Engineering Analysis/Matrices

Derivatives

Consider the following set of linear equations:

$a=b{x}_1+c{x}_2$

$d=e{x}_1+f{x}_2$

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

$A=\left[\begin{array}{cc}b& c\\ e& f\end{array}\right]$

$B=\left[\begin{array}{c}a\\ d\end{array}\right]$

$x=\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]$

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 × 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:

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

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 $A^{†}$

Left Pseudo-Inverse

Consider the following matrix:

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

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 $A^{‡}$