Just wanted some input to see if my proof is satisfactory or if it needs some cleaning up.
Here is what I have.
Proof
Suppose A is square matrix and invertible and, for the sake of contradiction, let 0 be an eigenvalue. Consider, (A−λI)⋅v=0 with λ=0
⇒(A−0⋅I)v=0
⇒(A−0)v=0
⇒Av=0We know A is an invertible and in order for Av=0, v=0, but v must be non-trivial such that det(A−λI)=0. Here lies our contradiction. Hence, 0 cannot be an eigenvalue.
Revised Proof
Suppose A is square matrix and has an eigenvalue of 0. For the sake of contradiction, lets assume A is invertible.
Consider, Av=λv, with λ=0 means there exists a non-zero v such that Av=0. This implies Av=0v⇒Av=0
For an invertible matrix A, Av=0 implies v=0. So, Av=0=A⋅0. Since v cannot be 0,this means A must not have been one-to-one. Hence, our contradiction, A must not be invertible.
Answer
Your proof is correct. In fact, a square matrix A is invertible if and only if 0 is not an eigenvalue of A. (You can replace all logical implications in your proof by logical equivalences.)
Hope this helps!
Attribution
Source : Link , Question Author : Derrick J. , Answer Author : Amitesh Datta