$E(x) = \mathbb{I}_{n} - xx^{*} $ is nonsingular if and only if $x^{*}x \neq 1$

Question

We define the matrix $E(x) := \mathbb{I}_{n} - xx^{*} $ , where $ \mathbb{I}_{n} $ is the identity matrix of $n \times n$ and $x \in \mathbb{C}^{n}$ ($x$ is a column matrix) and $x^{*}$ is the conjugate transpose of $x$. Here's my attempt

$(\leftarrow )$
suppose that $x^{*}x \neq 1$ and also that $E(x)$ is singular, that is, $\det(E(x)) = 0$. Since the $\det(E(x))$ is the product of the eigenvalues of $E(x)$ , then we have that $\lambda_{1} = 0$ is an eigenvalue of $E(x)$ associated to an nonzero eigenvector $v$. This means that:

$$ E(x)v = \lambda_{1}v = 0 $$ then $$ (\mathbb{I}_{n}-xx^{*})v = v -x x^{*}v = 0$$ this implies that

$$ xx^{*}v = v $$ which means that $xx^{*}$ has eigenvalue $\lambda=1$. I'm stuck and I don't know how to start the other direction. How would you do it? ...thank you!!

Answer 1

If $I-xx^*$ is singular, then as you say $xx^*$ has $1$ as an eigenvalue; it follows that $x^*x=1$. Indeed, if $xx^*v=v$, then $x^*x(x^*v)=x^*v$; and $x^*v\ne0$ since $x^*v=0$ implies $v=xx^*v=0$.

Conversely, if $x^*x=1$, then $xx^*x=x$, so $(I-xx^*)x=0$ and thus $I-xx^*$ is singular.

Answer 2

Your question recalls the equivalence of a square matrix $M$ being singular and $\,\det M =0\,$.

When relying on Sylvester's determinant identity $$\det\big(\mathbb{I}_n + AB\big)\; =\;\det\big(\mathbb I_m + BA\big)\qquad\forall\: A\in M_{n\times m}\,\text{ and }\,B\in M_{m\times n}$$ then with $x\in\mathbb C^n = M_{n\times 1}(\mathbb C)\,$ you obtain $$\begin{align} \det\big(\mathbb{I}_n-xx^*\big)\; & =\;\det\big(\mathbb I_1-x^*x\big) \\[1ex] & =\;1-x^*x\;=\; 1-\|x\|^2 \end{align}$$ which gives the claim: $\,\mathbb{I}_n-xx^*\,$ is singular iff $x$ is a unit vector (in the 2-norm).