Question

Let \(A(t)\) be an ( \(n \times n\) ) matrix function that is both differentiable and invertible on some \(t\)-interval of interest. It can be shown that \(A^{-1}(t)\) is likewise differentiable on this interval. Differentiate the matrix identity \(A^{-1}(t) A(t)=I\) to obtain the following formula: $$ \frac{d}{d t}\left[A^{-1}(t)\right]=-A^{-1}(t) A^{\prime}(t) A^{-1}(t) $$ [Hint: Recall the product rule, equation (9). Notice that the formula you derive is not the same as the power rule of single-variable calculus.]

Short answer

Question: Determine the formula for the derivative of the inverse of a matrix function, \(\frac{d}{dt}\left[A^{-1}(t)\right]\), given the matrix identity \(A^{-1}(t)A(t) = I\). Answer: \(\frac{d}{dt}\left[A^{-1}(t)\right] = -A^{-1}(t) \cdot \frac{d}{dt}[A(t)] \cdot A^{-1}(t)\)

Step-by-step solution

Differentiate the matrix identity

To find the derivative of \(A^{-1}(t) A(t)=I\), we will first need to differentiate both sides of the equation with respect to \(t\). Using the product rule for matrix functions, we will differentiate \(A^{-1}(t)A(t)\) on left hand side.

Apply the product rule

Applying the product rule, we get: \(\frac{d}{dt}\left[A^{-1}(t)A(t)\right] = \frac{d}{dt}[A^{-1}(t)] \cdot A(t) + A^{-1}(t) \cdot \frac{d}{dt}[A(t)]\) Since the right hand side of the matrix identity is \(I\), the derivative with respect to \(t\) will be a zero matrix of size (\(n \times n\)): \(\frac{d}{dt}[I] = 0\) So we have: \(\frac{d}{dt}[A^{-1}(t)] \cdot A(t) + A^{-1}(t) \cdot \frac{d}{dt}[A(t)] = 0\)

Isolate the term of interest

Now, our goal is to find an expression for \(\frac{d}{d t}\left[A^{-1}(t)\right]\). To do this, we will first multiply both sides of the equation by \(A^{-1}(t)\) from the right: \( \frac{d}{dt}[A^{-1}(t)] \cdot A(t) \cdot A^{-1}(t) + A^{-1}(t) \cdot \frac{d}{dt}[A(t)] \cdot A^{-1}(t) = 0 \cdot A^{-1}(t)\) Since \( A(t) \cdot A^{-1}(t) = I\), we get: \(\frac{d}{dt}[A^{-1}(t)] + A^{-1}(t) \cdot \frac{d}{dt}[A(t)] \cdot A^{-1}(t) = 0\)

Obtain the formula for \(\frac{d}{d t}\left[A^{-1}(t)\right]\)

Finally, we can write the formula for \(\frac{d}{d t}\left[A^{-1}(t)\right]\) by isolating the term we are interested in: \(\frac{d}{d t}\left[A^{-1}(t)\right] = -A^{-1}(t) \cdot \frac{d}{dt}[A(t)] \cdot A^{-1}(t)\) Thus, we have derived the formula for the derivative of the inverse of a matrix function with respect to \(t\).

Key concepts

Matrix Differentiation

Matrix differentiation extends the concept of differentiation from single variables to matrices. It's crucial when handling functions that output matrices based on inputs that can also be variables or matrices. Just like differentiating a regular function, we're interested in finding how changes in the input (e.g., time \( t \)) affect the output matrix.
For a matrix function like \( A(t) \), which changes with time, we calculate its derivative \( \frac{dA}{dt} \) by differentiating each element of \( A(t) \) individually. This element-wise differentiation means that each element in the resulting matrix \( \frac{dA}{dt} \) corresponds to its respective position in the original matrix \( A(t) \).
Often, understanding matrix differentiation requires familiarity with other matrix operations such as matrix addition, subtraction, and especially multiplication. Grappling with these fundamentals eases the journey into analyzing varying matrices in dynamic systems.

Inverse Function Theorem

The Inverse Function Theorem provides essential insights, especially when working with differentiable matrix functions. In essence, it assures us that if a function is invertible and differentiable at a specific point, then its inverse is also differentiable.
This stability of both invertibility and differentiability is pivotal when considering functions like \( A(t) \) and its inverse \( A^{-1}(t) \). If \( A(t) \) is continuously invertible and differentiable over an interval, the theorem guarantees that \( A^{-1}(t) \) is also differentiable.
This holds because the theorem not only checks that the operations are reversible at a point but extends its assurances over an interval where the conditions are consistently met. This concept significantly eases the process of deriving inverses when differentiating matrix identities such as \( A^{-1}(t)A(t)=I \). By ensuring differentiability, manipulating derivatives of inverse matrices remains orderly and consistent.

Product Rule for Matrices

Differentiating products of matrices is more complex than for single-variable calculus. Here, the product rule for matrices comes to the rescue. It's a tool we use when we need to differentiate products of matrices, such as \( A^{-1}(t)A(t) \).
When two functions are multiplied in single-variable calculus, the product rule states: \((uv)' = u'v + uv'\). For matrices, however, the rule adapts to the non-commutative nature of matrix multiplication:

• \( \frac{d}{dt}[A^{-1}(t)A(t)] = \frac{d}{dt}[A^{-1}(t)] \cdot A(t) + A^{-1}(t) \cdot \frac{d}{dt}[A(t)] \)

This expression shows that we consider the contribution of each product component's derivative, respecting the order of multiplication.
Thus, the product rule for matrices is crucial because it allows us to extract insights about how entire systems of equations or transformations evolve with respect to a parameter like time.

Matrix Algebra

Matrix algebra forms the backbone of working with matrices daily, encompassing operations like addition, subtraction, multiplication, and finding inverses.
A core aspect of matrix algebra involves understanding how matrices can interact with each other via multiplication. Matrix multiplication isn't commutative, meaning \( AB eq BA \) in most cases, and this non-commutative property shows up in differentiation tasks, making order vital in expressions like \( A(t) \cdot A^{-1}(t) \).
To dive deeper, one must become comfortable with identity matrices \( I \), where multiplication by \( I \) leaves a matrix unchanged, similar to multiplying by one in basic algebra. Another key operation is finding an inverse \( A^{-1} \), which reversibly "undoes" a matrix transformation.
Proper application of matrix algebra hinges on recognizing these elements and leveraging their properties in scenarios like deriving the formula for \( \frac{d}{dt}[A^{-1}(t)] \). In such derivations, understanding how these basic operations work together is fundamental to successfully maneuver through complex matrix equations.