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Let $P_i=V_{i}V_{i}^{\top}\in\mathbb{R}^{m\times m}$ where $\forall i\in[T]: V_{i}\in\mathbb{R}^{m\times n}$ is a “tall” matrix (i.e., $m \ge n$) with orthonormal columns. Note that these matrices are symmetric PSD.

Is the product of all these matrices, i.e., $P_T P_{T-1}\cdots P_1$, necessarily diagonalizable?

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No. Take $$v_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \qquad v_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \qquad v_3 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}.$$ Then $$P_1=v_1 v_1^T = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} \qquad P_2=v_2 v_2^T = \begin{bmatrix} 1&1 \\ 1&1 \end{bmatrix} \qquad P_3=v_3 v_3^T = \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix}$$ so $$P_1 P_2 P_3 = \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}$$ which is not diagonalizable. (Replace $v_2$ by $\tfrac{1}{\sqrt{2}} \left[ \begin{smallmatrix} 1 \\ 1 \end{smallmatrix} \right]$ if you wanted orthonormal columns instead of just orthogonal ones.)

Regarding the random eigenvalues: For $(x,y,z)$ in $\mathbb{R}^3$, let $$f(x,y,z) = (x^2+y^2+z^2) \mathrm{Id}_3 - \begin{bmatrix} x\\y\\z \end{bmatrix} \begin{bmatrix} x&y&z \end{bmatrix}.$$ So $f(x,y,z)/(x^2+y^2+z^2)$ is orthogonal projection onto $\left[ \begin{smallmatrix} x\\y\\z \end{smallmatrix}\right]^{\perp}$.

I computed $f(1,0,0) f(u,v,w) f(x,y,z) f(1,0,0)$ for randomly chosen $(u,v,w,x,y,z)$ and, on my third try, I found the following example: $$f(1,0,0) f(2,-3,-2) f(2,-3,1) f(1,0,0) = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 58 & -66 \\ 0 & 33 & 143 \\ \end{bmatrix}.$$ The characteristic polynomial is $$\lambda (\lambda^2 - 201 \lambda + 10472),$$ which has complex roots.

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  • $\begingroup$ Thanks much! Follow-up question: is it perhaps true that in those cases the (non-diagonalizable) product must be nilpotent (like in your example)? $\endgroup$
    – Itay
    Oct 21, 2021 at 18:03
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    $\begingroup$ If $m=1$, then the product will be rank $\leq 1$, so the only way to be non-diagonalizable is to be nilpotent. I can take the $v_i$ above and embed them into $3 \times 2$ matrices as $\left[ \begin{smallmatrix} v_i &\vec{0} \\ 0&1 \end{smallmatrix} \right]$, and get the product $\left[ \begin{smallmatrix} P & 0 \\ 0 & 1 \end{smallmatrix} \right]$, where $P$ is the product above, so I can get a nilpotent direct sum the identity. It is not obvious to me whether we can get complex generalized eigenvalues. My bet is that the answer is yes. $\endgroup$
    – David E Speyer
    Oct 21, 2021 at 18:15
  • $\begingroup$ Why don't you add ($\oplus$) some other matrix instead of just 1. $\endgroup$
    – markvs
    Oct 21, 2021 at 18:21
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    $\begingroup$ Yes, the more general example is that I can direct sum any constructions of this sort. Just trying to stay as concrete as I can while pointing out that, clearly, I can get something which is neither diagonalizable nor nilpotent. $\endgroup$
    – David E Speyer
    Oct 21, 2021 at 18:34

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