To prove:
Suppose A is an n x n symmetric matrix with n distinct eigenvalues, λ1, ..., λn.
If x1;...,xn are (normalized) eigenvectors corresponding to the eigenvalues λ1;..., λn, then the matrix B, such that Bi, the ith column of
B, is the vector xi, is an orthogonal matrix.

Question

To prove:
Suppose A is an n x n symmetric matrix with n distinct eigenvalues, λ1, ..., λn. 
If x1;...,xn are (normalized) eigenvectors corresponding to the eigenvalues λ1;..., λn, then the matrix B, such that Bi, the ith column of
B, is the vector xi, is an orthogonal matrix.

anonymous · Answer

ok, are you still there?

anonymous · Answer

yup.

anonymous · Answer

A is symmetric so it is equal to its transpose

anonymous · Answer

the normalized vectors all have length 1

zarkon · Answer

$$Ax_i=\lambda_ix_i$$

$$x_j'Ax_i=x_j'\lambda_ix_i\,, \hspace{.5cm} i\neq j$$

$$(A'x_j)'x_i=\lambda_ix_j'x_i$$

$$(Ax_j)'x_i=\lambda_ix_j'x_i$$

$$(\lambda_jx_j)'x_i=\lambda_ix_j'x_i$$

$$\lambda_jx_j'x_i=\lambda_ix_j'x_i$$

$$\lambda_jx_j'x_i-\lambda_ix_j'x_i=0$$

$$(\lambda_j-\lambda_i)x_j'x_i=0\Rightarrow x_j'x_i=0\text{ since }\lambda_j\neq\lambda_i$$

anonymous · Answer

wow, how did he do that?

anonymous · Answer

Well, i guess thats a good answer.
Thank you both.