# Orthogonality of eigenvectors of a Hermitian matrix [closed]

Are the eigenvectors of a Hermitian matrix(operator) always orthogonal?

If they are not always orthogonal, please explain when they are and when not.Thank you

• For the finite case always. It's the Spectral theorem. See Wiki. – Dog_69 May 27 '18 at 5:50
• Consider the identity operator. – WimC May 27 '18 at 5:51

Eigenvectors corresponding to distinct eigenvalues of a Hermitian matrix are always orthogonal, to wit:

Suppose

$$H^\dagger = H, \tag 0$$

and that

$$H \vec v_1 = \mu_1 \vec v_1, \tag 1$$

$$H \vec v_2 = \mu_2 \vec v_2; \tag 2$$

with

$$H \vec v = \mu \vec v, \; v \ne 0, \tag 3$$

we have

$$\mu \langle v, v \rangle = \langle v, \mu v \rangle = \langle v, H v \rangle = \langle H^\dagger v, v \rangle$$ $$= \langle Hv, v \rangle = \langle \mu v, v \rangle = \overline{\langle v, \mu v \rangle} = \bar \mu \overline{\langle v, v \rangle} = \bar \mu \langle v, v \rangle; \tag 4$$

since $$v \ne 0$$, $$\langle v, v \rangle \ne 0$$ so we may divide it out of (4) and see that

$$\mu = \bar \mu \tag 5$$

for any operator satisfying (0); this means that the eigenvalues of any operator satisfying (0) are real; therefore we may write

$$\mu_1 \langle v_1, v_2 \rangle = \langle \mu_1 v_1, v_2 \rangle = \langle H v_1, v_2 \rangle$$ $$= \langle v_1, H^\dagger v_2 \rangle = \langle v_1, H v_2 \rangle = \langle v_1, \mu_2 v_2 \rangle = \mu_2 \langle v_1, v_2 \rangle, \tag 6$$

or

$$(\mu_1 - \mu_2) \langle v_1, v_2 \rangle = 0, \tag 7$$

whence, assuming $$\mu_1 \ne \mu_2$$,

$$\langle v_1, v_2 \rangle = 0, \tag 8$$

and thus the vectors $$v_1$$, $$v_2$$ are orthogonal.

If $$\mu_1 = \mu_2$$ but $$v_1$$ and $$v_2$$ are linearly independent, then any vector in $$\text{span}\{v_1, v_2 \}$$ is an eigenvector for $$\mu$$:

$$H(av_ 1 + bv_2) = aHv_1 + bHv_2 = a\mu v_1 + b \mu v_2) = \mu(a v_1 + b v_2); \tag 9$$

in this case, we won't in general have $$\langle v_1, v_2 \rangle = 0$$, but we are free to choose $$v_1$$, $$v_2$$ suchly if we so desire; this is often the wise choice, since a set of orthogonal eigenvectors is often convenient for applications. But here the orthogonality is a choice, not a necessity.

The preceding remarks bind over any complex inner product space whenever (0), (1) and (2) hold; $$H$$ needn't even be bounded provided its action is restricted to a subspace for which (0), (1) and (2) are meaningful.

Note Addded in Edit, Tuesday 5 May 2020 3:06 PM PST: Данило Клименко asked, in a comment to this answer, if the analogous result holds for skew-Hermitian operators, that is for operators $$\Sigma$$ such that

$$\Sigma^\dagger = -\Sigma; \tag{10}$$

this query my be answered in the affirmative via an argument parallel to that presented above; we first show that with

$$\Sigma v = \sigma v, \tag{11}$$

$$\sigma$$ is purely imaginary. For in light of (11) we have

$$\sigma \langle v, v \rangle = \langle v, \sigma v \rangle = \langle v, \Sigma v \rangle = \langle \Sigma^\dagger v, v \rangle = \langle -\Sigma v, v \rangle = -\langle \Sigma v, v \rangle$$ $$= -\langle \sigma v, v \rangle = -\overline{\langle v, \sigma v \rangle} = -\bar \sigma \overline{\langle v, v \rangle} = -\bar \sigma \langle v, v \rangle; \tag{12}$$

with $$\langle v, v \rangle \ne 0$$ we find

$$\sigma = -\bar \sigma, \tag{13}$$

i.e., $$\sigma$$ is a purely imaginary number. Now if

$$\Sigma v_1 = \sigma_1 v_1 \tag{14}$$

and

$$\Sigma v_2 = \sigma_2 v_2, \tag{15}$$

in parallel with (6)-(8):

$$\sigma_2 \langle v_1, v_2 \rangle = \langle v_1, \sigma_2 v_2 \rangle$$ $$= \langle v_1, \Sigma v_2 \rangle = \langle \Sigma^\dagger v_1, v_2 \rangle = \langle -\Sigma v_1, v_2 \rangle$$ $$= \langle -\sigma_1 v_1, v_2 \rangle = -\langle \sigma_1 v_1, v_2 \rangle = -\bar \sigma_1 \langle v_1, v_2 \rangle = \sigma _1 \langle v_1, v_2 \rangle\tag{16}$$

or

$$(\sigma_1 - \sigma_2)\langle v_1, v_2 \rangle = 0, \tag{17}$$

and now if we assume

$$\sigma_1 \ne \sigma_2 \tag{18}$$

we conclude that

$$\langle v_1, v_2 \rangle = 0, \tag{19}$$

the desired result; that is, eigenvectors corresponding to distinct eigenvalues of skew-Hermitian operators are in fact orthogonal.

This may in fact be see directly from the above ((0)-(9)) discussion concerning Hermitian operators if we observe that (10) yields

$$(i\Sigma)^\dagger = \bar i \Sigma^\dagger = -i(-\Sigma) = i\Sigma, \tag{20}$$

that is, $$i\Sigma$$ is Hermitian; then (14) and (15) imply

$$i\Sigma v_1 = i\sigma_1 v_1, \tag{21}$$

and

$$i\Sigma v_2 = i\sigma_2 v_2, \tag{22}$$

and thus invoking our previous result we infer that

$$\langle v_1, v_2 \rangle = 0 \tag{23}$$

provided $$\sigma_1 \ne \sigma_2$$.

Lastly, the analog of (9) binds; indeed, (9) and its surrounding text apply for an operator $$H$$; the assertion is not restricted to Hermitian, skew-Hermitian, or any other class of matrices or operators. End of Note.

• does it work for skew-Hermitian matrix? – Dan Klymenko May 5 '20 at 12:55
• @ДанилоКлименко: I modified my answer to address your question. Cheers! – Robert Lewis May 5 '20 at 23:55

If $Ax = \lambda x$ and $Ay = \mu y$ then $$\lambda\langle x,y\rangle = \langle Ax,y\rangle = \langle x,Ay\rangle = \mu\langle x,y\rangle$$ which leads for $\lambda \neq \mu$ to $\langle x,y\rangle = 0$. (Remember that eigenvalues of the Hermitian operator are always real).

• very concise, but it requires that no two eigenvalues are equal to each other. – Quantum Guy 123 May 12 at 16:40