What is the determinant of a matrix with a zero column?

[Kate2010]

Homework Statement

A is an nxn matrix. Suppose A has the form ($$^{U}_{W}$$$$^{V}_{X}$$) in which U, V, W, X are n1xn1, n1x n2, n2xn1 and n2xn2 matrices respectively, such that n1 + n2 = n. If V=0, show that detA = detUdetX

Homework Equations

detA := $$\sum$$ $$_{\rho\in sym(n)}$$ sign ($$\rho$$)$$\Pi$$ a_{i i$$\rho$$}

The Attempt at a Solution

I don't really know how to go about this. If I expand along the 1st row I will get each of the u entries of the first row multiplied by their minor and sign summed together, and the v coefficients will all be zero. I don't know how to write this using the correct notation or where I would go from here.

 

[VeeEight]

I would try showing this using cofactor expansion, specifically along a row/column with a lot of zeros (so you are only dealing with one term).

 

[Kate2010]

Is this true?

Let $$\rho$$ act on (1 2 ... n1) and $$\sigma$$ act on (n1+1 ... n1 + n2)

So detU = $$\sum$$ $$_{\rho\in sym(n)}$$ sign ($$\rho$$)$$\Pi$$ a_{i i$$\rho$$}
and detX $$\sum$$ $$_{\sigma\in sym(n)}$$ sign ($$\sigma$$)$$\Pi$$ a_{i i$$\sigma$$}

detA = $$\sum$$ $$_{\rho\sigma\in sym(n)}$$ sign ($$\rho\sigma$$)$$\Pi$$ a_{i i$$\rho\sigma$$}
= $$\sum$$ $$_{\rho\in sym(n)}$$ sign ($$\rho$$)$$\Pi$$ a_{i i$$\rho$$} x$$\sum$$ $$_{\sigma\in sym(n)}$$ sign ($$\sigma$$)$$\Pi$$ a_{i i$$\sigma$$}
=detUdetX