The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

This discussion is now closed.

Check out other Related discussions

Linear algebra - adjoint matrix = 0

Ok so the question is:

Let A be an nxn matrix where n ≥\geq 2. Show A* = 0 (A* is the adjoint matrix) if and only if rank(A)≤\leq n-2.

I'm really struggling with answering this. These are my thoughts:

If rank(A) ≤\leq n-2 then A contains at least 2 dependent rows. Does this automatically make A* = 0?

If A*=0 then (-1)i+jdet(Aij) = 0 for all i and j. So det(Aij) = 0 for all i and j. Having a zero determinant implies dependence?
Reply 1
Avatar for Zhen Lin
Zhen Lin
12
I think you mean the adjugate matrix. I don't know the proof of the result, but I can imagine that it is true - almost every minor will obviously have 2 dependent rows or columns; the ones that don't obviously have dependent rows or columns lie on the diagonal.

Yes, if the determinant of a matrix is zero then there is a dependent row or column.

Related discussions

Latest

Trending

Trending

Articles for you