# If $A + A^t = 2I$ then $\det(A) \geq 1$

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

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

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

• Thanks! That's exactly what I just though about now, once I tried to write out the matrices and compute the sum. – C_M Dec 24 '18 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.