Matrices as vectors

Before leaving the subject of mappings, recall that Linear(Space, Space) is a vector space itself, fo it ought to have a basis, inner products, a bilinear functional, a dual, and so on. A FourierFourier pair Pair has already been exhibited as such a basis, at least for any matrix with a complete set of eigenvectors. The basis which is most directly connected with Linear(Space, Space) is the collection of matrices $e_{ij}$, all of whose matrix elements are zero except for the one at the intersection of the $i^{th}$ row and $j^{th}$ column. The rule of multiplication is

$$\begin{eqnarray*} e_{ij} e_{kl} & = & \delta(j, k) e_{il}, \end{eqnarray*}$$

The expansion of a matrix in this basis is just

$$\begin{eqnarray*} M & = & \sum{ m_{ij} e_{ij} }. \end{eqnarray*}$$

The trace of a matrix is the sum of its diagonal elements, and is the coefficient of $(-\lambda)^{n-1}$ in its characteristic polynomial. Consider the bilinear mapping of two matrices to the scalar coefficients,

$$\begin{eqnarray*}[M, N]& = & Trace(M N). \end{eqnarray*}$$

It suffices to define a dual space, whilst the use of a transpose,

$$\begin{eqnarray*} (M, N) & = & [M^T, N], \end{eqnarray*}$$

not only provides a positive definite inner product, but tells us that

$$\begin{eqnarray*} (M, M) & = & \sum{m_{ij}^2} \end{eqnarray*}$$

which is eucliden length, squared, for a vector which is just a list of all the matrix elements. Another interpretation is that theis inner product is just the Gram Matrix of the columns of $M$, and that two matrices are orthogonal to one another if $Trace( M^TN) = 0$.

Why isn't the determinant of a matrix product a candidate for an inner product? Because it is not bilinear. Are other inner products feasible? Consider $Trace( M^TQN)$ for a positive definite matrix $Q$, but that is just like introducing a metric matrix into the ordinary inner product for vectors.

How do we get a basis for matrices compatible with the eigenvector basis for vectors? Consider the column-by=row products (column i)(row j); their multiplication table is similar to the multiplication table for standard vectors; the trace relationships are also verifiable.

How do we get a basis for matrices consisting of monomials $M^iN^j$ when $M$ and $N$ don't commute and so $MN$ isn't $NM$ and $MNM$ isn't $M^2N$, and so on? Evidently it is sufficient to know how to rewrite $NM$, even though the calculation could be tedious. The best thing is to let the eigenvectors speak for themselves, which they do through the diagonalizing matrices and the column by row idempotents.

Suppose that $M$ has a family of idempotents $G_i$ which are column by row products of eigenvectors, and that $N$ has a similar family $H_i$. That leaves us with four bases which can be collected into matrices: $U$ whose columns are the eigenvectors of $M$, $U^{-1}$ whose rows are also eigenvectors of $M$; similarly $V$ and $V^{-1}$ for $N$. All the column eigenvectors of $N$ are linear combinations of column eigenvectors of $M$ since $V = U U^{-1} V (U^{-1} V)$ stands to the right since the matrix product wants to combine columns of $U$ to get $V$). That means that the products $G_i H_j$ form a basis for square matrices, since they are just multiples of a column eigenvector of $M$ by a row eigenvector of $N$, and columns by row already created bases, both for $M$ and for $N$, just as they do in the standard basis. Also, $U^{-1} = U^{-1} V V^{-1}$, making a particular row eigenvector of $M$ satisfy $e_i U^{-1} = (e_i U^{-1}V) V^{-1}$.

The point of this exercise is that $M$ and $N$, as well as any other matrix, can be written in this mixed basis; in that form their powers and products can then be calculated in terms of the matrix $U V^{-1}$ and its relatives.

If we truly belive everything we have said, then we ought to find

$$\begin{eqnarray*} Q & = & \sum{ Trace(G_i H_j, Q) G_i H_j}. \end{eqnarray*}$$