# Is the product of projection matrices diagonalizable?

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?

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.

• Thanks much! Follow-up question: is it perhaps true that in those cases the (non-diagonalizable) product must be nilpotent (like in your example)?
– Itay
Oct 21, 2021 at 18:03
• 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. Oct 21, 2021 at 18:15
• Why don't you add ($\oplus$) some other matrix instead of just 1. Oct 21, 2021 at 18:21
• 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. Oct 21, 2021 at 18:34