# Given a Hermitian matrix $A$, prove that $(A-iI)$ is nonsingular

The exercise is to prove that, given $$A$$ a Hermitian matrix, then $$(A-iI)$$ is nonsingular. I tried to think about what it meant to be nonsingular, like $$(A-iI)X=0$$ have not only the trivial solution, but was unable to prove it in any way.

• Hint: all the eigenvalues of such a matrix are real numbers. Aug 24 '19 at 20:46

Hint: The matrix $$A-i I$$ is singular iff $$i$$ is a singular value of $$A$$. Now, the spectral theorem tells us that all the singular values of a hermitian matrix are...

• I think eigenvalue is the correct term. Aug 25 '19 at 16:27

Since

$$A = A^\dagger, \tag 1$$

there exists a unitary matrix $$U$$,

$$UU^\dagger = U^\dagger U = I, \tag 2$$

such that

$$UAU^\dagger = \text{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n), \tag 3$$

where the $$\lambda_i$$ are the eigenvalues of $$A$$; of course (1) implies that

$$\lambda_i \in \Bbb R, \; 1 \le i \le n, \tag 4$$

as is well-known.

It follows that

$$U(A - iI)U^\dagger = UAU^\dagger - i UIU^\dagger = UAU^\dagger - iI$$ $$= \text{diag}(\lambda_1 - i, \lambda_2 - i, \ldots, \lambda_n - i); \tag 5$$

since the $$\lambda_i$$ are real,

$$\lambda_i - i \ne 0, \; 1 \le i \le n; \tag 6$$

thus the matrix $$U(A - iI)U^\dagger$$ is non-singular, hence so is

$$A - iI = U^\dagger \text{diag}(\lambda_1 - i, \lambda_2 - i, \ldots, \lambda_n - i) U. \tag 7$$

$$OE\Delta.$$