I have several questions on an assignment that I just can’t seem to figure out.
1) Let $A$ be $2\times 2$ matrix. $A$ is nilpotent if $A^2=0$. Find all symmetric $2\times 2$ nilpotent matrices.
It is symmetric, meaning the matrix $A$ should look like $A=\begin{bmatrix} a & b \\ b & c\end{bmatrix}$. Thus, by working out $A^2$ I find that
$a^2 + b^2 = 0$ and
$ab + bc = 0$.
This tells me that $a^2 = – b^2$ and $a = -c$.
I’m not sure how to progress from here.
2)Suppose $A$ is a nilpotent $2\times 2$ matrix and let $B = 2A$ – I. Express $B^2$ in terms of $B$ and $I$. Show that $B$ is invertible and find $B$ inverse.
To find $B^2$ can I simply do $(2A -I)(2A – I)$ and expand as I would regular numbers?
This should give $4A^2 – 4A + I^2$.
Using the fact that $A^2$ is zero and $I^2$ returns $I$, the result is $I – 4A$. From here do I simply use the original expression to form an equation for $A$ in terms of $B$ and $I$ and substitute it in? Unless I am mistaken $4A$ cannot be treated as $2A^2$ and simplified to a zero matrix.
3) We say that a matrix $A$ is an idempotent matrix if $A^2 = A$. Prove that an idempotent matrix $A$ is invertible if and only if $A = I$.
I have no idea how to begin on this one.
4) Suppose that $A$ and $B$ are idempotent matrices such that $A+B$ is idempotent, prove that $AB = BA = 0$.
Again, I don’t really have any idea how to begin on this one.
You should also find that $b^2 + c^2 = 0$ from the $(2,2)$-entry of $A^2$. Are you working with real matrices? If so, what do $a^2 + b^2 = 0$ and $b^2 + c^2 = 0$ tell you about $a$, $b$, and $c$?
Your work so far is alright. Now just note
$$B^2 = I – 4A = -2(2A – I) – I = -2B – I.$$
Clearly $I$ is idempotent and invertible. If $A$ is idempotent and invertible, then we can multiply both sides of $A^2 = A$ by $A^{-1}$ to get $A = I$.
Look at
$$A + B = (A + B)^2 = A^2 + AB + BA + B^2 = A + AB + BA + B,$$
where we used that $A$, $B$, and $A+ B$ are idempotent. Hence
$$AB = – BA. \tag{$\ast$}$$
Now compare the results of multiplying $(\ast)$ on the right by $B$ and on the left by $B$ to find that
$$AB = BA. \tag{$\ast\ast$}$$
Then it follows from $(\ast)$ and $(\ast\ast)$ that $AB = BA = 0$.
For #1, you should also have $b^2 + c^2 = 0$. If you’re working over the real numbers, note that the square of a real number is always $\ge 0$, and is $0$ only if the number is $0$.
If complex numbers are allowed, you could have $a = -c = \pm i b$.