Question

Let \(A\) and \(B\) be square matrices of the same size. a. Show that \((A B)^{2}=A^{2} B^{2}\) if \(A B=B A\). b. If \(A\) and \(B\) are invertible and \((A B)^{2}=A^{2} B^{2},\) show that \(A B=B A\). c. If \(A=\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right]\) and \(B=\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right],\) show that \((A B)^{2}=A^{2} B^{2}\) but \(A B \neq B A\).

Short answer

(a) If \(AB=BA\), then \((AB)^2=A^2B^2\). (b) Invertible and \((AB)^2=A^2B^2\) leads to \(AB=BA\). (c) For given \(A, B\), \((AB)^2=A^2B^2\), but \(AB \neq BA\).

Step-by-step solution

Use the Commutative Property for Matrices

To solve part (a), we start with the assumption that matrices \(A\) and \(B\) commute, that is, \(AB = BA\). To show \((AB)^2 = A^2 B^2\), we first expand \((AB)^2\):\[(AB)^2 = (AB)(AB) = A(BA)B.\]Since \(AB = BA\), we have \(A(BA)B = A(AB)B = A^2 B^2\). Therefore, if \(AB = BA\), then \((AB)^2 = A^2 B^2\).

Assume and Use Invertibility

For part (b), assume that matrices \(A\) and \(B\) are invertible and that \((AB)^2 = A^2 B^2\). Expand and transform both sides:\[(AB)(AB) = A(BA)B.\]Using invertibility, multiply both sides on the left by \(A^{-1}\) and on the right by \(B^{-1}\):\[A^{-1}(AB)(AB)B^{-1} = A^{-1}A(BA)BB^{-1}.\]This simplifies to:\[(AB) = (BA),\]which implies \(AB = BA\). Thus, if \(A\) and \(B\) are invertible and \((AB)^2 = A^2 B^2\), then \(AB = BA\).

Compute AB and BA for Given Matrices

For part (c), compute \(AB\) and \(BA\) using the given matrices \(A\) and \(B\):\[A = \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 1 \ 0 & 0 \end{bmatrix}\]\[AB = \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 \ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \ 0 & 0 \end{bmatrix},\]\[BA = \begin{bmatrix} 1 & 1 \ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix}.\]

Verifying (AB)^2 and A^2B^2

Next, compute \((AB)^2\) and \(A^2B^2\) for matrices \(A\) and \(B\):\[(AB)^2 = \begin{bmatrix} 1 & 1 \ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 \ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \ 0 & 0 \end{bmatrix},\]\[A^2 = \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix},\]\[B^2 = \begin{bmatrix} 1 & 1 \ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 \ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \ 0 & 0 \end{bmatrix},\]\[A^2B^2 = \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 \ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \ 0 & 0 \end{bmatrix}.\]This verifies that \((AB)^2 = A^2B^2\) and also shows \(AB eq BA\).

Key concepts

Commutative Property

Understanding the commutative property is important in mathematics, particularly in algebra. It describes a situation where the order of operation does not affect the result. For instance, in simple arithmetic, addition and multiplication follow this property. Numbers like 3 and 5 satisfy this property in addition and multiplication because 3 + 5 equals 5 + 3, and similarly, 3 * 5 equals 5 * 3, respectively.
However, when dealing with matrices, things get more complex. Generally, for matrices, the commutative property does not hold true. This means that for two matrices A and B, it is typically not the case that AB equals BA.

• The commutative property in matrix algebra is reserved for very special cases.
• This is critical to understand, as not every operation can be re-ordered in matrix algebra.

Situations where matrix multiplication is commutative are unusual and need certain conditions to be satisfied.

Invertible Matrices

An invertible matrix is one that has an inverse. Similar to how a number like 5 has an inverse in multiplication (which is 1/5), a matrix A has an inverse, denoted as **A^{-1}**, such that when it is multiplied by A, the result is the identity matrix.
This concept is crucial because only invertible matrices have inverses, and not all matrices are invertible.

• A matrix is invertible if its determinant is not zero.
• The identity matrix acts as 1 in matrix multiplication, so **A * A^{-1} = I**, where I is the identity matrix.

Understanding whether a matrix is invertible helps in solving equations and understanding matrix transformations.

Matrix Multiplication

Matrix multiplication is the process of multiplying two matrices by each other. But, it's not as straightforward as regular multiplication with numbers. Matrix multiplication involves a row-by-column rule that forms each element of the resulting matrix.
This operation is not commutative, which means that AB does not generally equal BA.

• To multiply two matrices, the number of columns of the first matrix must equal the number of rows of the second.
• The resulting matrix will have dimensions determined by the rows of the first matrix and the columns of the second.

Thinking of matrices as transformations helps understand why order matters: multiplying A by B applies B's transformation first, then A's.

Matrix Commutativity

Matrix commutativity is a special condition in matrix algebra where two matrices A and B can be multiplied in any order to yield the same product, i.e., AB = BA.
This phenomenon is not common in general matrix operations but can occur under certain conditions where matrices share properties or are specially constructed.

• Symmetry and certain diagonal matrices might display this behavior.
• If matrices are part of the same algebraic structure or have particular eigenvalue constraints, commutativity might be observed.

In many practical scenarios, matrices do not commute, making it a noteworthy exception when encountered. Recognizing when matrix commutativity occurs is essential in simplifying and solving matrix equations.