# How to solve $M=SU=US?$

This is from Berkeley Problems in Mathematics, Spring 1979:

Let $M$ be a $3\times 3$ non-singular real matrix. Prove that there are real matrices $S$ and $U$ such that $M=SU=US$, all the eigenvalues of $U$ equal 1, and $S$ is diagonalizable over $\mathbb{C}$.

I am thinking that since $U$'s eigenvalues are all 1, we have $(U-I)^{3}=0$. Thus up to Jordon canonical form $U$ should be either a whole Jordan block or has a rank 2 Jordan block. But I do not know how to use this to approach this problem. On the other hand $S$ can be expressed in $PDP^{-1}$, and we are supposed to solve $A=PDP^{-1}U=UPDP^{-1}$. Again no familiar formula can be obtained this way. So I venture to ask for a hint. This may be related to Schur decomposition but I do not know it well enough to solve it.

-
The original Problem 7.7.13 (Sp79) requires $M$, $S$, $U$ real and $M$ non singular: tinyurl.com/brr37v7. – enzotib Aug 1 '12 at 17:38
sorry for the typo. shall fix it. – Bombyx mori Aug 1 '12 at 17:48
I feel you can use the polar decomposition of a matrix, but could not say exactly how. – enzotib Aug 1 '12 at 18:26
The main catch (for me) is the real requirement. – copper.hat Aug 1 '12 at 18:48

Here is a rather tedious proof:

Since $M$ is real, either the eigenvalues are all real, or there is one real and a pair of conjugate eigenvalues.

The key result here is that a real matrix has a real Jordan form, not quite upper triangular, but 'almost'. In the invertible $3 \times 3$ case, this means that there exists a real $V$ such that $J = V^{-1}MV$ has one of the following forms:

$$J = \begin{bmatrix} \lambda_1 & M_{12} & 0 \\ 0 & \lambda_2 & M_{23} \\ 0 & 0 & \lambda_3\end{bmatrix} \ \ \ \ \ \ \ \ J = \begin{bmatrix} \lambda_1 & 0 & 0 \\ 0 & a & b \\ 0 & -b & a\end{bmatrix},$$

where all the entries are real, $M_{12}, M_{23} \in \{0,1\}$, $\lambda_i \neq 0$ and $b \neq 0$.

In the second case, since all eigenvalues are distinct ($b \neq 0$), $S$ is diagonalizable over $\mathbb{C}$. We can take $S= V J V^{-1}$, $U= I$ and we are finished.

Now for the first case. Since this is the Jordan form, we have $(\lambda_1 \neq \lambda_2) \Rightarrow M_{12} = 0$, and similarly $(\lambda_2 \neq \lambda_3) \Rightarrow M_{23} = 0$.

Let $\Sigma = \mathbb{diag}(\lambda_1, \lambda_2, \lambda_3)$, and let $W = I + \alpha E_{12} + \beta E_{23}$, where $E_{12}$ is the matrix with zeros everywhere except a one in the $1,2$ position, and similarly for $E_{23}$. We want to determine constants $\alpha, \beta$ so that $J = \Sigma W = W \Sigma$. We have: $$\Sigma W = \Sigma + \alpha \lambda_1 E_{12} + \beta \lambda_2 E_{23}\\ W \Sigma = \Sigma + \alpha \lambda_2 E_{12} + \beta \lambda_3 E_{23}$$

This shows that $\Sigma W = W \Sigma$ iff $\alpha( \lambda_1 - \lambda_2) = 0$ and $\beta( \lambda_2 - \lambda_3) = 0$. Furthermore, we want J$= \Sigma W$, and this will be true iff $M_{12} = \alpha \lambda_1$, and $M_{23} = \beta \lambda_2$.

Let $\alpha = \frac{M_{12}}{\lambda_1}$ and $\beta = \frac{M_{23}}{\lambda_2}$, then it is easy to verify that all the conditions are satisfied. Then we can take $S = V \Sigma V^{-1}$, and $U = V W V^{-1}$, and we are finished.

-
excuse me for my low level, but where can I find information related to the real normal form you are talking about? Is it just the rational canonical form? – Bombyx mori Aug 2 '12 at 6:57
It is real Jordan form, unfortunately my library is off the shelves at the moment so I can't give you a book reference, but wikipedia has one en.wikipedia.org/wiki/Jordan_normal_form#Real_matrices. It follows from the usual (possibly complex) Jordan form. – copper.hat Aug 2 '12 at 7:00