Show that $\det(I-xx{^T})=0$ if $x{^T}x=1$

Question

Show that

$$\det(I-xx{^T})=0$$

if $x{^T}x=1$.

Answer 1

A standard result is $\det (I+AB) = \det (I+BA)$.

Take $A=-x^T, B = x$, then we have $\det (I - x x^T) = 1- x^T x = 0$.

There are various ways to show the identity above, here is one: The non zero eigenvalues of $AB$ and $BA$ are the same, hence we have $\det (I+AB) = \prod_{\lambda \in \sigma(AB) \setminus \{0\}} (1+\lambda) \prod_{\lambda \in \sigma(BA) \setminus \{0\}} (1+\lambda)= \det(I+BA) $.

Answer 2

If $\alpha$ is an eigenvalue of $xx^T$, then $\det(\alpha I-xx^T)=0$. So it suffices to show that $1$ is an eigenvalue of $xx^T$. But this is trivial: $$ (xx^T)(x)=x(x^Tx)=x. $$