Inverse matrix operations

Question

We have a $3$ by $3$ matrix$$A=\begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$$

1) Show that $A^3=0$.

my answer:

I reduced A to RREF and I got $2$ pivots so the third variable is a free variable which means $A^3=0$. I am not sure if this is the right answer.

2) Find $I - A$ and $I + A + A^2$ and find the product of $(I - A)(I + A + A^2)$. Then find the inverse of $I - A$

my answer:

$$I - A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}-\begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$$ which gives $$\begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix}$$

I'm not sure if this is correct.

I don't know how to do next parts of question $2 (I + A + A^2)$ and $(I - A)(I + A + A^2)$ and inverse of $I - A$. I would appreciate if someone can help me with this. Thanks!

Answer 1

1. I would just compute $A^3$ and see that it is $0$.
2. Note that$$(\operatorname{Id}-A)(\operatorname{Id}+A+A^2)=\operatorname{Id}-A^3=\operatorname{Id}.$$Therefore,\begin{align}(\operatorname{Id}-A)^{-1}&=\operatorname{Id}+A+A^2\\&=\operatorname{Id}+A+\begin{bmatrix}0&0&0\\0&0&0\\1&0&0\end{bmatrix}\\&=\begin{bmatrix}1&0&0\\1&1&0\\1&1&1\end{bmatrix}.\end{align}

Answer 2

Another way to find $A^3$ is to observe its action on the vector $\mathbf{x}=\begin{bmatrix}x\\y\\z\end{bmatrix}$, i.e. we want to consider $A\mathbf{x}$. Observe that $$\begin{bmatrix}x\\y\\z\end{bmatrix} \xrightarrow{A}\begin{bmatrix}0\\x\\y\end{bmatrix} \xrightarrow{A}\begin{bmatrix}0\\0\\y\end{bmatrix}\xrightarrow{A}\begin{bmatrix}0\\0\\0\end{bmatrix}.$$ Thus $A^3\mathbf{x}=\mathbf{0}$ for all $\mathbf{x} \in \Bbb{R}^3$. Thus $A^3=\mathbf{0}$.

Rest of them are already answered.

Answer 3

You can check that $\det(\lambda I-A)=\lambda^3$. Therefore, by the Cayley-Hamilton Theorem, $A^3=0$.

Alternatively, let $e_1,e_2,e_3$ be the standard vectors $$ e_1=\left[\begin{array}{c}1 \\ 0 \\ 0\end{array}\right],\;\; e_2=\left[\begin{array}{c}0 \\ 1 \\ 0\end{array}\right],\;\; e_3=\left[\begin{array}{c}0 \\ 0 \\ 1\end{array}\right]. $$ Note that $Ae_1=e_2$, $Ae_2=e_3$, and $Ae_3=0$. It follows that $A^3=0$ because $A^3$ applied to any standard basis vector is $0$.