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

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..... May 30 at 6:44
• 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 $(\lambda_1 - \lambda_2)$. May 30 at 7:14
• 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) May 30 at 7:41

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 $$(Aw_1) . (Aw_2) = 0.$$