$AA^*A=A$ with eigenvalues $1$ and $0$, prove that $A$ is unitarily diagonalizable.

Question

Let $A$ be a $2$ by $2$ complex matrix such that $AA^*A=A$, and has eigenvalues of $1$ and $0$. Prove that $A$ is unitarily diagonalizable.

Well, if a matrix is Hermitian, then it is unitary diagonalizable. It also has real eigenvalues which is another property of Hermitian matrices. Hence, I want to prove Hermitian.

I want to prove it by showing that $\langle Az , w\rangle = \langle z, Aw\rangle$ but I am unable to find a clear way to showing this given the information. Is this the right approach?

Answer 1

Here's a proof that works:

Every matrix is unitarily triangularizable. So, there exists a unitary $U$ such that $A = UTU^*$, where $$ T = \pmatrix{1&t\\0&0} $$ If we show that $t$ is necessarily zero, then we may conclude that $A$ is unitarily diagonalizable (since $T$ would then be diagonal).

With that goal in mind, we note that $$ AA^*A = A \implies\\ UTU^*UT^*U^*UTU^* = UTU^* \implies\\ TT^*T = T $$ Moreover, we compute $$ TT^*T = T \pmatrix{1&t\\ \bar t & |t|^2} = \pmatrix{1 + |t|^2 & t(1 + |t|^2)\\0 & 0} $$ We see that $TT^*T = T$ indeed implies that $t = 0$, which means that $A$ is unitarily diagonalizable.

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It is interesting to see how this proof generalizes to the $n \times n$ case:

If $A$ has eigenvalues $0$ and $1$, then it is unitarily similar to the block matrix $$ T = \pmatrix{I & Q\\0&N} $$ where $N$ is strictly upper triangular (it is indeed true that the upper-left entry must be $I$; we note that the block associated with the eigenvalue $1$ is an invertible matrix satisfying $M = MM^*M$). As before, we may conclude that $$ AA^*A = A \implies TT^*T = T $$ and finally, we compute (with block-matrix multiplication) $$ TT^*T = \pmatrix{I & Q\\0&N} \pmatrix{I & Q\\Q^* & Q^*Q + N^*N} = \pmatrix{I + QQ^* & (I + QQ^*)Q + QN^*N\\NQ^*&N(Q^*Q + N^*N)} $$ noting that $QQ^* = 0 \iff \operatorname{tr}(QQ^*) = 0 \iff Q = 0$, we see that $TT^*T = T$ implies that $Q = 0$.

However, we have failed to prove that $N = 0$; we merely know that $N$ is strictly upper triangular with $N = NN^*N$. This does not imply that $N$ is zero, as seen with the counterexample $$ N = \pmatrix{0&1\\0&0} $$

Answer 2

The problem statement can be generalised as follows:

Let $A\in M_n(\mathbb C)$. If both $A$ and $AA^\ast$ are idempotent, $A$ must be Hermitian.

Proof. By unitary triangularisation, we may assume that $$ A=\pmatrix{P&X\\ 0&N} $$ where all eigenvalues of $P$ are $1$ and all eigenvalues of $N$ are zero. As $A$ is idempotent, $P$ and $N$ must be idempotent too. Thus $P=I,\ N=0$ and $$ AA^\ast = \pmatrix{I+XX^\ast&0\\ 0&0}. $$ Since $I+XX^\ast\succeq I$, in order that $AA^\ast$ is idempotent, $X$ must be zero. Thus $A$ is unitarily similar to $I\oplus0$ and it is Hermitian.