Linear transformations, linear algebra

The last linear algebra article ended just right after we introduced the concept of a matrix and justified why the matrix vector multiplication has the form it has. Now, we extend it to matrix-matrix multiplication.

In the past article we discovered that it we can identify the vectors of one basis as a linear combination of vectors of another one (we always can do this), then we can quickly determine the representation of any vector in the new basis. If the vectors in the new basis (the \(\mathbf{e'}\) vectors in the last article) are represented as columns in terms of the old basis, then we have something like

\[\mathbf{x_{new}} = \begin{pmatrix} \vert & \vert & \ldots & \vert \\ \mathbf{e_{old\rightarrow new,1}} & \mathbf{e_{old\rightarrow new,2}} & \ldots & \mathbf{e_{old\rightarrow new,n}} \\ \vert & \vert & \ldots & \vert \\ \end{pmatrix}\mathbf{x_{old}}\]

Where each of the columns is representing the old basis vectors in the new one.

The way we defined matrix vector multiplication, this is exactly the same as a linear composition, where the multipliers for each \(\mathbf{e_{old\rightarrow new,i}}\) vector are the corresponding entries of \(\mathbf{x_{old}}\):

\[\mathbf{x_{new}} = \sum_{i=1}^n{\mathbf{e_{old\rightarrow new,i}}x_{old,i}}\]

Now, let’s consider this scenario: we already converted from one base to another but the application we are solving requires us to convert again. For example, imagine first translating the grid and then rotating it.

We need to find the coordinates of the new centers of the squares after both transformations. If we denote by \(A\) the matrix to convert the basis in the first transformation and by \(B\) the matrix for the second conversion, then the product \(AB\) is the product that goes directly from the original to the new basis.

The fact that we want this to be true regardless of any (linear) transformation is the only reason why matrix multiplication is defined the way it is: multiply every value in a row with a corresponding value in a column and sum all these products together to get the value at the intersection of the row and the column.