Prove that a linear operator T on a finite dimensional vector space is invertible if and only if zero is not an eigenvalue of T
??
i'm a bit rusty in linear algebra (even though i took it just last semester) so here's your answer in a nutshell, you have to transform this into a rigorous proof.
Recall that eigen values and eigen vectors are "unchanged" under linear transformations. so the matrix resulting from applying a linear transformation on a finite demensional matrix will have the same eigenvalues and eigenvectors as the first matrix.
now, if we have an eigenvalue = 0, then det(I*lambda - A) = 0, this means that detA = 0, which means that the matrix is not invertible, since for a matrix A to be invertible, we must have detA not= 0
so that should get you started. sorry i can't help more, i was lost for most of that class
I prove it one way.
Let the standard matrix for T be A, that is [T]=A.
We say that A is invertible by hypothesis, meaning
det(A) not = 0.
The eigenvalues k are solutions to the polynomial,
det(k*I-A)=0
Now, if k is zero, then,
det(k*I-A)=det(0*I-A)=det(-A)=(-1)^n * det(A) !=0.
Because det(A)!=0.
Hence k=0 is not and eigenvalue.