Why is this matrix invertible

Question

I was wondering if there is a way to see why $(1+A)$ invertible, if $A$ is a skew symmetric matrix. and I know that all eigenvalues of $A$ have zero real part and $A$ is unitarily diagonalisable.

Answer 1

That's because $A$ does not have eigenvalues -1. If it did, $\det(A+I)$ would be 0.

Answer 2

The eigenvalues of a skew symmetric (real) matrix are all purely imaginary: of the form $\;bi\;,\;\;0\neq b\in\Bbb R\;$ , and they appear in conjugate pais, so we can list them as $\;bi\,,\,-bi\;,\;\;0\neq b\in\Bbb R\;$

If $\;A\;$ is diagonalizable and $\;J_A\;$ is its JCF , then $\;J_A+I\;$ has elements $\;1\pm bi\;,\;\;0\neq b\in\Bbb R\;$ on its main diagonal, and this matrix's determinant is non zero.

Fill in details now.

Answer 3

Write $A=P\Lambda P^{-1}$ where $$ \Lambda=\DeclareMathOperator{diag}{diag}\diag(\beta_1i,\dotsc,\beta_ni) $$ with $\beta_j\in\Bbb R$. Then $$ I+A=I+P\Lambda P^{-1}=P(I+\Lambda)P^{-1}=P\diag(1+\beta_1i,\dotsc,1+\beta_ni)P^{-1} $$ Thus the eigenvalues of $I+A$ are $\lambda_k=1+\beta_ki$. In particular, $I+A$ has nonzero eigenvalues. Hence $I+A$ is invertible.