Question

Show that if \(\mathrm{A}\) and \(\mathrm{B}\) are matrices which don't commute, then \(e^{\mathrm{A}+\mathrm{B}} \neq e^{\mathrm{A}} e^{\mathrm{B}},\) but if they do commute then the relation holds. Hint: Write out several terms of the infinite series for \(e^{\mathrm{A}}, e^{\mathrm{B}},\) and \(e^{\mathrm{A}+\mathrm{B}}\) and do the multiplications carefully assuming that \(A\) and \(B\) don't commute. Then see what happens if they do commute.

Short answer

If \( \text{A}\) and \( \text{B} \) don't commute, \( e^{\text{A} + \text{B}} eq e^{\text{A}} e^{\text{B}} \. If they do commute, \( e^{\text{A} + \text{B}} = e^{\text{A}} e^{\text{B}} \.

Step-by-step solution

- Understand Exponential of Matrices

The matrix exponential of a matrix \(\text{A}\) is defined as \[ e^{\text{A}} = \text{I} + \text{A} + \frac{\text{A}^2}{2!} + \frac{\text{A}^3}{3!} + \ldots \] where \(\text{I}\) is the identity matrix.

- Write Infinite Series for \(e^{\text{A}}\), \(e^{\text{B}}\), and \(e^{\text{A} + \text{B}}\)

Express the series for matrix exponentials: \[ e^{\text{A}} = \text{I} + \text{A} + \frac{\text{A}^2}{2!} + \ldots \] \[ e^{\text{B}} = \text{I} + \text{B} + \frac{\text{B}^2}{2!} + \ldots \] \[ e^{\text{A} + \text{B}} = \text{I} + (\text{A} + \text{B}) + \frac{(\text{A} + \text{B})^2}{2!} + \ldots \]

- Simplify \(e^{\text{A}+\text{B}}\)

Expand \((\text{A} + \text{B})^2\) to see the non-commuting property: \[ (\text{A} + \text{B})^2 = \text{A}^2 + \text{A}\text{B} + \text{B}\text{A} + \text{B}^2 \] If \(\text{A}\) and \(\text{B}\) don't commute, \(\text{AB} eq \text{BA}\).

- Compare Terms

Compare the series expansion of \(e^{\text{A}} e^{\text{B}}\) with \(e^{\text{A} + \text{B}}\). For non-commuting matrices, \[ e^{\text{A}} e^{\text{B}} = ( \text{I} + \text{A} + \frac{\text{A}^2}{2!} + \ldots )( \text{I} + \text{B} + \frac{\text{B}^2}{2!} + \ldots ) \] contains mixed terms such as \( \text{AB} + \text{BA} \). But in \( e^{\text{A} + \text{B}}\), these terms appear as \( \text{A}\text{B} \) and \( \text{B}\text{A} \) which will not cancel each other out.

- Show Commutative Case

If \(\text{A}\) and \(\text{B}\) commute (i.e., \(\text{AB} = \text{BA}\)), it simplifies the expansion: \((\text{A} + \text{B})^n = \text{A}^n + \binom{n}{1} \text{A}^{n-1} \text{B} + \ldots \) which means \[ e^{\text{A} + \text{B}} = e^{\text{A}} e^{\text{B}} \]

Key concepts

Non-Commutative Matrices

Matrices can be non-commutative, meaning that for two matrices \(\text{A}\) and \(\text{B}\), it is possible that \(\text{A}\text{B} eq \text{B}\text{A}\). This non-commutativity plays a crucial role in matrix exponentials. In many cases, the order in which matrices are multiplied affects the result, unlike simple numbers where multiplication is commutative (e.g., 3 \times\ 4 is the same as 4 \times\ 3).
Non-commutativity creates unique complexities, especially in areas such as quantum mechanics and linear algebra.
Let's see how non-commutative behavior can influence the computation of matrix exponentials.

Matrix Multiplication

Matrix multiplication involves combining two matrices to produce another matrix. To multiply two matrices \(\text{A}\) and \(\text{B}\), you take the dot product of the rows of \(\text{A}\) with the columns of \(\text{B}\).
Here's the general formula for matrix multiplication:
Continuing the pattern for higher powers of matrix exponentials, \[(\text{A} + \text{B})^2 = \text{A}^2 + \text{A}\text{B} + \text{B}\text{A} + \text{B}^2\]
If \(\text{A}\) and \(\text{B}\) don't commute, notice that the cross-terms \(\text{A}\text{B}\) and \(\text{B}\text{A}\) are different. Consequently, expanding \(\text{e}^{\text{A} + \text{B}}\) will include mixed terms different from individual expansions of \(\text{e}^{\text{A}}\) and \(\text{e}^{\text{B}}\).
This non-commutative property makes matrix multiplication interesting and complex. It is especially crucial in understanding matrix exponentials.

Series Expansion

Matrix exponentials can be expressed using a series expansion. For a matrix \(\text{A}\), the exponential is given by:\[\text{e}^{\text{A}} = \text{I} + \text{A} + \frac{\text{A}^2}{2!} + \frac{\text{A}^3}{3!} + \text{...}\]
This expansion is infinite, similar to the Taylor series for the exponential function of a scalar.
When dealing with two matrices \(\text{A}\) and \(\text{B}\), the series expansion for \(\text{e}^{\text{A} + \text{B}}\) involves terms from both matrices. If \(\text{A}\) and \(\text{B}\) commute, simplifying the series becomes easier and the relationship holds: \(\text{e}^{\text{A} + \text{B}} = \text{e}^{\text{A}} \text{e}^{\text{B}}\).
If they don't commute, the series terms involving \(\text{A}\text{B}\) and \(\text{B}\text{A}\) do not cancel each other, resulting in a different product. This is a key concept when dealing with matrix exponentials in linear algebra and various applications.