# Are Jordan “Formable” matrices closed under multiplication?

After learning about Jordan Canonical Form, I began thinking about if you have two matrices $$A, B \in M_n(\mathbb{R}),$$ whether or not their product $$AB$$ will also have a Jordan Canonical Form. Trivially, if we consider polynomials in $$M_n(\mathbb{C})$$ then our polynomial will always have roots in the field. So, from what I understand, this is equivalent to asking whether or not the characteristic polynomial of $$AB$$ will have real roots.

I began playing around with the $$2 \times 2$$ case and if we have that $$AB$$ has a negative determinant then we can't have that the eigenvalues of $$AB$$ are in $$\mathbb{C},$$ since they must multiply to the determinant, but since the complex eigenvalues always come in a conjugate pair, that would mean that their product is always positive. I am not sure how I can break down the case if the determinant is positive.

Finally, my own intuition says matrices which have Jordan form will be closed under multiplication. This is because if we think about what matrices with strictly, real eigenvalues do, they simply reflect and stretch space. So I don't believe it is possible to find two matrices whose product matrix will be a rotation in the plane.

Also, I couldn't find this question asked elsewhere, but if it has an answer, I'd be happy to read it.

• Consider 2 different reflections. Each of them has eigenvalues 1 and -1. However, their composition is a rotation with complex eigenvalues. – I like Serena Nov 15 '18 at 22:33
• @IlikeSerena That's a particularly nice way to resolve the question---would you consider promoting your comment to an answer? – Travis Nov 16 '18 at 0:13
• Thank you @Travis. Done. – I like Serena Nov 16 '18 at 0:20

This is false already in the $$2 \times 2$$ case. Consider the matrices

$$A = \left[ \begin{array}{cc} -1 & 2 \\ -2 & 3 \end{array} \right], B = \left[ \begin{array}{cc} 3 & 0 \\ 0 & 1 \end{array} \right]$$

$$B$$ has eigenvalues $$3, 1$$, and the characteristic polynomial of $$A$$ is $$t^2 - 2t + 1 = (t - 1)^2$$, so $$A$$ has eigenvalues $$1, 1$$. But

$$AB = \left[ \begin{array}{cc} -3 & 2 \\ -6 & 3 \end{array} \right]$$

has characteristic polynomial $$t^2 + 3$$, so has eigenvalues $$\pm i \sqrt{3}$$.

I found these matrices by first taking $$B$$ to be diagonal with diagonal entries $$s, 1$$ and then looked at the effect multiplication by $$B$$ had on the characteristic polynomial of a generic $$A$$. It's possible to choose $$s$$ so that $$AB$$ has trace zero, so then we just need its determinant to be positive to guarantee that it has complex eigenvalues.

Consider 2 reflections. Each of them has eigenvalues 1 and -1. However, their composition is generally a rotation with complex eigenvalues.

So the product $$AB$$ of two real matrices $$A$$ and $$B$$ with Jordan Canonical Forms does not necessarily have a Jordan Canonical Form as well.

• Well, in the generic case, yes - or more precisely, when the lines of reflection aren't coincident or perpendicular. For example, the reflections about the $x$-axis and $y$-axis compose to rotation by $\pi$, or $-I$ which is diagonal. – Daniel Schepler Nov 16 '18 at 1:35
• I've clarified that it's 'generally'. – I like Serena Nov 16 '18 at 10:28

Every square matrix $$A$$ over any field is similar to its own transpose via a non-singular symmetric matrix $$S$$ (see Taussky and Zassenhaus, On the similarity transformation between a matrix and its transpose).

If $$A=SA^TS^{-1}$$ for a symmetric $$S$$, then $$SA^T=AS=(SA^T)^T$$, i.e. $$SA^T$$ is symmetric. Similarly for $$A^TS^{-1}$$. Consequently, every square matrix can be written as a product of two symmetric matrices (because $$A=(SA^T)S^{-1}=S(A^TS^{-1})$$) and one of them can be chosen to be non-singular.

It follows immediately that the answer to your question is negative unless $$n=1$$.