What does non-singularity have to do with lower triangular matrices?

Question

Let $A$ be a lower triangular matrix of order $n$. I am asked to verify the following assertion:

If $ A $ is nonsingular, then $ A^k $ is lower triangular for all integers k.

Well, if you want to define $A^k$ for $k<0$ you need $A$ to be non-singular... — Arnaud D., Sep 20 '16 at 11:30
Can you see why positive powers $A^k$ must be lower triangular? Can you prove that the inverse must be lower triangular? Once you've shown these two things, you are done. — Derek Allums, Sep 20 '16 at 12:17

score 0 · Answer 1 · edited Apr 13 '17 at 12:21

Let $A \in \mathbb R^{n \times n}$ be lower triangular.

Case 1: $k=0$

I think we are supposed to have

$A^0 = I$

while $A^0 = AA^{-1}$

So we need invertibility.

Case 2: $k < 0$

Obviously we need invertibility.

As to why $A^k$ is still lower triangular, note that

Hence $\underbrace{A^{-1}A^{-1} \cdots A^{-1}}_{k \ times}$ is a lower triangular matrix.

Case 3: $k > 0$

Invertibility is not needed here. Consider

$$A = \begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix}$$

or even

$$A = \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix}$$

$$\det(A) = \prod_{i=1}^{n} a_{ii} \ \text{if A is triangular}$$

$$\det(A^2) = (\det(A))^2$$

$$\det(A^{-1}) = (\det(A))^{-1}$$

1 Answers1