Questions related to nilpotent and idempotent matrices

Question

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.

@Kev Do the matrices have all their entries over the real numbers or are you considering complex numbers as well? — Git Gud, Jan 28 '13 at 00:20
@GitGud The entries are all real numbers, complex numbers are something above my current level. — Kev, Jan 28 '13 at 00:24
@Kev Then take a look at Henry's hint in the comment on his answer. — Git Gud, Jan 28 '13 at 00:25

score 3 · Answer 1 · answered Jan 27 '13 at 23:59

3

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 $AB = BA = 0$.

answered Jan 27 '13 at 23:59

Henry T. Horton

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Thank you, I know understand 2), 3) and 4). – Kev Jan 28 '13 at 00:15
For 1)
a^2 + b^2 = 0 and b^2 + c^2 = 0 tell me that a^2 = c^2 and that b^2 = -a^2 = -c^2.

However I am still unsure how to use this information to find all the matrices I am asked for.
– Kev Jan 28 '13 at 00:17
Hint: The square of a real number is always greater than or equal to zero. – Henry T. Horton Jan 28 '13 at 00:19
Since I am only dealing with real numbers, the only way for b^2 = -a^2 = -c^2 to be fulfilled would be for a=b=c=0 is that right? – Kev Jan 28 '13 at 00:46
Additionally, for 2) I now know that B^2 = -2B - I.
I sense that I am supposed to apply this information to show that B is invertible and to find B inverse.

I am stuck again.
– Kev Jan 28 '13 at 00:50
@Kev Regarding 1. you are correct: $a=b=c=0$. – Git Gud Jan 28 '13 at 07:01
@Kev About 2., you can deduce $B(-B-2I)=I$. From this you can conclude that $B$ is invertible (justify it with determinants) and also you can imediately find $B^{-1}$. I suggest you take a look at this lemma in my answer to another question: http://math.stackexchange.com/questions/280884/prove-that-matrix-a-have-inverse/280892#280892 – Git Gud Jan 28 '13 at 07:32

score 1 · Answer 2 · answered Jan 27 '13 at 23:56

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 $a = -c = \pm i b$.

Questions related to nilpotent and idempotent matrices

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