Relation of eigenvalues and invertibilty of a matrix

Question

Proof that if a matrix has no eigenvalues, it'll not be invertible:

$$ \det(A-\lambda I) = 0 \Rightarrow \begin{pmatrix} a_{1,1}-\lambda & a_{1,2} & \cdots & a_{1,n} \\ a_{2,1} & a_{2,2}-\lambda & \cdots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \cdots & a_{n,n}-\lambda \end{pmatrix} $$

If all $\lambda=0$, we would get that our $\det(A) = 0$, hence, it would not be invertible.

Is this proof correct?!

Saying that $A$ has no eigenvalues is different from saying that $A$ has zero as its only eigenvalue — Ben Grossmann, Oct 25 '16 at 14:34
Sure, modulo some correct wording. $\lambda$ is an eigenvalue of $A$ iff $\det(A-\lambda I) = 0$. Hence $0$ is an eigenvalue of $A$ iff $\det A = 0$ iff $A$ is not invertible. — copper.hat, Oct 25 '16 at 14:41

score 1 · Accepted Answer · edited Apr 13 '17 at 12:19

1

Hint:

the determinant of a matrix is the product of its eigenvalues. For a simple proof see Show that the determinant of $A$ is equal to the product of its eigenvalues..

So, if one (not all) eigenvalue is zero the determinant is zero and the matrix is not invertible.

edited Apr 13 '17 at 12:19

Community

1

answered Oct 25 '16 at 14:46

Emilio Novati

62,675

Perfect, Thanks!!! – Bruno Reis Oct 25 '16 at 14:49

score 0 · Answer 2 · answered Oct 25 '16 at 14:42

0

If det (A)=0, then A has the eigenvalue 0

answered Oct 25 '16 at 14:42

Fred

77,394

Relation of eigenvalues and invertibilty of a matrix

2 Answers2