Question

If \(A_{1}\) and \(A_{2}\) commute (i.e. \(\left.A_{1} A_{2}=A_{2} A_{1}\right)\), then \(e^{\left(A_{1}+A_{2}\right) t}=\) \(e^{A_{1} t} \cdot e^{A_{2} t} .\) Prove this. Give a counterexample to this equality if \(A_{1}\) and \(A_{2}\) do not commute.

Short answer

If matrices commute, the exponential of their sum equals the product of exponentials. Non-commuting matrices provide a counterexample.

Step-by-step solution

Understanding Matrix Exponentials

First, we need to understand what it means to take the exponential of a matrix. The exponential of a matrix \(A\), written as \(e^{At}\), is defined as the power series \( I + At + \frac{A^2 t^2}{2!} + \frac{A^3 t^3}{3!} + \ldots \). This is similar to the exponential function for real numbers but extended to matrices.

Apply Properties of Commuting Matrices

When two matrices commute, their order of multiplication does not matter, i.e., \(A_1 A_2 = A_2 A_1\). Due to this property, when expanding both \(e^{(A_{1}+A_{2})t}\) and \(e^{A_1 t} \cdot e^{A_2 t}\) as power series and multiplying powers, the terms will align correctly, showing \(e^{(A_1+A_2)t} = e^{A_1 t} \cdot e^{A_2 t}\).

Power Series Expansion of Each Exponential

Expand each exponential in its respective series form. For \( e^{(A_1 + A_2)t} \), it's \( I + (A_1 + A_2)t + \frac{(A_1 + A_2)^2 t^2}{2!} + \cdots \). For \( e^{A_1 t} \cdot e^{A_2 t} \), use \( (I + A_1 t + \frac{A_1^2 t^2}{2!} + \cdots)(I + A_2 t + \frac{A_2^2 t^2}{2!} + \cdots) \). The commutativity simplifies the expansion, showing they are equal.

Counterexample for Non-Commuting Matrices

Consider two 2x2 matrices that do not commute. Take \( A_1 = \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix} \) and \( A_2 = \begin{bmatrix} 0 & 0 \ 1 & 0 \end{bmatrix} \). Calculate \( A_1 A_2 \) and \( A_2 A_1 \), showing they are not the same. Then, calculate the exponentials and show that \( e^{(A_1+A_2)t} eq e^{A_1 t} \cdot e^{A_2 t} \) due to the lack of commutativity.

Key concepts

Commutative Matrices

Matrices are commutative when their multiplication is independent of the order, meaning if you take two matrices \(A_1\) and \(A_2\), they commute if \(A_1A_2 = A_2A_1\). This is a crucial property because it simplifies calculations in linear algebra. Commutative matrices enjoy some nice mathematical properties which can be useful in various applications like quantum mechanics and systems theory.

When matrices are commutative, you can break down some complex matrix operations into simpler steps. For example:

• Combining their exponentials is possible. It becomes easy to prove that the combined exponential of two matrices \(e^{(A_1+A_2)t}\) equals the product of their individual exponentials \(e^{A_1t} \cdot e^{A_2t}\).
• The commutative property helps align terms correctly when you use the power series method to calculate matrix exponentials.

This property is fundamental in understanding why, when matrices commute, we can treat the exponential operations nearly as comfortably as we do with numbers.

Power Series Expansion

A significant technique in mathematics is the use of power series expansions. This is especially useful when extending functions from the realm of real numbers to matrices. The matrix exponential \(e^{At}\) is defined with a power series, similar to how we calculate the exponential for scalar values.

The power series allows us to write \(e^{At}\) as \[I + At + \frac{A^2 t^2}{2!} + \frac{A^3 t^3}{3!} + \cdots.\]Here, \(I\) is the identity matrix. Each term in the series becomes progressively smaller for small \(t\), mimicking the exponential growth with matrices.

This method breaks down complex matrix operations into manageable algebraic tasks:

• When matrices commute, substituting in their expansions shows clear alignment.
• This alignment makes it possible to combine separate matrices into one, proving equations like \(e^{(A_1+A_2)t} = e^{A_1t} \cdot e^{A_2t}\) for commuting matrices.

The power series offers a bridge between simple algebra and complex matrix dynamics.

Non-Commuting Matrices

When matrices do not commute, all bets are off in terms of certain simplifications. Non-commuting matrices mean reorder them and you get a different product, \(A_1A_2 eq A_2A_1\). This complicates many operations in linear algebra, especially when dealing with matrix exponentials.

For non-commuting matrices:

• You cannot simply separate products of their exponentials like \(e^{(A_1+A_2)t} eq e^{A_1t} \cdot e^{A_2t}\).
• Calculations require careful treatment of each matrix’s interaction with one another.

A simple example with 2x2 matrices can demonstrate this. Consider:
\[A_1 = \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix}, \quad A_2 = \begin{bmatrix} 0 & 0 \ 1 & 0 \end{bmatrix}.\]Here, \(A_1A_2\) is not the same as \(A_2A_1\). Calculating their exponentials separately, then trying to multiply them, will not yield the same result as calculating \(e^{(A_1+A_2)t}\) directly. This illustrates how sensitive matrix operations can be when commutativity isn’t present, demonstrating why each matrix's behavior must be individually considered in non-commuting scenarios.