This seems like a very basic question that I should know the answer to, but in my image processing class, my teacher explained that a basis set of images(matrices) are orthonormal. He said that the DOT product between two basis images (in this case two 2x2 matrices) is 0. so, for example \begin{equation} \begin{bmatrix} a & b\\ c & d \end{bmatrix} \cdot \begin{bmatrix} e & f\\ g & h \end{bmatrix} =0 \end{equation} I don't understand how this can be. I always thought it gave another matrix, and not a direct value: \begin{equation} \begin{bmatrix} a & b\\ c & d \end{bmatrix} \cdot \begin{bmatrix} e & f\\ g & h \end{bmatrix} = \begin{bmatrix} ae+bg & af+bh\\ ce+dg & cf+dh \end{bmatrix} \end{equation} Can someone help me out? It would be unbelieveably helpful, Thanks! Owen.
The only possibility I can think of is to take a 2x2 matrix and write it out in the form ##a e_{11} + b e_{12} + c e_{21} + d e_{22}##, ie as a four dimensional vector space. Then the e's form an orthonormal basis.
That's the matrix product, not the dot product. A dot product (inner product) is a scalar. Always. For matrices, the typical definition of the dot product is the Frobenius inner product. Simply compute as if the matrix was a vector. For real matrices, \begin{equation} A\cdot B \equiv \sum_i \sum_j A_{ij} B_{ij} \end{equation} For your pair of 2x2 matrices, \begin{equation} \begin{bmatrix} a & b\\ c & d \end{bmatrix} \cdot \begin{bmatrix} e & f\\ g & h \end{bmatrix} = ae + bf + cg + dh\end{equation}