How to compute this determinant, without the Sylvester determinant theorem,

Question

The problem statement is:

Show that there exists numbers $a$ and $b$ such that

$$det (A + sxy^*)= a+bs$$

here $A$ is an $nxn$ matrix with real entries, and $x,y\in R^n$.

I've been using brute force and using multi-linearity of the determinant several times, and the computations are getting longer, but I see no pattern yet.

Any ideas are welcome.

Thanks,

Answer 1

$xy^*$ is a rank one matrix. Try to choose a basis where the associated map is given by a one column matrix...

Edit: (Assume for simplicity you are working with the standard scalar product on $\mathbb{R}^n$ and) Assume $x\neq 0\neq y$. $xy^* (v) = \langle y, v\rangle x $ is a multiple of $x$, so one option is to choose a basis consisting of $\frac{x}{||x||}$ and and ONB of the the orthogonal complement of the space spanned by $x$. Then, after a corresponding change of basis, $x$ is just $e_1$ and the Matrix representation of $x^*y$ with respect to this base is a one row matrix. So it's not a one column matrix as suggested, but you should know that $\det$ is multilinear with respect to both rows and columns. If you want to write it as a one column matrix you need to work with $y$ and have to exend $y^*$ to a base of $\mathbb{R^n}^*$. That's more or less the same, but possibly less intuitive.

Answer 2

If $A$ is invertible, we have $$\det(A+sxy^*) = \det(A) \det(I+sA^{-1}xy^*) = \det(A) (1+sy^*A^{-1}x)$$ where the last step is using the Sylvester's determinant theorem. Hence, we have $a= \det(A)$ and $b=y^*A^{-1}x\det(A)$.

If $A$ has rank less than $n-1$, then $\det(A+sxy^*)$ can have rank at most $n-1$, which means the determinant is $0$ for all $s$.

If $A$ has rank $n-1$ and $x$ belongs to the range of $A$, then again the determinant is $0$ for all $s$.

If $A$ is rank $n-1$ and $x$ doesn't belong to the range of $A$, then the determinant is non-zero. Now write $A+sxy^*$ as $A+xy^* + (s-1)xy^*$ and apply the Sylvester determinant theorem again.

Hence,to summarise, we have \begin{align} a = \det(A) \text{ and }b = \det(A+xy^*) - \det(A) = y^*A^{-1}x \det(A) \end{align} where the last equality is true whenever $A$ is invertible.