Searching for a proof or reference of a statement.

Question

I am looking for a proof or reference of the follwoing statement:

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If a rank one $n$ by $n$ matrix $M$ acts on $\mathbb{C}^n$ as $Mx=w(x)u$ for some linear form $w$,
one has $det(Id+M)= 1 +w(u)$.

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Thanks for any help.

Answer 1

By Riesz representation, there is a unique $v \in \mathbb{C}^n$ such that $w(x) = (x, v)$ for all $x \in \mathbb{C}^n$, where $(\cdot, \cdot)$ denotes the Euclidean inner product. Now let $\{v/|v|, v_2, \dots, v_n\}$ be an orthonormal basis of $\mathbb{C}^n$. Consider the matrix representation of $M$ with respect to this basis: we have $M(v/|v|) = |v|u$, and $Mv_2 = \dots = Mv_n = 0$. Thus $I + M$ has first column $1 + |v|u$ and ones on the diagonal. Using a few column operations, we can kill all the non diagonal elements of $I + M$, so $\det(I + M) = 1 + |v|u_1$, where $u_1$ is the first coordinate of $u$ in this basis. We have $u_1 = (u, v/|v|)$, so $\det(I + M) = 1 + (u, v) = 1 + w(v)$.

Answer 2

Here's a sketch of a proof using some advanced machinery, in case it's helpful. There's probably a simpler proof.

We know that $$\det(M+I)=\sum_{p=0}^n\textstyle\mathop{\mathrm{tr}}(\bigwedge^pM\mathop{\square}\bigwedge^{n-p}I)\tag{1}$$ where $\square$ denotes the box product (generalization of the exterior power to more than one matrix). Since $M$ has rank $1$, $\bigwedge^pM=0$ for $p\ge 2$, so (1) reduces to $$\det(M+I)=\textstyle\mathop{\mathrm{tr}}(\bigwedge^n I)+\mathop{\mathrm{tr}}(M\mathop{\square}\bigwedge^{n-1}I)\tag{2}$$ But $\mathop{\mathrm{tr}}(\bigwedge^n I)=1$, and direct computation shows that $\mathop{\mathrm{tr}}(M\mathop{\square}\bigwedge^{n-1}I)=\mathop{\mathrm{tr}}(M)$, so (2) becomes $$\det(M+I)=1+\mathop{\mathrm{tr}}(M)\tag{3}$$ which gives your result since for $M=w\otimes u$, $\mathop{\mathrm{tr}}(M)=w(u)$.