Engineering Analysis/Matrices

Derivatives
Consider the following set of linear equations:


 * $$a = bx_1 + cx_2$$
 * $$d = ex_1 + fx_2$$

We can define the matrix A to represent the coefficients, the vector B as the results, and the vector x as the variables:


 * $$A = \begin{bmatrix}b & c \\ e & f\end{bmatrix}$$
 * $$B = \begin{bmatrix}a \\ d\end{bmatrix}$$
 * $$x = \begin{bmatrix}x_1 \\ x_2\end{bmatrix}$$

And rewriting the equation in terms of the matrices, we get:


 * $$B = Ax$$

Now, let's say we want the derivative of this equation with respect to the vector x:


 * $$\frac{d}{dx}B = \frac{d}{dx}Ax$$

We know that the first term is constant, so the derivative of the left-hand side of the equation is zero. Analyzing the right side shows us:

Pseudo-Inverses
There are special matrices known as pseudo-inverses, that satisfies some of the properties of an inverse, but not others. To recap, If we have two square matrices A and B, that are both n &times; n, then if the following equation is true, we say that A is the inverse of B, and B is the inverse of A:


 * $$AB = BA = I$$

Right Pseudo-Inverse
Consider the following matrix:


 * $$R = A^T[AA^T]^{-1}$$

We call this matrix R the right pseudo-inverse of A, because:


 * $$AR = I$$

but


 * $$RA \ne I$$

We will denote the right pseudo-inverse of A as $$A^\dagger$$

Left Pseudo-Inverse
Consider the following matrix:


 * $$L = [A^TA]^{-1}A^T$$

We call L the left pseudo-inverse of A because


 * $$LA = I$$

but


 * $$AL \ne I$$

We will denote the left pseudo-inverse of A as $$A^\ddagger$$