Least value of $\mid\{I + \lambda A^2\}\mid$ 
Let $A$ be a skew symmetric $n \times n$ matrix, then what is the least possible value of $\mid{I + \lambda A^2}\mid$, for any real value of $\lambda$?

I verified that its least value is $0$, by taking a $2 \times 2$ skew symmetric matrix with off diagonal elements as $1$, $-1$. But I need a formal proof for same. Any help will be appreciated.
 A: There are two components to this proof:


*

*There exists $\lambda,A$ such that $|I + \lambda A^2| = 0$

*For every $\lambda,A$, $|I + \lambda A^2| \geq 0$.


For the first part, your idea should suffice.  In particular, we can always take
$$
\lambda = 1, \quad A = 
\pmatrix{
0&-1&0&\cdots\\
1&0&0&\cdots\\
0&0&0\\
\vdots&\vdots&&\ddots} .
$$
Note that $I + \lambda A^2$ is diagonal with zeros on the diagonal and therefore has determinant zero.
For the second part, one approach is to use the spectral theorem.  In particular, since all eigenvalues of $A$ are either $0$ or of the form $\lambda = \pm \mu i$, we can deduce that $A^2$ has real eigenvalues, and all negative eigenvalues of $A^2$ have even multiplicity.  It follows that $I + \lambda A^2$ has real eigenvalues, and its negative eigenvalues of also have even multiplicity. So, $|I + \lambda A^2|$ (which is the product of these eigenvalues) is necessarily non-negative. 

Another approach is as follows: note that for $\lambda \leq 0$, we have
$$
(I + \sqrt{|\lambda|}A)^T(I + \sqrt{|\lambda|}A) = 
(I - \sqrt{|\lambda|}A)(I + \sqrt{|\lambda|}A) = I - |\lambda| A^2 = I + \lambda A^2.
$$
It follows that 
$$
|I + \lambda A^2| = |(I + \sqrt{|\lambda|}A)^T|\cdot |I + \sqrt{|\lambda|}A| = |I + \sqrt{|\lambda|}A|^2 \geq 0.
$$
