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Show that if a symmetric matrix $A \in \mathbb{R}^{n \times n}$ satisfies $A = A^{2}$, then all its eigenvalues must be either 1 or 0.

I think I'm able to show that the eigenvalue is equal to 1 through the following:

$$ Av - \lambda v = 0$$ $$ v(A - \lambda I) = 0 $$ $$ A-\lambda I = 0 $$ $$ A = \lambda I $$ $$ AA = \lambda I A$$ $$ A^{2} = \lambda A$$ $$ A = \lambda A $$ $$ \lambda = 1 $$

How can I show that the eigenvalue can also be equal to 0?

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idkmath28
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  • Hint: if a matrix satisfies this, what are its possible minimal polynomials? Note that the characteristic polynomial and the minimal polynomial have the same zero set. – Daniel Sep 19 '22 at 02:46
  • Also, there are some incorrect steps in your calculation. First, just because $(A - \lambda I)v = 0$ we cannot conclude that $A - \lambda I = 0 $. In fact, there are matrices $A$ satisfying $A^2 = A$ which are not a scalar multiple of the identity. – Daniel Sep 19 '22 at 02:50
  • $\lambda(\lambda -1) = 0$ has only two solutions. – copper.hat Sep 19 '22 at 03:04
  • "Av−λv=0

    v(A−λI)=0" Multiplication is not communitive.

     – Acccumulation Sep 19 '22 at 04:05

1 Answers1

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Let $\lambda$ be an eigenvalue of the matrix $A$ and $\mathbf{x}$ an eigenvector corresponding to the eigenvalue $\lambda$

We have

$$A\mathbf{x}=\lambda \mathbf{x}$$,$\mathbf{x}\neq0$.

Then we have $$A^2\mathbf{x}=A\mathbf{x}\stackrel{(*)}{=} \lambda \mathbf{x}$$.

Also we have \begin{align*} A^2\mathbf{x}=A(A\mathbf{x})\stackrel{(*)}{=}A(\lambda \mathbf{x})=\lambda (A\mathbf{x})\stackrel{(*)}{=}\lambda (\lambda \mathbf{x})=\lambda^2\mathbf{x}. \end{align*} Comparing these last $2$ we get $\lambda \mathbf{x}=\lambda^2 \mathbf{x}$.

By solving the last equation and since $\lambda$ is our eigenvalue we get that $\lambda=0$ or $\lambda=1$

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