# Let $A_1,A_2,\dots,A_d$ be $d\times d$ matrices that are strictly upper triangular. Then, the product of $A_1,A_2,\dots,A_d$ is the zero matrix.

SOLUTION: Let $$B_i$$ be given by $$A_1A_2\dots A_i$$. It can be shown inductively that $$B_i$$ is also strictly triangular, but with at least $$i$$ zero rows and columns. Therefore, $$B_d$$ will be the zero matrix.

I can't quite wrap my head around this. I would expect that in fact, the product of any two strictly triangular matrices would be the $$0$$ matrix. And therefore any other matrix (can be non-zero) that is multiplied with it will be the $$0$$ matrix. But the answer seems to say that it only has at least $$i$$ zero rows and columns after multiplying $$i$$ strictly trangiangular matrices. It might be that I am misunderstanding the entire answer in general but any simplification explanation is welcomed.

SOURCE: Linear Algebra and Optimization for Machine Learning: A Textbook (Problem 1.23)

• $\begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix}\times\begin{bmatrix}0&0&0\\0&0&1\\0&0&0\end{bmatrix}\ne O$ Dec 31, 2020 at 12:49
• Think in linear transformation like math.stackexchange.com/a/532459/290189, and you'd be convinced about this fact without the need to do induction. Dec 31, 2020 at 12:59

It is not true that the product of two upper triangular matrices is necessarily $$0$$. As an example, take $$d = 4$$ and $$A_1 = A_2 = A_3 = A_4 = A = \pmatrix{0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0}.$$ In a sense, this is the "classic" example of a strictly upper triangular matrix considered in the context of Jordan normal form. Notice what happens when we multiply these matrices: $$A^2 = \pmatrix{0&0&1&0\\0&0&0&1\\0&0&0&0\\0&0&0&0}, \quad A^3 = \pmatrix{0&0&0&1\\0&0&0&0\\0&0&0&0\\0&0&0&0}, \quad A^4 = 0.$$ In a sense, what happens here is what happens in general: multiplying two upper-triangular matrices "pushes" the first non-zero "diagonal" further up. In fact, I recommend that you prove the following:
We will say that $$A$$ is "upper triangular of order k" if its entries are such that $$a_{ij} = 0$$ whenever $$j < i+k$$. Prove that if $$A$$ is upper-triangular of order $$p$$ and $$B$$ is upper-triangular of order $$q$$, then $$AB$$ is upper-triangular of order $$p + q$$.
In a more abstract sense, what upper-triangularity tells us about the transformation associated with the matrices $$A_1,\dots,A_d$$ is that there are subspaces $$\{0\} = V_0 \subsetneq V_1 \subsetneq \cdots \subsetneq V_{n-1} \subsetneq V_n = \Bbb R^d$$ for which $$A(V_i) \subseteq V_{i-1}$$ for each $$i = 1,\dots, n$$. That is, for every $$v \in V_i$$, $$Av$$ is an element of $$V_{i-1}$$.
This gives us an alternative proof. For any vector $$v \in \Bbb R^d = V_n$$, $$A_d v$$ is an element of $$V_{n-1}$$. Similarly, $$A_{d-1}(A_d v)$$ is an element of $$V_{n-2}$$. Continuing in this fashion, we can conclude that $$A_2 \cdots A_d v \in V_1$$, so that $$A_1 A_2 \cdots A_d v = A_1(A_2 \cdots A_d v) \in V_0 = \{0\},$$ which is to say that $$A_1 A_2 \cdots A_d v = 0$$ for every vector $$v$$. It follows that $$A_1 A_2 \cdots A_d = 0$$.