Similar matrices for matrix with first column zero This is exercise in Artin's Algebra, chapter 4, exercise 3b: 

Let $A=\begin{pmatrix}a&b\\c&d\\ \end{pmatrix}$ be a real matrix. Which matrices with $c=0$ are similar to a matrix in which the "$a$" entry is zero?

I tried: Suppose $\begin{pmatrix}p&q\\ r&s\end{pmatrix}$ is any invertible matrix. Then a similar matrix is $$\frac1{ps-qr}\begin{pmatrix}s&-q\\ \:-r&p\end{pmatrix}\begin{pmatrix}0&b\\ 0&d\end{pmatrix}\begin{pmatrix}p&q\\ r&s\end{pmatrix}=\frac1{ps-qr}\begin{pmatrix}r\left(bs-dq\right)&s\left(bs-dq\right)\\ r\left(dp-br\right)&s\left(dp-br\right)\end{pmatrix}.$$ 
That's it? Is only this required for exercise? or can I do better?
 A: You want to find matrices of the form $A=\begin{pmatrix} x \quad y \\ 0 \quad z\end{pmatrix}$ which are similar to another matrix of the form $B=\begin{pmatrix} 0 \quad b \\ c \quad d\end{pmatrix}$. 
Certainly we must have the same determinant and trace. So $x+z = d$ and $xz = -bc$. Note that in the $2 \times 2$ case, this is enough to determine the characteristic polynomial of the given matrix, since this determines the roots of the characteristic polynomial, which is of degree $2$ and would thus have roots $x,z$. I erred in saying that two matrices with the same characteristic polynomial are similar in the $2  \times 2$ case. Thus, the eigenvalues of $B$ are also precisely $x,z$.
Now, by the JCF, $A,B$ are similar to only two kinds of matrices : one which is diagonal with entries $x,z$ (this is implied by, but not equivalent to $x \neq z$) and the other which is upper triangular with the same entries $x=z$ on the diagonal, and $1$ on the top right corner. If $A,B$ have the same JCF, then they will be similar.

Now, suppose $x \neq z$, then $A,B$ will have distinct eigenvalues, implying that both of them have JCF diagonal with entries $x,z$, showing that they are similar , provided we choose $d = x+z$ and $bc$ to be any entries such that $-bc = xz$.

Suppose $x = z$ now. We note that if $A$ has eigenvector $v= (v_1,v_2)$ then $xv_1+yv_2 = xv_1$ by comparing  first entries of $Av = xv$.
In similar fashion, $Bw =xw$ implies $bw_2 = xw_1$ and $cw_1 = (x-d)w_2 = -xw_2$. Note that $x = 0$ implies $A$ is already a matrix of the form that $B$ is, so we take $x \neq 0$, then note that $bw_2 = xw_1$ again implies that only one  dimension is spanned by an eigenvector of $B$. Thus, $B$ has JCF as a single Jordan block.
Suppose $y \neq 0$. Then, this forces $v_2 = 0$, so we get a unique eigenvector $(1,0)$. It follows that $A$ must have JCF as a single Jordan block with $1$ on the top right. Now, the JCF of $A$ matches the JCF of $B$ so $A,B$ are similar.
But if $y = 0$, then indeed $A$ is a diagonal matrix itself, so doesn't have the same JCF as $B$ anyway, showing that $A,B$ can't be similar.
Finally, the conclusion is that $y \neq 0$ or $x \neq z$ are necessary and sufficient for $A$ to be similar to a matrix of the form $B$.
A: Let the matrix be
$$
A = 
\begin{pmatrix}
a & b\\
0 & d
\end{pmatrix}
$$
Claim 1: If $b\neq 0$ then we can find the similar desired matrix.
If $b\neq 0$, then 
$$
\begin{pmatrix}
1 & 0\\
\frac{a}{b} & 1
\end{pmatrix}
\begin{pmatrix}
a & b\\
0 & d
\end{pmatrix}
\begin{pmatrix}
1 & 0\\
-\frac{a}{b} & 1
\end{pmatrix} = 
\begin{pmatrix}
0 & b\\
\frac{-ad}{b} & a+d
\end{pmatrix}
$$
This proves claim 1
Claim 2: If $a\neq d$ then we can find the similar desired matrix.
The characteristic polynomial is $p(t) = t^2 - (a+d)t + ad$ whose discriminant is 
$$
\Delta = (a+d)^2 - 4ad = (a-d)^2 \geq 0
$$ Now, if $a\neq d$ then $\Delta > 0$ and $p(t)$ will have two distinct real roots. In other words, $A$ will have two distinct real eigenvalues. So it has real independent eigenvectors. This means that $P^{-1}A P= \Lambda$ where $P$ is an invertible real matrix whose columns are the eigenvectors and $\Lambda$ is a diagonal matrix whose diagonal elements are the eigenvalues. Thus, any matrix $B$ that has the same trace and determinant as $A$ will have the same characterstic polynomial and the same eigenvalues so $B$ will be similar to $\Lambda$ and $A, B$ will be similar. Now let
$$
B = 
\begin{pmatrix}
0 & x\\
y & z
\end{pmatrix}
$$
Then by choosing its elements such that $z = a+d$ and $ad = -xy$ we ensure that $A, B$ will have the same trace/determinant and by the argument above, they'll be similar. This proves claim 2.
The only case left is when $a = d$ and $b = 0$. Then $A = aI$. So for any invertible $P$ we have
$$
P^{-1}A P= P^{-1}a I P = aI PP^{-1} = aI
$$
which is obviously not similar unless $a = 0$ in which case we are left with the zero matrix.
To sum it all up, any matrix that is not of the form $aI, a\neq 0$ will do the job.
