Using Gram-Schmidt to orthogonalize Matrix Bases

Question

$\newcommand\m[1]{\begin{pmatrix}#1\end{pmatrix}}$ I have the billinear form $f(A,B) = trace(A*B)$ on $V = Mat(2x2,R)$. If I just start from the standard basis of $V$ then I end up with my second matrix in the Gram-Schmidt process being $v_2$ Which is clearly not orthogonal. What am I doing wrong? $v_1 = \m{1&0\\0&0} \\ v_2 = \m{0&1\\0&0}-\frac{f(v_1,v_2)}{f(v_1,v_1)}*\m{1&0\\0&0} = \m{0&1\\0&0}$

Answer 1

Your bilinear form should be: $f(A,B)= \mbox{tr} \; A^T B = \sum_{i,j} a_{ij}b_{ij}$. Note that writing a 2 by 2 matrix as a 4-vector: $X_A=(a_{11}, a_{12},a_{21},a_{22})$ your (modified) bilinear form is simply the usual scalar product of $X_A$ with $X_B$. In your example, the standard matrix basis is in fact orthonormal for this scalar product.