Eigenvalues of a real orthogonal matrix lie on the unit circle and occur either as real values $\pm1$ or in conjugate pairs $e^{\pm i\theta}$ ( $\theta\neq0,\pi$ ), and every conjugate pair contributes $e^{i\theta}e^{-i\theta}=1$ to the product of eigenvalues. Since $\det A=-1$ is the product of all four eigenvalues, the product of the real eigenvalues alone (after discarding complex-conjugate pairs, which each contribute $+1$ ) must equal $-1$ .

If there are $0$ conjugate pairs, all $4$ eigenvalues are real $\pm1$ 's whose product is $-1$ : this forces an odd count of $-1$ 's among $4$ slots (one or three), hence an even, nonzero count of $+1$ 's (three or one) — so $+1$ occurs at least once. If there is $1$ conjugate pair, the remaining $2$ real eigenvalues must multiply to $-1$ , i.e. one is $+1$ and the other $-1$ — again $+1$ occurs. Two conjugate pairs is impossible here, since that would leave no real eigenvalues and force the total product to be $+1\neq-1$ .

In every admissible case $A$ has eigenvalue $1$ , so $I_4-A$ has eigenvalue $1-1=0$ , and therefore $\det(I_4-A)=\prod_i(1-\lambda_i)=0$ .