Show that no $n \times n$ real matrix $A^2$ can be a diagonal matrix with distinct negative entries

Question

Show that no $n \times n$ real matrix $A$ can be of the form $A^2=\begin{bmatrix} -a_1 \\ &\ddots \\ & & -a_n\end{bmatrix}$, where each $a_i$ is positive, real and distinct.

Suppose the contrary.

I see that in the case $n$ is odd, we have $0 <\det A < \prod_{i=1}^n (-a_i)= - \prod_{i=1}^n (a_i) $, which is a contradiction.

Now suppose $n$ is even.

Let $\rho(x) =\prod_{i=0}^n (-a_i-x)= \prod_{i=0}^n (a_i+x)$. Then $\rho(A^2)=0$. So $A$ is a zero of $\rho(x^2)$. So the minimal polynomial divides $\rho(x^2)$.

From here I'm not sure how to proceed.

Any help will be appreciated.

Answer 1

Let's suppose that $$A^2=\begin{pmatrix} -a_1 \\ &\ddots \\ & & -a_n\end{pmatrix}$$

Then the characteristic polynomial of $A^2$ is $P(X)= (-1)^n\prod_{k=1}^n (X+a_k)$, so by Cayley-Hamilton, the polynomial $Q(X)= \prod_{k=1}^n (X^2 + a_k ) $ is a vanishing polynomial for $A$. But $$Q(X) = \prod_{k=1}^n (X + i \sqrt{a_k})(X- i \sqrt{a_k})$$

So the eigenvalues of $A$ are among the $\pm i \sqrt{a_k}$. But since $A$ has real entries, if $i\sqrt{a_k}$ is eigenvalue of $A$, then $-i\sqrt{a_k}$ should also be an eigenvalue of $A$. This contradicts the fact that, since the $a_k$'s are distinct, $-a_k$ has multiplicity $1$ as an eigenvalue of $A^2$.

Answer 2

$A^2=\begin{bmatrix} -a_1 \\ &\ddots \\ & & -a_n\end{bmatrix}$

implies, working over $\mathbb C$, that each eigenvalue of $A$ satisfies $\lambda_k^2 \lt 0$, i.e. each eigenvalue must be purely imaginary. But non-real eigenvalues for real matrices come in conjugate pairs (i.e. $\lambda_k= x \cdot i$ for some $x\in \mathbb R-\{0\}$ and $\lambda_{k+1}=-\lambda_k$) hence $A^2$ has at most $\frac{n}{2}$ distinct entries on its diagonal, a contradiction.

Answer 3

Here is a simpler proof that requires only that $a_1$ be different from the other $a_j$'s.

Supposing that such a matrix $A$ exists, let $p$ be a polynomial such that $p(-a_1)=1$, and $p(-a_j)=0$, for all $j\geq 2$. Then $$ p(A^2)=E_{1, 1}:= \pmatrix{ 1 & 0 & \cdots & 0 \cr 0 & 0 & \cdots & 0 \cr \vdots & \vdots & \ddots & \vdots\cr 0 & 0 & \cdots & 0 \cr }. $$ From this it follows that $A$ commutes with $E_{1,1}$, which means that the range of $E_{1,1}$ is invariant under $A$. Since that range is spanned by $e_1:=(1,0,\cdots ,0)$, we see that $e_1$ is necessarily an eigenvector for $A$. Letting $\lambda $ be the corresponding eigenvalue, we have that $\lambda ^2=-a_1$, a contradiction.