Let $A$ be a $n \times n$ real matrix such that $$ A + A^t = 2I, $$ where $I$ is the $n \times n $ identity matrix.

Prove that $\det(A) \geq 1$.

It is obvious that tr$(A) = n$. Furthermore, we also have that $$A - 2I = -A^t, $$ so we get that $$\det(A-2I) = (-1)^n \cdot \det(A).$$

Now, we let $\lambda_1, \lambda_2, \cdots, \lambda_n \in \mathbb{C}$ be the eigenvalues of $A$, so we get that $$(\lambda_1 - 2)(\lambda_2 - 2) \cdots (\lambda_n - 2) = (-1)^n\lambda_1\lambda_2 \cdots \lambda_n, $$ but I couldn't derive anything about the product $\lambda_1\lambda_2 \cdots \lambda_n = \det(A)$ (the only known thing is $\lambda_1 + \cdots + \lambda_n = n$).

Also, I tried the same approach for $2 \times 2$ and $3 \times 3 $ matrices, but it didn't lead to anything (especially since the eigenvalues can be complex numbers).


2 Answers 2


Note that $A - I$ is a real skew-symmetric matrix, so all of its eigenvalues are pure imaginary (or zero) and appear in complex conjugate pairs. Therefore the eigenvalues of $A$ are either $1$ or of the form $1 + c i, 1-c i$ for some $c\in\mathbb{R}$. Note that $(1+ci)(1-ci) = 1+c^2 \geq 1$. Therefore

$$\det(A) = \prod (1 + c^2) \geq 1$$

  • $\begingroup$ Thanks! That's exactly what I just though about now, once I tried to write out the matrices and compute the sum. $\endgroup$
    – C_M
    Dec 24, 2018 at 18:44

Let $A = I+B.$ One can immediately notice that $B$ is skew-symmetric or $B = -B^T.$ This means the eigenvalues of $A$ are precisely $1+\lambda_j$, where $\lambda_j$ are the eigenvalues of $B.$ Now, it's well-known that the eigenvalues of a skew-symmetric matrix comes in pairs $\pm i\lambda_j.$ This means that: $$\det A = \prod(1+\lambda^2_j)\geq 1.$$ You are encouraged to fill in the details separating the cases where $n$ is odd or even.


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