How to realise the equality of a map in the quotient vector space

Question

Suppose $$A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i\end{pmatrix},$$ and $$v = \begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}$$ is an eigenvector for $A$ with eigenvalue $\lambda$. Here $V= \mathbb C^{ 3}$ be the vector space that $A$ acts on, and we get that $\mathbb C v \subseteq V$ is an invariant subspace for $A$. Therefore, $A$ acts on the quotient vector space $V/\mathbb C v$. If we take $e_1, e_2, e_3$ to be the standard basis for $V$, then one basis for $V/\mathbb C v$ is given by $e_1 + \mathbb C v, e_2 + \mathbb C v$(I have realised it just because it is killing one dimension). Then how to see that the matrix of the transformation on $V/\mathbb C v$ with respect to this basis is given by $$\begin{pmatrix}a-g & b-h \\ d - g & e-h\end{pmatrix}$$ i.e., subtracting the last row from the first two rows, taking only the first two columns.

I tried and then get baffled in the work of the main vector space and quotient space.

I have tried it in this way, we have to take $\overline {\begin{pmatrix}x \\ y \\ z\end{pmatrix}} \in V/\mathbb C v$ and identify $\begin{pmatrix}a-g & b-h \\ d - g & e-h\end{pmatrix}\overline {\begin{pmatrix}x \\ y \\ z\end{pmatrix}}$ with $\overline { \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}}$.

Now, $$ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}=\begin{pmatrix}ax+by+cz \\ dx+ey+fz \\ gx+hy+iz\end{pmatrix}=\begin{pmatrix}ax+by+(\lambda-a-b)z \\ dx+ey+(\lambda-d-e)z \\ gx+hy+(\lambda-g-h)z\end{pmatrix} = \\ \begin{pmatrix}ax+by+(-a-b)z \\ dx+ey+(-d-e)z \\ gx+hy+(-g-h)z\end{pmatrix}+\lambda z\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}$$. So, $$\overline { \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}} = \overline{\begin{pmatrix}ax+by+(-a-b)z \\ dx+ey+(-d-e)z \\ gx+hy+(-g-h)z\end{pmatrix}}$$

Now there are two questions

1. Is there any simpler way to prove the original question?
2. Is there any simpler method to justify my arguement?

Answer 1

Here's a way of framing the calculation that I like.

First of all, rather than going through the alphabet I will use the convention that $a_{ij}$ denotes the entry of $A$ in the $i$th row and $j$th column.

The only part of your argument that might need justification is that $\{\{e_1 + \Bbb C v, e_2 + \Bbb C v\}\}$ indeed forms a basis of $V/\Bbb Cv$; to see this, it suffices to note that that $e_1,e_2,v$ are linearly independent. Now, we want to find the matrix of the transformation that $A$ induces on $V/\Bbb C v$ with respect to the basis $\mathcal B = \{e_1 + \Bbb C v, e_2 + \Bbb C v\}$; I'll use $\tilde A$ to denote the induced map. We have $$ \tilde A(e_1 + \Bbb C v) = (a_{11} e_1 + a_{21} e_2 + a_{31} e_3) + \Bbb C v. $$ If we can write this output in the form $\alpha(e_1 + \Bbb C v) + \beta (e_2 + \Bbb C v)$ for some $\alpha,\beta,$ then $(\alpha,\beta)$ will be the first column of the linear transformation with respect to $\mathcal B$. To find this $\alpha,\beta$, we need to select a representative from the equivalence class $\tilde A(e_1 + \Bbb C v)$ whose $e_3$ coefficient is zero. To that end, we note that $$ (a_{11} e_1 + a_{21} e_2 + a_{31} e_3) + \Bbb C v = \\ (a_{11} e_1 + a_{21} e_2 + a_{31} e_3) - a_{13}v + \Bbb C v = \\ (a_{11} - a_{31})e_1 + (a_{21} - a_{31}) e_2 + \Bbb Cv. $$ Thus, we have $\alpha = a_{11} - a_{31}$ and $\beta = a_{21} - a_{31}$, which is to say that the matrix of $\tilde A$ with respect to $\mathcal B$ has the form $$ [\tilde A]_{\mathcal B} = \pmatrix{a_{11} - a_{31} & ?\\ a_{21} - a_{31} & ?}. $$ By calculating $\tilde A(e_2 + V)$, we similarly deduce that the second column is $(a_{12} - a_{32},a_{22} - a_{32})$.