Question

It can be shown that if \(Y\) is a differentiable and invertible square matrix function, then \(Y^{-1}\) is differentiable. (a) Show that \(\left(Y^{-1}\right)^{\prime}=-Y^{-1} Y^{\prime} Y^{-1}\), (Hint: Differentiate the identity \(Y^{-1} Y=I .\) ) (b) Find the derivative of \(Y^{-n}=\left(Y^{-1}\right)^{n},\) where \(n\) is a positive integer. (c) State how the results obtained in (a) and (b) are analogous to results from calculus concerning scalar functions.

Short answer

#Answer#: The derivative of the inverse of a square differentiable and invertible matrix Y is given by: \((Y^{-1})' = -Y^{-1}(Y')Y^{-1}\) The derivative of Y^(-n) for a positive integer n is given by: \(\frac{d}{dt}Y^{-n} = n\left[-Y^{-1}\left(\frac{d}{dt}Y\right)Y^{-1}\right](Y^{-1})^{n-1}\) These results are analogous to calculus concerning scalar functions, where the derivatives for functions like x^(-1) and x^(-n) also contain negative powers and a factor associated with the power.

Step-by-step solution

Differentiating the identity Y^-1 * Y = I

We are given the identity \(Y^{-1} Y=I\). To find the derivative of \(Y^{-1}\), we will differentiate both sides of the equation with respect to the variable (which might be time or another continuous variable). Using the chain rule, we get: \[\frac{d}{dt}\left(Y^{-1} Y\right) = \frac{d}{dt}(I) = 0\] Now, using the product rule for matrix functions, we get: \[\left(\frac{d}{dt}Y^{-1}\right)Y+Y^{-1}\left(\frac{d}{dt}Y\right)=0\]

Solving for (Y^-1)'

To find the derivative of \(Y^{-1}\), we will rearrange the equation we derived in Step 1: \[\left(\frac{d}{dt}Y^{-1}\right)Y = -Y^{-1}\left(\frac{d}{dt}Y\right)\] Now, we multiply both sides of the equation by \(Y^{-1}\) from the right to isolate the derivative of \(Y^{-1}\): \[\left(\frac{d}{dt}Y^{-1}\right) = -Y^{-1}\left(\frac{d}{dt}Y\right)Y^{-1}\]

Derivative of Y^(-n)

Now that we have the derivative of \(Y^{-1}\), we can find the derivative of \(Y^{-n} = \left(Y^{-1}\right)^n\). We can apply the chain rule for matrix functions and the result we derived in Step 2: \[\frac{d}{dt}Y^{-n} = n\left[Y^{-1}\frac{d}{dt}Y^{-1}\right](Y^{-1})^{n-1}\] Using the result from Step 2: \[\frac{d}{dt}Y^{-n} = n\left[-Y^{-1}\left(\frac{d}{dt}Y\right)Y^{-1}\right](Y^{-1})^{n-1}\]

Analogy with scalar functions in calculus

The results we obtained in parts (a) and (b) are analogous to the results from calculus concerning scalar functions. For instance, if we have a scalar function such as \(f(x)=x^{-1}\), its derivative is given by \(\frac{df}{dx}=-x^{-2}\). Similarly, if we have \(f(x)=x^{-n}\), its derivative is given by \(\frac{df}{dx}=-nx^{-n-1}\). These results are analogous to the results we obtained for the matrix functions in parts (a) and (b), where the derivatives include negative powers and a factor related to the power, respectively.

Key concepts

Differentiable Matrix Functions

When we explore the realm of matrix calculus, a concept that stands out is the 'differentiable matrix functions'. A matrix function is considered differentiable if, for a small change in the input, the output changes in a predictable way that can be approximated linearly. Similar to how a differentiable scalar function has a derivative at each point, a differentiable matrix function has a matrix of derivatives, commonly referred to as the Jacobian matrix.

To verify the differentiability of a matrix function, we often rely on the derivative of the identity matrix, as seen in the provided exercise. For a square matrix function Y, its inverse Y^-1 is also differentiable. This is illustrated by differentiating the identity Y^-1Y = I, which, through applying matrix calculus rules similar to regular calculus, allows us to find the derivative of the inverse function.

The ability to work with differentiable matrix functions expands our mathematical toolbox, making it possible to handle more complex multidimensional problems in subjects such as engineering, economics, and machine learning, where optimization and linear approximations are commonplace.

Product Rule for Matrices

In scalar calculus, the product rule is a fundamental property that allows us to differentiate expressions where two functions are multiplied together. The 'product rule for matrices' is an extension of this concept into the world of matrix calculus.

This rule states that if we have two differentiable matrix functions U(t) and V(t), their product's time derivative is given by

\[\frac{d}{dt}(U(t)V(t)) = \frac{dU(t)}{dt}V(t) + U(t)\frac{dV(t)}{dt}\]

In the exercise, applying the product rule was essential in deriving the derivative of the inverse of a matrix function Y^-1. Understanding and applying the product rule correctly is crucial as it ensures that the intricacies of matrix multiplication (which is not commutative) are accounted for, and provides a clear method for working through the differentiation of complex matrix expressions in various applications.

Chain Rule in Matrix Calculus

The chain rule is a powerful technique in regular calculus used to differentiate the composition of two or more functions. In matrix calculus, the 'chain rule' serves a similar purpose. It allows us to take derivatives of matrix functions that depend on other matrix functions.

When applying the chain rule in matrix calculus, one has to be mindful of the order of multiplication, as matrix multiplication is not commutative. For a matrix function Y that depends on a variable t, and another function Z(Y), the chain rule tells us that the derivative of Z with respect to t involves both the derivative of Z with respect to Y and the derivative of Y with respect to t.

The exercise showed this application in finding the derivative of Y^-n. By recognizing Y^-n as the composition of (Y^-1)ⁿ, one can use the chain rule in concert with the product rule to find the derivative efficiently. Mastering the chain rule in the matrix setting is vital in advanced fields such as control theory and econometrics where models are highly dependent on matrix functions.