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.

  • 2
    $\begingroup$ Hint: all the eigenvalues of such a matrix are real numbers. $\endgroup$
    – DonAntonio
    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...

  • $\begingroup$ I think eigenvalue is the correct term. $\endgroup$
    – Heedong Do
    Aug 25 '19 at 16:27


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

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

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



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