Determinant of an identity matrix multiplied by a unit complex number

Question

I have read on solution to homework question that for $I\epsilon M_n(\mathbb{C})$, $$ \det(e^{-i\theta/n}I) = e^{-i\theta} $$ But I do not understand, why is the complex number not treated like a constant and taken out front, somehow it is multiplied by n?

Answer 1

The determinant function is multi-linear which means it's linear in each row rather than for the whole matrix. In general for an $n \times n$ matrix A then $\det (kA)=k^n \det(A)$ because we have to factor out the $k$ from each row separately.

Answer 2

Or rather, it's raised to the $n$th power. Take the determinant of the following:

$$ \det \left [ \begin{array}{cccc} e^{-i\theta/n} & 0 & \ldots & 0 \\ 0 & e^{-i\theta/n} & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & e^{-i\theta/n}\\ \end{array} \right ]. $$

Answer 3

The determinant of a diagonal matrix is the product of the diagonal entries. So, $\det(cI) = c^n$. Your misconception is that you seem to be thinking of $\det$ as a linear function, when it is actually a multi-linear function of the columns (or rows) of a matrix.

In particular, if $v_1, \dots, v_n$ are vectors, and $(v_1, \dots, v_n)$ is the $n \times n$ matrix with $v_i$ as the $i^{th}$ column, then for any constants $c_1, \dots, c_n$, \begin{align} \det(c_1v_1, \dots, c_n v_n) = c_1\cdots c_n \det (v_1, \dots, v_n) \end{align}