Is a matrix AA with an eigenvalue of 00 invertible?

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

Here is what I have.


Suppose A is square matrix and invertible and, for the sake of contradiction, let 0 be an eigenvalue. Consider, (AλI)v=0 with λ=0

We know A is an invertible and in order for Av=0, v=0, but v must be non-trivial such that det(AλI)=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=0vAv=0

For an invertible matrix A, Av=0 implies v=0. So, Av=0=A0. Since v cannot be 0,this means A must not have been one-to-one. Hence, our contradiction, A must not be invertible.


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!

Source : Link , Question Author : Derrick J. , Answer Author : Amitesh Datta

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