Just wanted some input to see if my proof is satisfactory or if it needs some cleaning up.

Here is what I have.

ProofSuppose 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 ProofSuppose 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*