If $w_1, w_2$ are eigenvectors of $A^TA$ associated with different eigenvalues, then $Aw_1$ is orthogonal to $Aw_2$.

Question

if $w_1, w_2$ are eigenvectors of $A^TA$ associated with different eigenvalues. How to show $Aw_1$ is orthogonal to $Aw_2$

I would like to use the SVD factorization to prove this statement. Is it correct to say if $w_1,w_2$ are eigenvectors of $A^TA$ then exists a SVD $A=USV^T$ such that $w_1,w_2$ are normalized in any of the columns of $V$ ?

Note that $A^TA$ is a symmetric matrix so spectral theorem gives you..... — Anurag A, May 30 '23 at 06:44
If you would consider proving this without using SVD, then a (vague) hint would be to write down the definition of orthogonality between $Aw_1$ and $Aw_2$ and manipulate the expression so you get something involving $(\lambda_1 - \lambda_2)$. — Chee Han, May 30 '23 at 07:14
Does this answer your question? Prove that the multiples of two orthogonal eigenvectors with a matrix are also orthogonal (together with this wellknown fact alluded to by @Anurag) — Anne Bauval, May 30 '23 at 07:41

score 2 · Accepted Answer · answered May 30 '23 at 07:24

To spell it out, suppose that $w_1$ and $w_2$ are eigenvectors of $A^T A$, and suppose that the corresponding eigenvalues are $\lambda_1$ and $\lambda_2$, with $\lambda_1 \neq \lambda_2$.

Then $$ \lambda_1 w_1^T w_2 = \left( A^T A w_1 \right)^T w_2 = w_1^T A^T A w_2 = w_1^T \left( A^T A w_2 \right) = \lambda_2 w_1^T w_2.$$ Since $\lambda_1 \neq \lambda_2$, we must have $$ w_1^T w_2 = 0.$$

Now $$ (Aw_1) . (Aw_2) = (Aw_1)^T (Aw_2) = w_1^T A^T A w_2,$$ but by the calculation above, we have $$ w_1^T A^T A w_2 = \lambda_1 w_1^T w_2 = 0, $$ so $$ (Aw_1) . (Aw_2) = 0.$$

If $w_1, w_2$ are eigenvectors of $A^TA$ associated with different eigenvalues, then $Aw_1$ is orthogonal to $Aw_2$.

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