If A is a symmetric $n\times n$ matrix, prove that the eigenvectors associated to distinct eigenvalues are orthogonal

Question

This one is a little confusing to me, but I'll show what I've done so far.

I know that a symmetric matrix is a matrix that is equal to its transpose, like the identity matrix.

I also know that for a vector to be an eigenvector of some matrix $A$, the following must be true

$Av = \lambda v$,

and for the eigenvectors to be orthogonal, their dot product must be $0$.

Does this mean that the only eigenvectors for a symmetric $n\times n$ matrix are the zero vectors?

Any help will be appreciated.

Hint: Let $v_1$ and $v_2$ be eigenvectors corresponding to eigenvalues $\lambda_1$ and $\lambda_2$ respectively. Now consider the product $v_1^TAv_2$. — Malcolm, Dec 17 '17 at 01:44
Could you explain why you think this? The $n \times n$ identity matrix is symmetric and every nonzero $x \in \mathbb{R}^n$ is an eigenvector. — MathematicsStudent1122, Dec 17 '17 at 01:44
speeeeectral theorem. you can find an orthonormal basis of eigenvectors for $\mathbb{R}^n$ (or $\mathbb{C}^n$) from a symmetric/hermitian matrix $A$ such that eigenvectors corresponding to distinct eigenvalues are orthogonal. — Batman, Dec 17 '17 at 01:46
I'm still confused, maybe i need another hint or another explanation. — Soon_to_be_code_master, Dec 17 '17 at 02:31
Does this answer your question? Eigenvectors of real symmetric matrices are orthogonal — , Jul 25 '20 at 22:53

Robert Lewis · Accepted Answer · 2018-01-01T20:28:07.263

1

Suppose $\vec x_1$ and $\vec x_2$ are eigenvectors of the matrix $A$, corresponding to eigenvalues $\mu_1 \ne \mu_2$; then

$A \vec x_1 = \mu_1 \vec x_1, \tag 1$

and

$A \vec x_2 = \mu_2 \vec x_2; \tag 2$

then, using the angle-bracket notation for the inner or "dot" product, we have

$\mu_1 \langle \vec x_1, \vec x_2 \rangle = \langle \mu_1 \vec x_1, \vec x_2 \rangle = \langle A \vec x_1, \vec x_2 \rangle$ $= \langle \vec x_1, A^T \vec x_2 \rangle = \langle \vec x_1, A \vec x_2 \rangle = \langle \vec x_1, \mu_2 \vec x_2 \rangle = \mu_2 \langle \vec x_1, \vec x_2 \rangle; \tag 3$

thus,

$(\mu_1 - \mu_2) \langle \vec x_1, \vec x_2 \rangle = 0; \tag 4$

since $\mu_1 \ne \mu_2$, $\mu_1 - \mu_2 \ne 0$, forcing

$\langle \vec x_1, \vec x_2 \rangle = 0. \tag 5$

edited Jan 01 '18 at 20:28

answered Dec 17 '17 at 02:49

Robert Lewis

71,180

I have a quick question here, On step (2) why did you use the equation $Ax_2 = u_2x_1$? why didn't you use the equation $Ax_2 = u_2x_2$ – Soon_to_be_code_master Jan 01 '18 at 20:03
@Soon_to_be_code_master: Typo. Will correct. Thanks. – Robert Lewis Jan 01 '18 at 20:27

If A is a symmetric $n\times n$ matrix, prove that the eigenvectors associated to distinct eigenvalues are orthogonal

1 Answers1