Can the concatenation of projection operators be nilpotent with an index k>=3?

Question

Let $\boldsymbol{V}_{1},\dots,\boldsymbol{V}_{n}\in\mathbb{R}^{d\times m}$ be $n$ “tall” matrices (where $d\ge m$) with orthonormal columns.

And let $\boldsymbol{P}_{1},\dots,\boldsymbol{P}_{n}\in\mathbb{R}^{d\times d}$ be the orthogonal projection matrices defined as $\boldsymbol{P}_{i}=\boldsymbol{V}_{i}\boldsymbol{V}_{i}^{\top}$.

Finally, let $\boldsymbol{T}$ be an operator defined the concatenation of the projection matrices $\boldsymbol{T}=\boldsymbol{P}_{n}\cdots\boldsymbol{P}_{1}$.

Question: does there exists a sequence $\boldsymbol{V}_{1},\dots,\boldsymbol{V}_{n}$ such that the concatenation operator $\boldsymbol{T}$ is nilpotent with index $k\ge3$? That is,

$$\boldsymbol{T}^{k-1}=\left(\boldsymbol{P}_{n}\cdots\boldsymbol{P}_{1}\right)^{k-1}\neq\boldsymbol{0}_{d\times d},\,\,\,\,\,\,\,\boldsymbol{T}^{k}=\left(\boldsymbol{P}_{n}\cdots\boldsymbol{P}_{1}\right)^{k}=\boldsymbol{0}_{d\times d}$$

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Special case (example): when $n=3$ and $d=2$, we can choose projections such that $\boldsymbol{T}$ is a nilpotent operator with an index of $k=2$.

Choose $\boldsymbol{v}_{1}=\left[0,1\right]^{\top}, \boldsymbol{v}_{2}=\frac{1}{\sqrt{2}}\left[1,1\right]^{\top},\boldsymbol{v}_{3}=\left[1,0\right]^{\top}$.

Then, the projection matrices are $$\boldsymbol{P}_{1}=\boldsymbol{v}_{1}\boldsymbol{v}_{1}^{\top}=\left[\begin{array}{cc} 0 & 0\\ 0 & 1 \end{array}\right],\,\,\,\boldsymbol{P}_{2}=\frac{1}{2}\left[\begin{array}{cc} 1 & 1\\ 1 & 1 \end{array}\right],\,\,\,\boldsymbol{P}_{3}=\left[\begin{array}{cc} 1 & 0\\ 0 & 0 \end{array}\right]$$ And the concatenation of these matrices is $$\boldsymbol{T}=\boldsymbol{P}_{3}\boldsymbol{P}_{2}\boldsymbol{P}_{1}=\frac{1}{2}\left[\begin{array}{cc} 0 & 1\\ 0 & 0 \end{array}\right]$$ which is a nilpotent matrix with an index of $k=2$ (i.e., $\boldsymbol{T}^{2}=\boldsymbol{0}_{d\times d}$).

Answer 1

I think your example is easily generalisable for any index. For example, let $$ Q_1=P_1\oplus(1), Q_2=P_2\oplus(1), Q_3=P_3\oplus(1)=(1)\oplus P_1, Q_4=(1)\oplus P_2, Q_5=(1)\oplus P_3. $$ Then $Q_5Q_4Q_3Q_2Q_1$ is nilpotent of index 3, and I think that a similar construction with $2n-1$ matrices of size $n$ gives nilpotence of index $n-1$.

Answer 2

The paper Products of orthogonal projections shows in Theorem 1 that any singular matrix with norm less than $1$ is a product of projection matrices and gives an estimate on how many projection matrices you need. You can always scale a nilpotent matrix to have norm less than $1$ and so you can obtain a nilpotent matrix of any index that is a product of orthogonal projections. I'm not sure why the tall thing is needed here since you allow $d=m$ and you can just take $V_i=P_i$ for any $d\times d$ projection matrix $P_i$.