Question

 Show that the inverse of a square matrix A exists if and only if the

eigenvalues λ1

,λ2

,··· ,λn of A are different from zero. If A

−1

exists

show that its eigenvalues are 1

λ1

,

1

λ2

,···

1

λn

.

Answer 1

If $\lambda=0$ is an eigenvalue of the matrix A, and $X\ne0$ is a corresponding column-eigenvector, then the inverse matrix $A^{-1}$ doesn't exist, since $X=IX=(A^{-1}A)X=A^{-1}(AX)=A^{-1}0=0$.

Conversely, if $\lambda=0$ is not an eigenvalue of the matrix A, then $\lambda=0$ is not a root of the characteristic polynomial $p_A(\lambda)=\det(A-\lambda I)$ of the matrix A. Therefore, $\det A\ne 0$ and the matrix A is invertible.

Assuming A is invertible, consider the characteristic polynomial $p_{A}(\lambda)=\det(A-\lambda I)$ of the matrix A.

$p_{A}(\lambda)=\det(A-\lambda I)=\det A\det(I-\lambda A^{-1})=(-\lambda)^n\det A \det(A^{-1} - \lambda^{-1}I)=(-\lambda)^n\det A \cdot p_{A^{-1}}(1/\lambda)$

From this we can see that $\lambda$ is an eigenvalue of the matrix A if and only if $1/\lambda$ is an eigenvalue of the matrix $A^{-1}$. Moreover, one can see that multiplicity of the eigenvalue $\lambda$ of the matrix A is the same as multiplicity of the eigenvalue $1/\lambda$ of the matrix $A^{-1}$.

So, if $\lambda_1,\dots,\lambda_n$ are eigenvalues of the matrix A, then $1/\lambda_1,\dots,1/\lambda_n$ are eigenvalues of the matrix $A^{-1}$.

The assertion is proved.