Proof for determinant of a scalar multiplied by a vector

[3.141592654]

Homework Statement

Let A be an n x n matrix and $\alpha$ a scalar. Show that $det(\alpha A) = \alpha^{n}det(A)$

Homework Equations

$det(A) = a_{11}A_{11} + a_{12}A_{12} + \cdots + a_{1n}A_{1n}$

where $A_{ij} = (-1)^{i+j}det(M_{ij})$

The Attempt at a Solution

$det(A) = a_{11}A_{11} + a_{12}A_{12} + \cdots + a_{1n}A_{1n}$

$det(\alpha A) = \alpha a_{11}A^{\alpha}_{11} + \alpha a_{12}A^{\alpha}_{12} + \cdots + \alpha a_{1n}A^{\alpha}_{1n}$

$det(\alpha A) = \alpha (a_{11}A^{\alpha}_{11} + a_{12}A^{\alpha}_{12} + \cdots + a_{1n}A^{\alpha}_{1n})$

I can see that as I go through and calculate the cofactors I will continue to get an additional alpha coefficient each time, so that I will end up with $det(\alpha A) = \alpha^{n}det(A)$, but I am having trouble formalizing it. Thank you in advance for any help.

 

[spamiam]

I haven't tried it, but induction on $n$ might work. Alternatively, I think it might be easier to think of the determinant as a multilinear function of the columns of the matrix rather than using cofactor expansion.

 

[lanedance]

or do you know the expression for determinant using the e-permutation symbol, see:
http://www.math.odu.edu/~jhh/part2.PDF
example 1.1-9
should follow straight form there

However, it should follow straight from your work though, note that if $C_{ij}$ is a cofactor of $A$, then $\alpha C_{ij}$ is a cofactor of $\alpha A_{ij}$
$$C_{ij} = (-1)^{i+j}M_{ij}$$