im trying to prove that 0 is an eigenvalue of A iff A is non invertible. may i know why for a homogeneous solution Ax = 0 to have a non trivial solution, it means that A in non-invertible?
cant it be that x is a zero vector?
if det of A is non-zere, where A is invertible, then there doesnt exist a non trivial solution in the homogenous solution Ax=0?
i think im confusing myself:/
"may i know why for a homogeneous solution Ax = 0 to have a non trivial solution, it means that A in non-invertible?"
Certainly, let us suppose A is an nxn matrix. Having a non trivial solution for the homogeneous equation means that there is a redundancy in the one of the column vectors, namely that one of the column vectors can be written as a linear combination of the others. To find an invertible matrix for an nxn matrix we can solve
[ A I ] and try to get [ I A^-1 ] , for example if i had
to find its inverse i can do
using our row operations i can make this matrix into
so the inverse of is
this can also be done by solving : Ax=b where b are the column vectors of the identity for nxn matricies.
for which the solution is so this x is the first column of our inverse matrix, so u can do with all the column vectors of the identity.
Since the matrix u asked has a non trivial soln to Ax = 0. it means that even though A is nxn that it can really be modified into a matrix with 1 less column vector and still span as much space as the original nxn matrix. clearly a nx(n-1) matrix cannot span R^n which is the space spanned by the identity nxn matrix. so u will come upon a column vector from the nxn identity for which Ax=b dosent have a solution, so there can't exist an inverse matrix for ur nxn matrix. i hope this view point clear it up. if you have any question plz feel free to ask.
Simpler, I think:
If A is invertible, then if Ax= v, we have or so that Ax= v has a unique solution.
If A is invertible, Ax= 0 has the unique solution . Therefore, if Ax= 0 has a non-trivial (non-zero) solution then A is not invertible.
Yes, x can be 0- in fact, x= 0 is always a solution to Ax= 0. The question is whether that (trivial) solution is unique or not. If it is not unique, if Ax= 0 has other, non-zero, solutions, then A must not have an inverse.