# 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?

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.

• Easier proof for Claim 2: If $b = 0$, then it follows from Claim 1. Otherwise, $A = \begin{pmatrix} a & 0 \\ 0 & d \end{pmatrix} = U^{-1} \begin{pmatrix} 0 & a \\ -d & a+d \end{pmatrix} U$, where $U = \begin{pmatrix} 1 & a \\ 1 & d \end{pmatrix}$. (The invertibility of $U$ follows from $\det U = d-a \neq 0$.) Commented Jan 26, 2020 at 21:49

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 $$B$$.

• Your argument seems to imply that any 2×2 matrices with the same trace\determinant are similar. This is not true. Finally, the matrix you mentioned above has trace $1+\det{A}$ but since it's similar to $A$ its trace must be equal to $tr {A}$ so $tr{A} = 1+\det{A}$ which is not true in general. Commented Jan 26, 2020 at 21:28
• The similarity class of a $2\times 2$-matrix is not uniquely determined by its tr and det. Keep in mind that there can be a nontrivial Jordan block, which neither tr nor det can distinguish from the identity matrix. Commented Jan 26, 2020 at 21:42
• I see the point. I will make the correction. Commented Jan 27, 2020 at 2:03