Determinant of matrix of linear transformation in complex vector space

Question

Let $V$ be finite complex vector space, $a\not= 0$ an element of $V$, and $f$ linear functional on space $V$. $A: V \to V$ has definition: $A(x)= x - f(x)*a$. Find determinant of $A$.

Answer 1

Give $V$ a basis $\{b_1,\dots,b_n\}$, where $b_1 = a$. Then the matrix of $A$ looks like $$ \begin{bmatrix} 1 - f(b_1) & -f(b_2) & -f(b_3) & \cdots & -f(b_n) \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & 0 \\ 0 & 0 & 0 & \cdots & 1 \end{bmatrix} $$ and hence the determinant is $1-f(b_1) = 1-f(a)$.

Answer 2

As $f$ is a linear functional on $V$, there is $v \in V$ such that $f(x) = v^Tx$. Note that

$$A(x) = x - f(x)a = x - af(x) = x - a(v^Tx) = x - (av^T)x.$$

Therefore $A = I_n - av^T$ where $n = \dim V$. Now by Sylvester's determinant theorem, we have

$$\det A = \det(I_n - av^T) = \det(I_1 - v^Ta) = \det(1 - f(a)) = 1 - f(a).$$