How would you show it? I went all over the book: looked at Linear Transformation definition again, looked at orthogonality of bases for symmetric matrices.
I got to the point:
Since L is linear transformation it must have coefficient matrix, lets call it B.
But I got stuck here. If B were symmetric, I would say that B^T = B and then it has an eigenvalue and therefore I can pull eigenvalue out.
Does anybody know how to solve the problem? Thank you very much.
P.S. Sorry, didn't have time to learn to format equations here, so I did it in Word and past it in.
Note that $B$ is a $n \times m$ matrix, so it is not square unless $m = n$. However, the product $B^T B$ is a square matrix and, as stated in the hint, it is symmetric: $$(B^T B)^T = B^T (B^T)^T = B^T B$$ Consequently, there exists an orthonormal basis $\{v_1,\dots,v_m\}$ of $\mathbb{R}^m$ and real numbers $\lambda_1,\dots,\lambda_m$ such that $B^T B v_i = \lambda_i v_i$ ($1 \le i \le m$). This implies that $$L(v_i) \cdot L(v_j) = v_i^T B^T B v_j = v_i^T (\lambda_j v_j) = \lambda_j (v_i^T v_j) = 0$$ whenever $i \neq j$.