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 × 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

$RA\neq I$

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

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

$AL\neq I$

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