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Suppose that $M$ is a $3 \times 3$ matrix over an algebraically closed field and suppose further that $M$ is not a scalar multiple of the identity matrix. I have a feeling the following statement is obtainable by considering the Jordan normal form of $M$ but I am not so comfortable with the notion of rational canonical forms and am looking for a nice explanation to the following question.

How do we show $M$ is similar to a matrix whose diagonal is $(0, 0, \operatorname{Tr}(M))$?

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I thought about it a bit; if the Jordan canonical form has a single $3\times 3$ block, then the rational canonical form will do it since the matrix is similar to the companion matrix of its characteristic polynomial. But I'm not sure what to do in general. I'll keep thinking about it at the gym today, I just wanted to let you know the question is not forgotten. – Arturo Magidin Dec 6 '11 at 16:33
Thanks for your all your help. – user7980 Dec 6 '11 at 21:21
Okay, I can get it done in every case except when there is a single eigenvalue and the Jordan form has a $2\times 2$ and a $1\times 1$ block, and when the matrix has exactly two eigenvalues and is diagonalizable. – Arturo Magidin Dec 7 '11 at 20:20
Bit more progress; of the remaining case (diagonalizable, $\lambda$ a double eigenvalue, $\mu$ an eigenvalue) $\lambda=0$ is trivial, I can do $\mu=0$, and I can do $\lambda=-\mu\neq 0$. I have a system of equations that I need to solve in the remaining case, but the simplificating assumptions I tried didn't work. I wonder if there is a simple way of doing it in general... – Arturo Magidin Dec 15 '11 at 6:39
Okay, finally got it. – Arturo Magidin Dec 15 '11 at 20:17
up vote 3 down vote accepted

I can solve the problem (essentially by doing the Rational Canonical Form "behind the scenes) in all but two cases one case now, namely, the cases where the minimal and characteristic polynomials don't coincide: when the Jordan form is $$\left(\begin{array}{ccc} \lambda & 1 & 0\\ 0 &\lambda & 0\\ 0 & 0 & \lambda \end{array}\right),\qquad\lambda\neq 0,$$ (Added: Done now; see portion after the horizontal line below); and when the Jordan form is $$\left(\begin{array}{ccc} \lambda & 0 & 0\\ 0 & \lambda & 0\\ 0 & 0 & \mu \end{array}\right),\qquad \lambda\neq\mu, \lambda\neq 0.$$

Added: All done now. There were a couple of cases where the rational canonical form didn't do the trick, and I kept messing up the resulting equations. Finally did them carefully enough. See at the end.

When the minimal polynomial and the characteristic polynomial of $M$ coincide, then the companion matrix works. The reason is that a $3\times 3$ matrix is completely determined by its characteristic and minimal polynomials: if $A$ and $B$ are $3\times 3$ matrices, and the characteristic polynomials of $A$ and $B$ are equal and the minimal polynomials of $A$ and $B$ are equal, then $A$ is similar to $B$. You can see this by considering the Jordan canonical form, which will have to be the same in all cases when this occurs (as long as you are working in a $1$-, $2$-, or $3$-dimensional space; it breaks down once you get to $4\times 4$, since $$\left(\begin{array}{cccc} \lambda & 1 & 0 & 0\\ 0 & \lambda & 0 & 0\\ 0 & 0 & \lambda & 1\\ 0 & 0 & 0 & \lambda\end{array}\right)\qquad\text{and}\qquad \left(\begin{array}{cccc} \lambda & 1 & 0 & 0\\ 0 & \lambda & 0 & 0\\ 0 & 0 & \lambda & 0\\ 0 & 0 & 0 & \lambda\end{array}\right)$$ have the same characteristic polynomial ($t-\lambda)^4$), the same minimal polynomial $(t-\lambda)^2$), but are not similar.

The companion matrix to $p(t)$ is a matrix whose characteristic and minimal polynomial are both equal to $p(t)$. Hence, if the characteristic and minimal polynomial of $M$ are equal, equal to $\chi_M(t)$, then the companion matrix of $\chi_M(t)$ has the desired form: because $$\begin{align*} \chi_M(t) &= (t-\lambda_1)(t-\lambda_2)(t-\lambda_3)\\ &= t^3 - (\lambda_1+\lambda_2+\lambda_3)t^2 + (\lambda_1\lambda_2+\lambda_1\lambda_3+\lambda_2\lambda_3)t - \lambda_1\lambda_2\lambda_3\end{align*}$$ where $\lambda_1,\lambda_2,\lambda_3$ are the eigenvalues of $M$, and the companion matrix is then $$\left(\begin{array}{ccr} 0 & 0 & \lambda_1\lambda_2\lambda_3\\ 1 & 0 & -\lambda_1\lambda_2 - \lambda_1\lambda_3 - \lambda_2\lambda_3\\ 0 & 1 & \lambda_1+\lambda_2+\lambda_3 \end{array}\right).$$ Since the trace equals the sum of the eigenvalues, and $M$ is similar to this matrix since they have the same characteristic and minimal polynomials, we are done.

To explicitly construct the similarity matrix, start with a Jordan canonical basis $\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3$ of $M$; then let $\mathbf{w}_1$ be the sum of the "heads" of each Jordan block (the vector corresponding to the lower right corner of the block), and let $\mathbf{w}_2=M\mathbf{w}_1$ and $\mathbf{w}_3=M\mathbf{w}_2$. If you work it out, you'll find that the matrix of $M$ relative to this basis (which will be basis) is exactly the one above.

For example, if the Jordan canonical form has $2\times 2$ block associated to $\lambda$ and a $1\times 1$ block associated to $\mu$, with $\lambda\neq \mu$, and $\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3$ is the corresponding Jordan canonical basis, then let $$\begin{align*} \mathbf{w}_1 &= \mathbf{v}_2 + \mathbf{v}_3\\ \mathbf{w}_2 &= M\mathbf{w}_1 = \mathbf{v}_1 + \lambda\mathbf{v}_2 + \mu\mathbf{v}_3\\ \mathbf{w}_3 &= M\mathbf{w}_2 = 2\lambda\mathbf{v}_1 + \lambda^2\mathbf{v}_2 + \mu^2\mathbf{v}_3. \end{align*}$$ Verify that the coefficient matrix of the above, $$\left(\begin{array}{rrr} 0 & 1 & 1\\ 1 & \lambda & \mu\\ 2\lambda & \lambda^2 &\mu^2 \end{array}\right)$$ is invertible, so that $\mathbf{w}_1,\mathbf{w}_2,\mathbf{w}_3$ is a basis, and that $$M\mathbf{w}_3 = \lambda^2\mu\mathbf{w}_1 - (\lambda^2+2\lambda\mu)\mathbf{w}_2 + (2\lambda+\mu)\mathbf{w}_3,$$ which will mean that the matrix of $M$ relative to $\mathbf{w}_1,\mathbf{w}_2,\mathbf{w}_3$ will have the desired form.

The result cannot hold for nonzero scalar multiples of the identity, since they are central so they are only similar to themselves.

But I'm not yet sure what to do with the two cases mentioned at the beginning. Trying to do the same thing as above leads to a set of linearly dependent vectors, and some attempts at tweaking the construction didn't work. I'll give it some more thought.

Okay, to deal with the case where the Jordan canonical form of $M$ is $$\left(\begin{array}{ccc} \lambda & 1 & 0\\ 0 & \lambda & 0\\ 0 & 0 & \lambda \end{array}\right),$$ I proceeded as follow: whatever $\mathbf{w}_3$ is, we need $M(\mathbf{w}_3)$ to have $3\lambda$ as its $\mathbf{w}_3$ component. One way to achieve this is to have $\mathbf{w}_3 = \mathbf{v}_1 + 2\lambda\mathbf{v}_2$, and then just make sure that $\mathbf{w}_1$ and $\mathbf{w}_2$ don't have any $\mathbf{v}_1$ components; then $M(\mathbf{w}_3) = 3\lambda\mathbf{v}_1 + 2\lambda^2\mathbf{v}_2$, and we will need $3\lambda\mathbf{w}_3$ to express it (since $\mathbf{w}_3$ is the only vector that has a $\mathbf{v}_1$ component). Trying with $\mathbf{w}_1=\mathbf{v}_2$ and $\mathbf{w}_2=\mathbf{v}_3$ cannot work (since $\mathbf{v}_3$ is an eigenvector). So I tried $$\begin{align*} \mathbf{w}_1 &= \mathbf{v}_2 + \mathbf{v}_3\\ \mathbf{w}_2 &= \mathbf{v}_2 - \mathbf{v}_3\\ \mathbf{w}_1 &= \mathbf{v}_1 + 2\lambda\mathbf{v}_2. \end{align*}$$ And that worked. We have $$\begin{align*} M\mathbf{w}_1 &= \mathbf{v}_1 + \lambda\mathbf{v}_2 + \lambda\mathbf{v}_3\\ &= (\mathbf{v}_1 + 2\lambda\mathbf{v}_2) - \lambda(\mathbf{v}_2 - \mathbf{v}_3)\\ &= -\lambda\mathbf{w}_2 + \mathbf{w}_3.\\ M\mathbf{w}_2 &= \mathbf{v}_1 + \lambda\mathbf{v}_2 - \lambda\mathbf{v}_3\\ &= -\lambda \mathbf{w}_1 + \mathbf{w}_3.\\ M\mathbf{w}_3 &= 3\lambda\mathbf{v}_1 + 2\lambda^2\mathbf{v}_2\\ &= 3\lambda(\mathbf{v}_1 + 2\lambda\mathbf{v}_2) -4\lambda^2\mathbf{v}_2\\ &= -2\lambda^2\mathbf{w}_1 - 2\lambda^2\mathbf{w}_2 + 3\lambda\mathbf{w}_3. \end{align*}$$ So the matrix of $M$ relative to the basis $\mathbf{w}_1,\mathbf{w}_2,\mathbf{w}_3$ is $$\left(\begin{array}{rrr} 0 & -\lambda & -2\lambda^2\\ -\lambda & 0 & -2\lambda^2\\ 1 & 1 & 3\lambda \end{array}\right),$$ which is of the form we needed.

Why did I try $\mathbf{v}_2+\mathbf{v}_3$ and $\mathbf{v}_2-\mathbf{v}_3$? I'll confess it simply seemed like a good idea at the time.

Now we're only missing the diagonalizable with two distinct eigenvalues case.

Okay, I finally solved it, essentially by brute force. I would be interested to know if there is a simpler solution to this.

We are down to the case in which the matrix $A$ is diagonalizable with two distinct eigenvalues $\lambda$ and $\mu$, with $\lambda$ a double eigenvalue. That is, the Jordan form of $M$ is $$\left(\begin{array}{ccc} \lambda & 0 & 0\\ 0 & \lambda & 0\\ 0 & 0 & \mu \end{array}\right).$$ Let $\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3$ be a basis of eigenvectors, with $\mathbf{v}_3$ corresponding to $\mu$.

Case 1. $\lambda=0$. Then the Jordan canonical form (and the Jordan canonical basis) solve the problem.

Case 2. $\mu=0$.

Let $\mathbf{w}_1 = \mathbf{v}_1+\mathbf{v}_3$, $\mathbf{w}_2=\mathbf{v}_2+\mathbf{v}_3$, and $\mathbf{w}_3 = \mathbf{v}_1+\mathbf{v}_2+\mathbf{v}_3$. Then $$\begin{align*} M\mathbf{w}_1 &= \lambda\mathbf{v}_1 = 0\mathbf{w}_1 -\lambda\mathbf{w}_2 + \lambda\mathbf{w}_3\\ M\mathbf{w}_2 &= \lambda\mathbf{v}_2 = -\lambda\mathbf{w}_1 + 0\mathbf{w}_2 + \lambda\mathbf{w}_3\\ M\mathbf{w}_3 &= \lambda\mathbf{v}_1+\lambda\mathbf{v}_2 = -\lambda\mathbf{w}_1 -\lambda\mathbf{w}_2 + 2\lambda\mathbf{w}_3. \end{align*}$$ So the matrix of $M$ relative to $[\mathbf{w}_1,\mathbf{w}_2,\mathbf{w}_3]$ is $$\left(\begin{array}{rrr} 0 & -\lambda & -\lambda\\ -\lambda & 0 & -\lambda\\ \lambda & \lambda & 2\lambda \end{array}\right).$$

(I'll note how I obtained that basis below).

Case 3. $\lambda=-\mu$. In this case, we can use another modification of the Rational canonical form. take $\mathbf{w}_1=\mathbf{v}_1+\mathbf{v}_3$, $\mathbf{w}_2 = M\mathbf{w}_1 = \lambda\mathbf{v}_1-\lambda\mathbf{v}_3$, $\mathbf{w}_3 = \mathbf{v}_2$.

Then $M\mathbf{w}_3 = \lambda^2\mathbf{v}_1 + \lambda^2\mathbf{v}_3 = \lambda^2\mathbf{w}_1$. So we have that the matrix of $M$ is given by $$\left(\begin{array}{rrr} 0 & \lambda^2 & 0\\ 1 & 0 & 0\\ 0 & 0 & \lambda \end{array}\right).$$

The key here is that the $M$-annihilator of $\mathbf{v}_1+\mathbf{v}_2$ is $(t-\lambda)(t+\lambda) = t^2-\lambda^2$, so the companion matrix has $0$s in the diagonal.

Case 4. Final case: $\lambda\neq 0$, $\mu\neq 0$, $\lambda\neq\pm\mu$.

Here's how I proceeded: let $E_{\lambda}$ and $E_{\mu}$ be the eigenspaces. Then $\mathbf{V}=E_{\lambda}\oplus E_{\mu}$.

Neither $\mathbf{w}_1$ nor $\mathbf{w}_2$ can be eigenvectors; $\mathbf{w}_1$ will have nonzero $E_{\lambda}$ and nonzero $E_{\mu}$ components. By scaling if necessary, we may take $\mathbf{w}_1 = \mathbf{v}_1+\mathbf{v}_3$.

The same argument holds for $\mathbf{w}_2$; but if the $E_{\lambda}$ component of $\mathbf{w}_2$ is just a multiple of $\mathbf{v}_1$, then we may assume that the $E_{\lambda}$ component of $\mathbf{w}_3$ is $\mathbf{v}_2$ (since that component will have to give a basis together with $\mathbf{v}_1$). But then conditions don't add up: we have $\mathbf{w}_1 = \mathbf{v}_1 + \mathbf{v}_3$, $\mathbf{w}_2 = \alpha\mathbf{v}_1 + \beta\mathbf{v}_3$, and $\mathbf{w}_3 = \mathbf{v}_2 + \gamma\mathbf{v}_3$. Expressing $M\mathbf{w}_3$ in terms of the basis will give a linear combination in which the coefficient of $\mathbf{w}_3$ is $\lambda$, but we need it to be $2\lambda+\mu$.

Therefore, we may assume that $\mathbf{w}_2 = \mathbf{v}_2+\alpha\mathbf{v}_3$.

Finally, $\mathbf{w}_3=\beta\mathbf{v}_1+\gamma\mathbf{v}_2 + \delta\mathbf{v}_3$. If $\mathbf{w}_3$ "works", then so will $\frac{1}{\delta}\mathbf{w}_3$, so we may assume that $\delta=1$. Thus, we will have: $$\begin{array}{rcccccc} \mathbf{w}_1 &=& \mathbf{v}_1 & & & + & \mathbf{v}_3\\ \mathbf{w}_2 &=& &&\mathbf{v}_2 &+&\alpha\mathbf{v}_3\\ \mathbf{w}_3 &=&\beta\mathbf{v}_1 &+&\gamma\mathbf{v}_2 &+&\mathbf{v}_3. \end{array}.$$

The conditions we want mean that we need to express $M\mathbf{w}_1$ using only $\mathbf{w}_2$ and $\mathbf{w}_3$; $M\mathbf{w}_2$ using only $\mathbf{w}_1$ and $\mathbf{w}_3$; and $M\mathbf{w}_3$ using $2\lambda+\mu$ times $\mathbf{w}_3$. From these, we quickly get the following equations: $$\begin{align*} M\mathbf{w}_1 &= \frac{\lambda}{\beta}(\mathbf{w}_3) - \frac{\lambda\gamma}{\beta}(\mathbf{w}_2)\\ M\mathbf{w}_2 &= \frac{\lambda}{\gamma}(\mathbf{w}_3) - \frac{\lambda\beta}{\gamma}(\mathbf{w}_1)\\ M\mathbf{w}_3 &= -(\lambda+\mu)\beta\mathbf{w}_1 - (\lambda+\mu)\gamma\mathbf{w}_2 + (2\lambda+\mu)\mathbf{w}_3 \end{align*}$$ Substituting the values of $\mathbf{w}_1,\mathbf{w}_2,\mathbf{w}_3$, we get the following equations for $\alpha$, $\beta$, and $\gamma$: $$\begin{align*} \frac{\lambda}{\beta}(1-\alpha\gamma) &= \mu\\ \frac{\lambda}{\gamma}(1-\beta) &= \mu\\ \lambda + (\lambda+\mu)(1-\beta-\alpha\gamma) &= \mu. \end{align*}$$ (When $\mu=0$, it follows that $\beta=1$, $\alpha\gamma=1$, so we can take $\alpha=\beta=\gamma=1$, which is how I got the basis in Case 2 above).

Substituting $1-\alpha\gamma = \frac{\beta\mu}{\lambda}$ into the third equation, we get $$\lambda + (\lambda+\mu)\left(\frac{\mu}{\lambda}-1\right)\beta = \mu.$$ Solving for $\beta$ we obtain $\beta=\frac{\lambda}{\mu+\lambda}$ (hence the need to separate Case 3).

Plugging into $\frac{\lambda}{\gamma}(1-\beta) = \mu$ we obtain $\gamma =\frac{\lambda(\mu+\lambda)}{\mu}$.

Plugging both into $\frac{\lambda}{\beta}(1-\alpha\gamma) = \mu$ we obtain the value for $\alpha$. In summary, $$\alpha = \frac{\mu}{(\mu+\lambda)^2},\qquad \beta=\frac{\lambda}{\mu+\lambda},\qquad \gamma=\frac{\lambda(\mu+\lambda)}{\mu}.$$

So: $$\begin{align*} \mathbf{w}_1 &= \mathbf{v}_1 + \mathbf{v}_3\\ \mathbf{w}_2 &=\mathbf{v}_2 + \frac{\mu}{(\mu+\lambda)^2}\mathbf{v}_3\\ \mathbf{w}_3 &= \frac{\lambda}{\mu+\lambda}\mathbf{v}_1 + \frac{\lambda(\mu+\lambda)}{\mu}\mathbf{v}_2 + \mathbf{v}_3 \end{align*}$$ This is a basis, since $$\det\left(\begin{array}{ccc} 1 & 0 & 1\\ 0 & 1 & \frac{\mu}{(\mu+\lambda)^2}\\ \frac{\lambda}{\mu+\lambda} & \frac{\lambda(\mu+\lambda)}{\mu} & 1 \end{array}\right) = 1 - \frac{\lambda}{\mu+\lambda} - \frac{\mu\lambda(\mu+\lambda)}{\mu(\mu+\lambda)^2} = 1 - \frac{2\lambda}{\mu+\lambda} = \frac{\mu-\lambda}{\mu+\lambda}.$$ Since $\mu\neq\lambda$, this is nonzero. And it is now straightforward to verify that the matrix of $M$ relative to this basis is $$\left(\begin{array}{ccc} 0 & -\frac{\lambda\mu}{(\mu+\lambda)^2} & -\lambda\\ -\frac{\lambda(\mu+\lambda)^2}{\mu} & 0 & -\frac{\lambda(\mu+\lambda)^2}{\mu}\\ \mu+\lambda & \frac{\mu}{\mu+\lambda} & 2\lambda+\mu \end{array}\right)$$ which is of the desired form.

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Wow thanks for such a great explanation and taking the time to detail useful examples – user7980 Dec 8 '11 at 3:47
@user7980: Well, one more case down. Now I only need to figure out what trick to use in the diagonalizable with two distinct eigenvalues case. Maybe today while I'm proctoring a calculus final. – Arturo Magidin Dec 8 '11 at 17:50

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