# How to prove that 2 vectors in C(A) corresponding to orthogonal vectors in row-space of A are orthgonal?

I was watching Singular Value Decomposition Lecture by Gilbert Strang. He takes two orthonormal vectors $v_1$, $v_2$. Let $\sigma_1$$u_1=Av_1 and \sigma_2$$u_2$=A$v_2$. He takes $u_1$ is orthogonal to $u_2$. How? How to prove this?

Edit1: Adding link to video lecture https://www.youtube.com/watch?v=Nx0lRBaXoz4 (Start from 3:30 mins)

• You're assuming $\sigma_1\ne \sigma_2$ here. Consider $\langle Mv_1,u_2\rangle$, and use $Mv_1=\sigma_1 u_1$ and $M^*u_2 = \sigma_2 v_2$. – Ted Shifrin May 12 '18 at 5:01
• @TedShifrin, I didn't get it. Can you elaborate? Also 1 more thing. Does $M^T$$u_2=v_2? – Nagabhushan S N May 12 '18 at 8:44 ## 2 Answers Recall that the SVD is given by A = U \Sigma V ^ T , hence A ^ T A = ( U \Sigma V ^ T) (U \Sigma V ^ T)^T = U \Sigma ^ 2U ^T. Namely, the columns of U are the eigenvectors of A ^ T A, and as A^TA is symmetric it is can be diagonilzed by an orthogonal matrix (U). • I ran into the problem while deriving SVD! – Nagabhushan S N May 13 '18 at 22:08 • u_1 and u_2 are the first and the second column of U, which is orthogonal, hence by definition u_1 ^ T u_2 = 0. And the proof (that U is orthogonal follows from my answer). – V. Vancak May 13 '18 at 22:12 • How do you say U is orthogonal? In your answer, you've used SVD. But as I said before, Gilbert Strang uses the fact that U is orthogonal to prove SVD. So, there should be another way to prove that U is orthogonal without using SVD. I'm stuck there. – Nagabhushan S N May 14 '18 at 14:29 I found an explanation here and here. It explains that we choose v_1 such that Av_1 is maximum. Then we choose v_2 such that v_2 \perp v_1 and Av_2 is maximum. In that case, the above referred answers prove that$$Av_1 \perp Av_2$$We choose$$u_1=Av_1/||Av_1||u_2=Av_2/||Av_2||$$Hence$$ u_1 \perp u_2$\$

I guess Gilbert Strang omitted/missed this explanation in that lecture.