Proving that matrix is invertible

Question

Let $I$ be an identity matrix of size $3$ and let $A$ be the matrix of size $3$ having three different nonzero eigenvalues.

Prove that matrix $A + A^{-1} + I$ is invertible.

I was trying to somehow show that determinant is different than zero, but I am not sure if this is good approach.

Answer 1

The matrix could be singular. For example let $A=\begin{pmatrix}0&1&0\\-1&-1&0\\0&0&1\end{pmatrix}$

Then $A+A^{-1}+I=\begin{pmatrix}0&0&0\\0&0&0\\0&0&3\end{pmatrix}$

Answer 2

Since you have three different eigenvalues, you can diagonalize $A = T D T^{-1} $, where $ D$ is the diagonal matrix with the three different eigenvalues on the diagonal. Now,

$$ A + A^{-1} + I = T (D + D^{-1} + I)T^{-1}.$$

Use this to say something about the eigenvalues of the matrix.

In response to your comment: The above formula gives you $$ \tilde\lambda_i = \lambda_i +\lambda_i^{-1} + 1 $$ for the new eigenvalues, where $ \lambda_i $ are the eigenvalues of $A$. Therfore

$$ \tilde\lambda_i = 0 \iff \lambda_i^2 + \lambda_i + 1 = 0 .$$

Since above polynomial does not have any real roots, the statement holds, if the eigenvalues of $A$ are required to be real. If the eigenvalues are allowed to be complex, the statement fails, since the polynomial then has zeros. An example is given in the other answer.