How to determine the eigenvalues of a $k$-order idempotent (symmetric) matrix $A$?

Question

I have a following question:

Let $A$ satisfying $A^{\prime} = A$, $A^k = A$ and $A^i \neq A$ for $i=2, \cdots, k-1$. In this case, $A$ is referred to as a $k$-order idempotent matrix. How to determine the eigenvalues of $A$?

I try to solve this question as follows:

$A^{\prime} = A$ means $A$ is a symmetric matrix, which is necessarily a square matrix. WLOG, we set that $A$ is an $n$-order square matrix . For simplicity, we assume that $k$ is a positive integer with $k\ge 3$ (is this assumption necessary?).

For any eigenvalue $\lambda$ and corresponding eigenvector $v\ne 0$ of $A$ (i.e., $A v = \lambda v$), when raising both sides to the power of $k$ (or by mathematical induction), we get:

$$A^k v = A^{k-1}(Av)= A^{k-1}(\lambda v)=\lambda A^{k-1} v=...= \lambda^k v.$$

But because of given $A^k = A$, this becomes:

$$A v = \lambda^k v.$$

Combined with $A v = \lambda v (v\ne 0)$ it follows that for any eigenvalue $\lambda$, it must satisfy $\lambda^k = \lambda$.

Then I solve $\lambda^k = \lambda$, where $k\in\mathbb z, k\ge 3$. The solutions to the equation $x^k = x$ include:

1. Real solution: $x = 0$
2. Complex solutions: $x = e^{2\pi i m/(k-1)}$, where $m= 0, 1, \ldots, k-2$

However, I am not sure whether my analytical approach and solution process are correct, as given conditions such as the symmetry of matrix $A$ as well as $A^i \neq A$ for $2 \leq i \leq k-1$ have not been used yet!

Any help?

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update

By the comments from @GBA@Afntu, the condition the symmetry of matrix $A$ can guarantee the eigenvalues are real, which are listed as follows:

Consider a real symmetric matrix $A$, whose eigenvalues and corresponding eigenvectors are denoted as $\lambda$ and $x$, respectively. Furthermore, let $\overline{A}$, $\overline{\lambda}$, and $\overline{x}$ represent the conjugate complex numbers of each element in $A$, $\lambda$, and $x$, respectively.

Firstly, we have: $$ \overline{x}^TAx = \overline{x}^T\overline{A}x = (\overline{A}^T\overline{x})^Tx = (\overline{A}\overline{x})^Tx = \overline{(Ax)}^Tx=\overline{(\lambda x)}^Tx=\overline{\lambda}\cdot\overline{x}^Tx \tag{1} $$

Additionally, $$ \overline{x}^TAx = \overline{x}^T\lambda x = \lambda \cdot\overline{x}^T x \tag{2} $$

Since expressions (1) and (2) are equal, it follows that: $$ (\overline{\lambda} - \lambda) \cdot\overline{x}^Tx = 0 $$

Given that the eigenvector $x \neq 0$, thus $\overline{x}^Tx > 0$, we can conclude: $$ \overline{\lambda} = \lambda $$

This demonstrates that the eigenvalues are real numbers.

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question

Still no idea for how to use the condition $A^i \neq A$ for $2 \leq i \leq k-1$!

Answer 1

I assume $k\gt 1$.

Note that symmetric real matrices have only real eigenvalues and, moreover, are diagonalizable (in fact, orthogonally diagonalizable, but we don't need that here).

In fact, your conditions can only be met for $k=2$ and $k=3$; no higher value of $k$ will work. That is:

Theorem. let $A$ be a symmetric $n\times n$ matrix with real coefficients. If $k\gt 1$ is such that

• $A^k=A$; and
• $A^i\neq A$ for $2\leq i\lt k$; then:

1. If $\lambda$ is an eigenvalue of $A$, then $\lambda\in\{0,1,-1\}$.
2. $k=2$ if and only if $-1$ is not an eigenvalue of $A$.
3. If $k=2$, then either $A=I$, $A=0$, or $A$ has both $0$ and $1$ as eigenvalues, and all three possibilities may occur.
4. $k=3$ if and only if $-1$ is an eigenvalue of $A$.
5. If $k=3$, then $A=-I$, or else $A$ also either $1$ or $0$ (or both) as eigenvalues, and all four cases may occur.

Because $A^k=A$, then $A$ satisfies the polynomial $x^k-x = x(x^{k-1}-1)$. That means that the minimal polynomial must divide $x(x^{k-1}-1)$, and thus all eigenvalues of $A$ are roots of this polynomial.

Because $A$ is symmetric, all eigenvalues are real.

The roots of $x^{k-1}$ are the $(k-1)$th complex roots of unity. The only real ones are $1$ when $k$ is even, and also $k-1$ when $k$ is odd.

Thus, the only possible eigenvalues of $A$ are $0$, $1$, and $-1$.

If $-1$ is not an eigenvalue of $A$, then $A$ is diagonalizable and its only eigenvalues are $0$ and $1$ (possibly just one of them); that means that $A$ is a projection, so $A^2=A$ and $k=2$. The condition "$A^i\neq A$ for $2\leq i\lt 2$" is vacuous. You can have $A=I$; you can have $A=0$, or you can have a matrix that has both $1$ and $0$ as eigenvalues. Conversely, if $k=2$, then $-1$ is not an eigenvalue. Thus, $-1$ is not an eigenvalue if and only if $k=2$.

If $-1$ is an eigenvalue, then $k$ is odd as noted above. Because $A$ is diagonalizable and its eigenvalues are among $0$, $1$, and $-1$, we must have that $k=3$ (because we will necessarily have $A^3=A$, which can be verified by thinking about the diagonal form of $A$).

In this situation, we could have any combination of eigenvalues from among $0$, $1$, and $-1$, that includes $-1$. We could have just $-1$, in which case $A=-I$; we could have $0$ and $-1$ as eigenvalues; we could have $1$ and $-1$; and we could have all of $0$, $1$, and $-1$. Diagonal matrices provide examples showing all the possibilities can be realized.

Finally, if $k=3$, then $-1$ must be an eigenvalue by item 2. $\Box$