Let $A ∈ M_{n×n}(\mathbb{R})$ differ from $I$ and $O$. If $A$ is idempotent, show that its Jordan canonical form is a diagonal matrix.

Question

Let $A ∈ M_{n×n}(\mathbb{R})$ differ from $I$ and $O$. If $A$ is idempotent, show that its Jordan canonical form is a diagonal matrix.

I'm not sure how to do this. Any solutions/hints are greatly appreciated.

Answer 1

Let $J_A$ denote the Jordan cannonical form of $A$; then, as is well-known,

$J_A = S^{-1}AS, \tag{1}$

where $S$ is some non-singular $n \times n$ matrix. It follows that

$J_A^2 = (S^{-1}AS)^2 = S^{-1}A^2S$ $= S^{-1}AS = J_A, \tag{2}$

and it follows from (2) that the eigenvalues of $J_A$, which are of course the same as those of $A$, lie in the set $\{ 0, 1 \}$; explicitly, if

$J_A \vec v = \lambda \vec v, \tag{3}$

then

$\lambda \vec v = J_A \vec v = J_A^2 \vec v$ $= J_A(\lambda \vec v) = \lambda J_A \vec v = \lambda^2 \vec v, \tag{4}$

whence

$(\lambda^2 - \lambda) \vec v = 0; \tag{5}$

thus, since $\vec v \ne 0$ (it being an eigenvector of $J_A$), we have $\lambda$ satisfying $\lambda = 0, 1$ as asserted.

We take it as well-known that $J_A$ is a block diagonal matrix, and that any non-diagonal block is of the form $\lambda I_m + N_m$, where $I_m$, $m \le n$, is the $m \times m$ identity matrix and $N_m$ is the $m \times m$ matrix having all ones on the superdiagonal and zeroes elsewhere. Since taking powers preserves the block structure of block-diagonal matrices, (2) implies each such (non-diagonal) block must satisfy

$(\lambda I_m + N_m)^2 = \lambda I_m + N_m; \tag{6}$

this implies

$\lambda^2 I_m + 2 \lambda N_m + N_m^2 = \lambda I_m + N_m; \tag{7}$

since $\lambda^2 = \lambda$ in the present case, we have

$\lambda I_m + 2 \lambda N_m + N_m^2 = \lambda I_m + N_m \tag{8}$

or

$2 \lambda N_m + N_m^2 = N_m, \tag{9}$

impossible! and why? well, if $m = 2$, $N_m^2 = 0$ by a simple computation; then with $\lambda \in \{ 0, 1 \}$ we have $N_m = 0$; for $m \ge 3$, $(N_m^2)_{i,i +2} = 1$ for $1 \le i \le m- 2$, but $(N_m)_{i,i + 2} = 0$; inspection of (9) reveals that this is contradictory, since it implies

$2\lambda (N_m)_{i,i + 2} +(N_m^2)_{i,i +2} = (N_m)_{i,i + 2}; \tag{10}$

substituting the known values if the matrix entries $(N_m)_{i,i + 2}$, $(N_m^2)_{i,i +2}$ yields

$1 = 0, \tag{11}$

an untenable conclusion which rules out the possibility that a Jordan block is not diagonal; since each Jordan block is in fact diagonal, so is $J_A$. QED.

Note that we have nowhere used the hypothesis $A \ne 0, I$; in any event, problem is trivial in these cases.

Answer 2

This is a weird question: you don't need anything about Jordan normal forms here, as you just need to show that $A$ is diagonalisable (the diagonal form then is the Jordan form). This follows immediately from the general result that any matrix annihilated by a polynomial that splits with simple roots (here that is the polynomial $X^2-X=X(X-1)$) is diagonalisable.

If you don't want to use that result, here is a proof specialised to the concrete situation.

Since $A^2=A$, any eigenvalue$~\lambda$ satisfies $\lambda^2=\lambda$, so $\lambda\in\{0,1\}$. Let $V_0=\ker(A)$ and $V_1=\ker(A-\mathbf I)$ be the two possible eigenspaces. As always their sum is direct (no vector can be eigenvector for two different eigenvalues); we need to show that $V_0\oplus V_1$ fills the whole space $\Bbb R^n$.

For every $v\in\Bbb R^n$ on has $Av\in\ker(A-\mathbf I)=V_1$ and $(A-\mathbf I)v\in\ker(A)=V_0$, because of $A\circ(A-\mathbf I)=0=(A-\mathbf I)\circ A$. Since $v=Av-((A-\mathbf I)v)\in V_1+V_0$ we have $V_1+V_0=\Bbb R^n$.