finding the determinant of a matrix involving a product

Question

Let $A = I + yx^H $. where $x,y \in \mathbb{C}^n $. I am trying to compute $\det A $. I know

$$ \det ( A ) = 1 + \det( yx^H )$$

But, how can I find $\det (y x^H) $ ?. According to my notes, it should be $x^H y $, but I dont see why

Answer 1

The formula that you claim you know is wrong. For a counterexample, consider the case that both $x$ and $y$ are $(1,1)^T$. See Wikipedia for the correct formula.

Anyway, $yx^H$ is at most rank-one, so its determinant is zero when $n>1$.

Answer 2

Suppose that $x\neq 0$, $y\neq 0$, $x^H y \neq 0$, and $n\geq 2$. Then the matrix $A = I + yx^H$ has two distinct eigenvalues. One is $\lambda_1 = 1$. To see this, observe that $\det(A - I) = \det(yx^H) = 0$ since $yx^H$ is at most rank $1$. The set $E_{\lambda_1} = \{z\in \mathbb{C}^n : x^Hz = 0\}$ is the eigenspace corresponding to this eigenvalue. Since $E_{\lambda_1}$ is the orthogonal complement of $\mathrm{span}\{x\}$, it has dimension $n - 1$. The other eigenvalue is $\lambda_2 = 1 + x^Hy$, which corresponds to the eigenvector $y$. On the other hand, if $x^Hy = 0$, then the only eigenvalue is $\lambda_1 = 1$. Since the determinant is just the product of the eigenvalues, it follows that $\det(A) = (1 + x^Hy)(1)^{n-1} = 1 + x^Hy$.

Answer 3

I think what you are looking for is the matrix determinant lemma, which states that: $$\det(B + uv^T) = (1 + v^TB^{-1}u)\det B$$ and you have $B = I$, so you get: $$\det(I + uv^T) = (1 + v^Tu) \det I = 1 + v^Tu.$$