prove that $\det(I + tA^2)\ge 0$ for all real $t$ 
The source of this problem is problem 10 from this Putnam training handout.


Let $A$ be a real skew-symmetric square matrix (i.e. $A^T = -A$). Prove that $\det(I + tA^2) \ge 0$ for all real $t$.

Let $n$ be the number of rows of A. Since $\det(A^T) = \det(A), \det(-A) = (-1)^n \det(A) = \det(A),$ which implies that either A is singular or A has an even number of rows (and hence columns). First assume A is invertible. Then $\det(I + tA^2) = \det(AA^{-1} + tA^2) = \det(A)\det(A^{-1} + tA).$ It doesn't seem clear why $\det(A^{-1} + tA)$ would be nonzero. I think it would be useful to use the fact that if $p$ is a polynomial in n variables $x_1,\cdots, x_n$ with real coefficients, then $p(\overline{x_1},\cdots, \overline{x_n}) = \overline{p(x_1,\cdots, x_n)},$ where $p$ denotes the complex conjugate. A useful idea seems to be to show that $\det(I + tA^2) = |c|^2$ for some complex number c. Taking the square of the determinant doesn't seem to be useful because the square root may be negative. Also, from the above discussion about polynomials, one can prove that $\det(A + iI) = \overline{\det(A - iI)}$ for any matrix A with real entries. I'm not sure if ideas from this post about an integer skew symmetric matrix could help.

Here is an important claim used in the answer to the duplicate:

If the canonical form of $A$ contains a Jordan block with dimension more than $2$, then there exist (orthonormal) $u,v$ such that $Au=\lambda u$ and $A v=\lambda v+u$.
Here's an attempt to justify the Jordan block claim:
Note that it suffices to show the claims hold for linearly independent $u$ and $v$ because in that case, we can let $v' := v- u\langle v,u\rangle/|u|^2.$ Then $\langle v', u\rangle = 0$ and if we let $x = \langle v,u\rangle/|u|^2,$ then $Av' = A(v-x u) = \lambda v + u - x(\lambda u) = \lambda (v-xu) + u = \lambda v' + u.$ So we can just replace $v$ with $v'$. Then we can normalize u and v by dividing them by $|u|$ and $|v|$ respectively.
Suppose the canonical form of $A$ has a Jordan block with dimension $m>2$, say $\begin{pmatrix}\lambda & 1 & 0 & 0 & \cdots & 0\\
0 & \lambda & 1 & 0 & \cdots & 0\\
\vdots & \vdots &\vdots & \ddots & \ddots &\vdots\\
0 & 0 & 0 & 0 & \cdots &1\\
0 & 0 & 0 & 0 & \cdots &\lambda\end{pmatrix}$
Then we may find a cycle of generalized eigenvectors $u_1,\cdots, u_m$ so that $(A - \lambda I)u_1 = 0,\cdots ,(A-\lambda I)u_{m} = u_{m-1}$. In general, $(A-\lambda I)^i u_i = 0,$ so $u_i \in Null(A-\lambda I)^i$. Pick $u=u_1, v=u_2$. Note that we can show these vectors are linearly independent by induction on n. Assume the result holds for $n-1$ (the base case of $n=1$ is trivial). If $\sum_{i=1}^n c_i u_i = 0,$ then applying $A - \lambda I$ to both sides gives $\sum_{i=1}^{n-1} c_{i+1} u_i = 0$. By induction, $u_1,\cdots, u_{n-1}$ are linearly independent and so $c_{i+1} = 0$ for $1\leq i\leq n-1.$ Thus $c_1 = 0$ after substituting these equations into the original equation.
Note that the above proof shows that no Jordan block can have size at least $2$. So every Jordan block has size 1, meaning that the matrix is diagonalizable.
 A: (Too long for a comment.)

It doesn't seem clear why $\det(A^{-1} + tA)$ would be nonzero.

It can be zero! Consider $A=\pmatrix{0&-1\\ 1&0}$ and $t=1$ for instance.

A useful idea seems to be to show that $\det(I + tA^2) = |c|^2$ for some complex number $c$.

Yes. However, when $t\ne0$, you need to differentiate between the cases $t>0$ and $t<0$. By absorbing $\sqrt{|t|}$ into $A$, we may assume that $t=\pm1$.
When $t=-1$,
$$
\begin{aligned}
\det(I-A^2)
&=\det\big((I-A)(I+A)\big)\\
&=\det\big((I+A)^T(I+A)\big)\\
&=\det\big((I+A)^T\big)\det(I+A)\\
&=\det(I+A)\det(I+A)\\
&=\big(\det(I+A)\big)^2\\
&\ge0.
\end{aligned}
$$
When $t=1$,
$$
\begin{aligned}
\det(I+A^2)
&=\det\big((I-iA)(I+iA)\big)\\
&=\det(I-iA)\det(I+iA)\\
&=\det(\overline{I+iA})\det(I+iA)\\
&=\overline{\det(I+iA)}\det(I+iA)\\
&=|\det(I+iA)|^2\\
&\ge0.
\end{aligned}
$$
This is true for all real square matrices $A$, not just the skew-symmetric ones. However, when $A$ is skew-symmetric, we have $\det(I-iA)=\det\big((I-iA)^T\big)=\det(I+iA)=\overline{\det(I-iA)}$. So, in this case we may obtain a stronger conclusion that $\det(I-iA)=\det(I+iA)=$ some $x\in\mathbb R$ and $\det(I+A^2)=x^2\ge0$.
