If $A$ is idempotent, then the eigenvalues of $A$ are $0$ or $1$

Question

If $A$ is idempotent (defined by $AA=A$) then the eigenvalues of $A$ are $0$ or $1$.

Proof: $$Ax = \lambda x$$ $$\Rightarrow Ax = AAx = \lambda Ax = \lambda^{2}x.$$

Then in the lecture notes, it says that the above implies that $\lambda^{2} = \lambda$ which implies $\lambda = 0$ or $\lambda = 1$.

I don't quite follow how it is deduced that $\lambda^{2} = \lambda$ from the above calculations.

Answer 1

In addition to my comment, let me suggest another proof that leads to a slightly larger insight.

For any matrix $M$, and $k \ge 1$, if $x$ is an eigenvector for $\lambda$ then $$ M^k x = \lambda^k x. $$

The proof is just induction, so I won't write it out.

Now start from $A^2 = A$. Rewrite this as $$ A^2 - A = 0 $$ and then let $x$ be an eigenvector for $\lambda$, an eigenvalue of $A$. We then get \begin{align} (A^2 - A)x &= 0 \\ A^2x - Ax &= 0 \\ \lambda^2x - \lambda x &= 0 & \text{by applying the lemma}\\ (\lambda^2 - \lambda) x &= 0 \\ \lambda^2 - \lambda &= 0 & \text{because $x$ is nonzero}\\ \end{align}

Now the polynomial that $A$ satisfied --- $A^2 - A = 0$ --- was nothing special. Suppose instead that we knew that $A^3 - 3A^2 - A + I = 0$. We can write this as $p(A)$, where $p(x) = x^3 - 3x^2 - x + 1$. By exactly the same kind of argument, we'd find that if $\lambda$ was an eigenvalue for $A$, then $p(\lambda) = 0$.

Summary: if a matrix $A$ satisfies $p(A) = 0$ for some polynomial $p$, then for any eignevalue $\lambda$ of $A$, we also have $p(\lambda) = 0$.

Answer 2

Because at the beginning you have $Ax$ which is equal to $\lambda x $ because x is an eigenvector.

Answer 3

By definition eigenvector is vector, which $\neq 0$: $Av = \lambda v $

So, we have: $\lambda^2v=\lambda v = > (\lambda^2 - \lambda)v = 0$.

For non zero $v_i$ you have (over field of C you don't have divisors of zero.)

$\lambda^2-\lambda = \lambda(\lambda - 1) = 0 $. And that's why only $\lambda = 0$ or $\lambda = 1$. And eigenvector $\neq 0$