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I'd love your help with this.

Let $A\in M_{n}(\mathbb{C})$ be a Hermitian matrix $(A^{*}=A)$.

How can I prove that $I-iA$ is Invertible?

Thank you

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up vote 9 down vote accepted

$I$ has only eigenvalue 1, and $A$ has only real eigenvalues, thus $iA$ will only have imaginary eigenvalues. Say $A$ has eigenvalues $\lambda_k$, where $\lambda_k$ could possibly be zero, with corresponding eigenvector $v_k$. Then:

$$(I-iA)v_k= v_k - i\lambda_k v_k = (1-i\lambda_k)v_k$$

Thus the eigenvalues for $(I-iA)$ will be on the form $1-i\lambda_k$, $\lambda_k \in \mathbb R$. Since all eigenvalues are non-zero, the matrix will be invertible.

Edit: Hermitian matrices have only real eigenvalues since, if $Av = \lambda v$, then $$v^* A v = \lambda v^*v$$ taking the hermitian conjugate of both sides yields: $$v^* A^* v = \lambda^* v^* v$$ Since $A = A^*$ we get $\lambda = \lambda^*$ from the two equations above, where $\lambda^*$ is the complex conjugate of $\lambda$. Thus $\lambda$ has to be real.

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Thank you for the answer.Why Cannot A have a complex eigenvalues? – user6163 May 20 '11 at 15:12
Dear Nir, Hermitian matrices have purely real eigenvalues, that's why they're used by the physicists to express all real observables such as position or angular momentum. Compute $b(v,Av)$ for any normalized eigenvector $v$ - it's real because of hermiticity which makes it equal to $b(Av,v)=b(v,Av)^*$. This inner product is the eigenvalue, too. So it must be real. – Luboš Motl May 20 '11 at 15:14
Thank you Lubos. – user6163 May 20 '11 at 16:07
This is a special case of the spectral mapping theorem, see e.g.… – mac May 20 '11 at 17:32
@Calle: do you need the assumption that $A$ is diagonalizable? – Fabian May 21 '11 at 6:25

Define $$B= (I+A^2)^{-1}(I +i A).$$ Note that $I+A^2$ is invertible, as $\det(I+A^2) = \det(I+A^\dagger A) \geq 1$ (a sum of a positive definite and a positive semidefinite matrix is invertible). Then $$B(I-iA) = (I+A^2)^{-1}(I +i A)(I-iA) = (I+A^2)^{-1} (I+ A^2) =I,$$ so $B$ is the inverse of $A$.

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Just a comment, $A^*A$ could be positive semidefinite. But the result still holds. – Calle May 20 '11 at 17:24
@Calle: thank you, I have edited the post. – Fabian May 21 '11 at 6:24

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