- #1
Mdhiggenz
- 327
- 1
Homework Statement
I have a quick question about the proof below.
Let A be an nxn matrix. Prove that A is singular if and only if λ=0
I searched the proof online, and they did it using Ax=0
However,
When I tried doing on my own my solution was this
If A is singular then the det(A)=0
However we know from the following relation ship that
(λ1*λ2...*λi)=det(A)
thus there must be at least one eigenvalue λi such that
(λ1*λ2...*λi)=0 end of proof
Is my reasoning correct?
Thank you