Does $A^2 = k^2 I$ imply $A$ is diagonalizable?

Question

For an $n\times n$ matrix $A$ over $\Bbb R$, given that $$A^2 = k^2 I_n$$ where $k > 0$, can one prove that $A$ is diagonalizable?

I have a strong feeling about this but I am not really sure as the original question does not explicitly state that this is true.

Answer 1

$\textbf{Case 1:---}$ $k=0$. Then $A^2=0$. Now we can't say, $A$ is diagonalizable. For example consider the matrix $$A:=\begin{bmatrix}0\ 1\\0\ 0\end{bmatrix}.$$ Here $A^2=0$ is zero implies all eigenvalues of $A$ are zero. So if $A$ were diagonalizable then, $A$ would be zero matrix.

$\textbf{Case 2:---}$ $k\not=0$ and $k\in \Bbb R$. In this case $A$ is diagonalizable over $\Bbb R$. As $A$ annihilates the polynomial $(x-k)(x+k)\in \Bbb R[x]$ i.e. minimal polynomial of $A$ is product of distinct linear polynomials.

$\textbf{Case 3:---}$ $k\not\in\Bbb R$ but $k^2\in \Bbb R$. In this case, $A$ annihilates the polynomial $x^2-k^2\in\Bbb R[x]$, and this polynomial does not split over $\Bbb R$. So $A$ can not be diagonalizable over $\Bbb R$ in this case also.

$$\text{ A real matrix is diagonalizable over real numbers}$$ $$\text{if and only if}$$ $$\text{it's minimal polynomial splits as product of distinct linear polynomials with real coefficients.}$$

$\textbf{Edit:---}$ You have assumed $k>0$. So $A$ annihilates the polynomial $(x-k)(x+k)\in \Bbb R[x]$. If there is no non-zero $v\in \Bbb R^n$ with $Av=kv$. Then, $A+kI=0$ as $A-kI$ is invertible. Hence $A$ is scalar matrix.

Similarly if there is no non-zero $v\in \Bbb R^n$ with $Av=-kv$ we can say, $A=kI$.

Next suppose, both $A\pm kI$ are not injective. And write $V=\text{ker}(A-kI)$ and $W=\text{ker}(A+kI)$. Clearly, $V\cap W=\{0\}.$ Also for any $v\in\Bbb R^n$ we have, $$v=\frac{1}{2k}\big(Av+kv\big)-\frac{1}{2k}\big(Av-kv\big),$$$$\frac{1}{2k}\big(Av+kv\big)\in V,$$$$\frac{1}{2k}\big(Av+kv\big)\in W.$$ So that, $V\oplus W=\Bbb R^n$. That is $A$ is diagonalizable.

Answer 2

You wrote in your comment that $k>0$. Hence the minimal polynomial of $A$ is

$$m_A( \lambda)=(\lambda -k)(\lambda+k).$$

$m_A$ splits over $ \mathbb R$, hence $A$ is diagonalzable.

Answer 3

From $A^2-k^2I=0$ we get

$$ \mathbb R^n= ker(A-kI) \oplus ker(A+kI).$$

Let $B_1$ be a basis of $ker(A-kI)$ and $B_2 $ be a basis of $ker(A+kI).$ Then $B:= B_1 \cup B_2$ is a basis of $ \mathbb R^n$, which consists of eigenvectors of $A$. Hence $A$ is diagonalizable.

Answer 4

Let $\rm A = P J P^{-1}$ be a Jordan decomposition of $\rm A$. Hence,

$$\mathrm P \,\mathrm J^2 \mathrm P^{-1} = k^2 \mathrm I_n$$

and, thus,

$$\mathrm J^2 = k^2 \,\mathrm P^{-1} \mathrm P = k^2 \mathrm I_n$$

which allows one to conclude that $\mathrm J = \mbox{diag} (\pm k, \dots, \pm k)$, i.e., all Jordan blocks are $1\ \times 1$ and, therefore, matrix $\rm A$ is diagonalizable.