# $A \in M_{n}(\mathbb{R})$ satisfies $A+A^{t}=I$, then does it imply $\text{det}(A)>0$

I am stuck on this simple question for a long time.

• If $A \in M_{n}(\mathbb{R})$ satisfies $A+A^{t}=I$, then does it imply $\text{det}(A)>0$?

I tried finding a counter-example as well as tried proving it. But couldn't succeed. One question, which I would like to ask the experts is: How does one Judge the "Truth/False ness" of the question by seeing it. Is it something which comes by experience.

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Yes. Since $A = A^2 + A^T A = A^2 + A A^T$, it follows that $A$ is normal, hence by the spectral theorem has an orthonormal basis of eigenvectors $v_1, ... v_n$. Moreover, the given condition says precisely that the corresponding eigenvalues $\lambda_i$ satisfy $A v_i + A^T v_i = \lambda v_i + \overline{\lambda} v_i = v_i$, hence all have real part exactly $\frac{1}{2}$. Since $A$ is real, the set of its eigenvalues is closed under complex conjugation, so every eigenvalue $\frac{1}{2} + iy$ is paired with its complex conjugate $\frac{1}{2} - iy$ (and the eigenvalue $\frac{1}{2}$ can occur with any multiplicity). The product of these is positive, so $\det A > 0$.

Here is an alternate argument which does not rely on the spectral theorem. The condition gives $$\langle v, v \rangle = \langle (A + A^T) v, v \rangle = \langle Av, v \rangle + \langle A^T v, v \rangle = 2 \langle Av, v \rangle$$

for $v \in \mathbb{R}^n$. It follows that any such matrix $A$ is invertible, since $Av = 0$ implies $\langle v, v \rangle = 0$. On the other hand, the set of all matrices $A$ satisfying this condition is an affine subspace - in particular, it is path-connected. A path-connected set of invertible matrices must lie in a path component of $\text{GL}_n(\mathbb{R})$, hence the determinant must be the same sign for every element of the set. And $\frac{1}{2} I$ is in this set.

Note that the set of matrices satisfying this condition is precisely a translate of the set of skew-symmetric matrices by $\frac{1}{2} I$.

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Really superb. – user9413 May 7 '11 at 8:43
I cannot agree with @Chandru1 more. – awllower May 7 '11 at 8:50
Could you please explain your thought procedure. That would be more helpful. – user9413 May 7 '11 at 9:41
You mean the set is an affine subspace, not a linear subspace (it doesn't contain 0). – Chris Eagle May 7 '11 at 9:55
@Chris: yes, thanks, that's what I meant. @Chandru1: the first solution just comes from being familiar with the spectral theorem. The second solution came from playing around with the condition using the defining property of the transpose. As you say, these things come with experience. – Qiaochu Yuan May 7 '11 at 19:32

Write $A$ as the form $\frac{1}{2}I+B$, where $B$ is an anti-symmetric matrix. We need to show that $\det(\frac{1}{2}I+B)>0$ for any such $B$. For that we'll look at the polynomial $P(\lambda)=\det(I-\lambda B)$. Clearly $P(0)=1$. To show it only take positive value we just need to show it has no real root. The roots of $P(\lambda)$ are exactly the reciprocal of the nonzero eigenvalues of $B$. Let's write $\langle\cdot, \cdot \rangle$ for the usual inner product on $\mathbb{R}^n$. Assume that $\mu$ is a real eigenvalue of $B$ with eigenvector $v$, then

$$\mu\langle v, v\rangle=\langle Bv, v\rangle = \langle v, B^T v\rangle=\langle v, -Bv\rangle= -\mu \langle v, v\rangle.$$

Since $\langle v, v\rangle\neq 0$, it follows that $\mu=0$. So $B$ has no nonzero real eigenvalues, and hence $P(\lambda)$ has no real roots.

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Thanks. – user9413 May 7 '11 at 9:45

It is enough to show that $\det(\lambda I+B)>0$ for any skew symmetric matrix $B$ and any $\lambda>0$. I will argue by induction on the dimesion $n$ of the matrix. For $n=1$ $$\det((\lambda))=\lambda>0.$$ To reduce the $n$-dimensional case to the $(n-1)$-dimensional, use Chio's pivotal condensation.

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