Determinant of an endomorphism (Proof)

Question

I Need to prove That for every a there is an endomorphism phi such that det(phi)=a.

I habe no idea how to do this. Can anybody help me?

Answer 1

Assuming you're talking of a $n$ dimensional vector space that is non trivial.

If $a \in \mathbb{K}$, and $\mathbb{K}^n = \mathrm{vect}(e_1,e_2,\ldots,e_n)$, what can you say about the determinent of the endomorphism $u$ defined by \begin{align} u(e_1) &= a\cdot e_1 \\ u(e_j) &= e_j, j>1 \end{align} ?

Answer 2

If we are in a $n$-dimensional vector space $V$ over a field $\mathbf F$, consider the endomorphism $\phi : V \to V$ such that, in some basis, its matrix representation is $$\begin{pmatrix} a & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{pmatrix}.$$