{\mathcal{B}} = \begin{bmatrix} 0\\0 \end{bmatrix}$
$[Te_1]_{\mathcal{B}'}=P^{-1} [Te_1]_{\mathcal{B}}  = \begin{bmatrix} 2\\-i \end{bmatrix}$ and similarly, $[Te_1]_{\mathcal{B}'}= \begin{bmatrix} 0\\0 \end{bmatrix}$. So, $A =\begin{bmatrix}2&0\\-i&0 \end{bmatrix} $
Also note, $Te_1 = \begin{bmatrix} 1 \\0 \end{bmatrix} = 2 \alpha_1 - i \alpha_2$ and $Te_2 = \begin{bmatrix} 0 \\0 \end{bmatrix} = 0 \alpha_1 - 0 \alpha_2$. So the answer is correct and this is the alternative way of finding the matrix.
b) $T\alpha_1 = \begin{bmatrix} 1 \\0 \end{bmatrix} = 1 e_1 + 0 e_2$ and $T\alpha_2 = \begin{bmatrix} -i \\0 \end{bmatrix} = -i e_1 - 0 e_2$. So, $A =\begin{bmatrix}1&-i\\0&0 \end{bmatrix} $
c) $[T]_{\mathcal{B}'}=P^{-1}[T]_{\mathcal{B}}P$
d) $\begin{bmatrix} \alpha_2 \\ \alpha_1 \end{bmatrix} = \begin{bmatrix} -i&i\\2&i \end{bmatrix} e$, So, $P=\begin{bmatrix} -i&i\\2&i \end{bmatrix}$ and $P^{-1} = \begin{bmatrix} -i&1\\2&i  \end{bmatrix}$.
$[T]_{\mathcal{B}''}=P^{-1}[T]_{\mathcal{B}}P$
all answers match the references.
