Question

Suppose \(Y\) is a differentiable square matrix. (a) Find a formula for the derivative of \(Y^{2}\). (b) Find a formula for the derivative of \(Y^{n},\) where \(n\) is any positive integer. (c) State how the results obtained in (a) and (b) are analogous to results from calculus concerning scalar functions.

Short answer

Short Answer: The derivative of a square matrix \(Y^n\) can be calculated using the formula \(\frac{d}{dt}(Y^n) = \sum_{i=1}^{n} Y^{i-1}Y'(t)Y^{n-i}\). This result is analogous to the product and power rules in scalar calculus. In the matrix case, however, we have to consider all possible ways to distribute the factor of \(Y'(t)\) in the product \(Y^n\).

Step-by-step solution

(a) Derivative of \(Y^2\)

We are given \(Y\) as a differentiable square matrix, and we must find the derivative of \(Y^2\). To do this, recall the product rule in calculus, which states that \({(fg)' = f'g + fg'}\). Here, \(f\) and \(g\) are functions. We need to find the derivative of \(Y^2 = YY\). Let \(Y'(t)\) denote the derivative of the matrix \(Y(t)\). $$ \frac{d}{dt}(Y^2) = \frac{d}{dt}(YY) = Y'(t)Y(t) + Y(t)Y'(t) $$ Thus, the derivative of \(Y^2\) is given by the formula \(Y'(t)Y(t) + Y(t)Y'(t)\).

(b) Derivative of \(Y^n\)

Now, let's find the formula for the derivative of \(Y^n\), where \(n\) is any positive integer. We'll do this by induction. Base case - When \(n=1\): $$ \frac{d}{dt}(Y^1) = \frac{d}{dt}(Y) = Y'(t) $$ Assume the formula holds for \(n=k\): $$ \frac{d}{dt}(Y^k) = \sum_{i=1}^{k} Y^{i-1}Y'(t)Y^{k-i} $$ Now, let's prove the formula holds for \(n=k+1\): $$ \frac{d}{dt}(Y^{k+1}) = \frac{d}{dt}(Y^k Y) \\ = \bigg(\sum_{i=1}^{k} Y^{i-1}Y'(t)Y^{k-i}\bigg) Y + Y^k Y'(t) \quad (\text{by the induction hypothesis}) \\ = \sum_{i=1}^{k} Y^{i-1}Y'(t)Y^{k-i+1} + Y^k Y'(t) \\ = \sum_{i=1}^{k+1} Y^{i-1}Y'(t)Y^{(k+1)-i} $$ So the formula for the derivative of \(Y^{n}\) is: $$ \frac{d}{dt}(Y^n) = \sum_{i=1}^{n} Y^{i-1}Y'(t)Y^{n-i} $$

(c) Analogy with scalar calculus

The results we derived in (a) and (b) are analogous to results from scalar calculus in the following ways: 1. The formula for the derivative of \(Y^2 = Y'(t)Y(t) + Y(t)Y'(t)\) is analogous to the product rule for scalar functions, \((fg)' = f'g + fg'\). 2. The formula for the derivative of \(Y^n = \sum_{i=1}^{n} Y^{i-1}Y'(t)Y^{n-i}\) is analogous to the power rule in scalar calculus, where the derivative of \(x^n\) is \(nx^{n-1}\) for a scalar function. In the matrix case, we have to consider all possible ways to distribute the factor of \(Y'(t)\) in the product \(Y^n\).

Key concepts

Differentiation of Matrices

Differentiating matrices involves understanding how changes in a matrix affect its derivatives over time. When dealing with matrices, we handle arrays of functions rather than single scalar functions. So, instead of straightforward differentiation, we employ techniques similar to those used in vector calculus. Although this might seem complex, it's manageable once you understand the core principles.

To differentiate a matrix, consider each element of the matrix as a function of time or some other variable. The derivative of the entire matrix is simply the matrix formed by differentiating each of its components. This means the differentiation process for matrices is an element-wise operation, just as it is for vectors in calculus.

This provides a foundation to apply further rules and simplify differentiation tasks involving matrix expressions, particularly when combined with the following concepts, such as product and power rules for matrices.

Matrix Product Rule

The matrix product rule is a generalization of the product rule from scalar calculus, which many are familiar with. When we deal with the product of two matrices, the differentiation must account for the matrix nature of the variables. In scalar calculus, the product rule is given by \( (fg)' = f'g + fg' \).

For matrices, if you have two matrices \(A(t)\) and \(B(t)\), the product rule extends to: \

• \(\frac{d}{dt}(AB) = A'B + AB'\)

This formula means you first differentiate the first matrix and multiply with the second, then add to that the product of the original first matrix and the derivative of the second.

Importantly, matrix multiplication is not commutative, so \(AB\) does not equal \(BA\). This non-commutative property requires careful attention in order to apply the correct differentiation sequence and maintain the correct order of matrix operations.

Matrix Power Differentiation

Matrix power differentiation is useful for finding derivatives of powers of a matrix, like \(Y^n\), where \(Y\) is a matrix and \(n\) is a positive integer. The intuitive method we use in scalar calculus does not directly apply here due to matrices being non-commutative.

The solution involves a formula reminiscent of the binomial expansion, constructed by induction. For example, the derivative of \(Y^n\) is given by the sum

• \(\frac{d}{dt}(Y^n) = \sum_{i=1}^{n} Y^{i-1}Y'(t)Y^{n-i}\)

This expression makes sense when you consider distributing the derivative function \(Y'(t)\) across different positions in the product expansion of \(Y^n\).

Think of it as counting all possible ways the derivative of \(Y\) can be slotted into the expression \(Y^n\). Each slotting position corresponds to a different term in the summation. This strategy ensures we capture all necessary derivatives to calculate the change in matrix power accurately.