Question

If \(A\) and \(B\) are two \(N \times N\) matrices, show that \((A B)^{T}=B^{T} A^{T}\) where \(A^{T}\) is the transpose of \(A\) defined by interchanging the rows and columns of \(A\).

Short answer

\((AB)^T = B^TA^T\) is proven by showing that the ij-th element in both matrices is the result of the same dot product of the i-th column of B and the j-th row of A.

Step-by-step solution

Understand the Transpose of a Product

The transpose of a product of two matrices, \((AB)^T\), is such that the elements are switched along its diagonal, meaning the row and column indices are swapped. The individual elements of the resulting matrix are given by \((AB)^T_{ij} = (AB)_{ji}\). We aim to prove that this is equal to \(B^TA^T\). To do this, we will show that the ij-th element in both \((AB)^T\) and \(B^TA^T\) are the same.

Express the Elements of \((AB)^T\)

The elements of the product of two matrices \((AB)_{ij}\) are the dot products of the i-th row of A and the j-th column of B. To transpose, we switch the rows and columns, so \((AB)^{T}_{ij} = (AB)_{ji}\).

Relate to Transposes of A and B

Consider the transposes of \(A\) and \(B\) separately: \(A^T\) and \(B^T\). The elements of \(B^TA^T\) are such that the i-th row of \(B^T\) is the i-th column of \(B\), and the j-th column of \(A^T\) is the j-th row of \(A\). So, the elements of \(B^TA^T\) are given by dot products of the j-th row of \(A\) and the i-th column of \(B\), or \((B^TA^T)_{ij} = \$\sum_k B^T_{ik} A^T_{kj} = \$\sum_k B_{ki} A_{jk}\), which is the same as the dot product for \((AB)_{ji}\).

Show Equality of Elements

Now we demonstrate that each element of \((AB)^T\) matches the corresponding element of \(B^TA^T\). We know that both \((AB)^T_{ij}\) and \((B^TA^T)_{ij}\) represent the sum of products of the corresponding elements of the i-th column of B and the j-th row of A, therefore, they are equal: \((AB)^T_{ij} = (B^TA^T)_{ij}\).

Conclude the Proof

Since we have shown that every element of both matrices \((AB)^T\) and \(B^TA^T\) match, it implies that the matrices themselves are equal. Thus, we have proven that \((AB)^T = B^TA^T\).

Key concepts

Transpose of a Product of Matrices

Matrix operations often come with their own set of unique properties, one of which is the behavior of transposing a product of matrices. The transpose operation flips a matrix over its diagonal, turning rows into columns and columns into rows. When two matrices, let's call them matrix A and matrix B, are multiplied together to form a matrix AB, transposing this product results in a new matrix. Interestingly, the transpose of the product is not simply the product of the transposes in the original order.

In mathematical terms, for matrices A and B, \( (AB)^T \) is not the same as \( A^T B^T \). Instead, one of the key properties is that \( (AB)^T \) equals \( B^T A^T \). To understand why this property holds, it helps to delve into the meaning of matrix multiplication and how individual elements of the resulting matrix are determined by the dot product. By recognizing these building blocks, we can grasp the proof in linear algebra that demonstrates this intriguing behavior.

Matrix Multiplication

Matrix multiplication is a cornerstone of linear algebra, and understanding it is vital when dealing with matrices. Unlike scalar multiplication, which is straightforward, matrix multiplication involves a series of dot products between the rows of the first matrix and the columns of the second matrix. The result of these dot products forms the elements of the resulting matrix.

Specifically, if we're multiplying matrix A by matrix B, then the element in the i-th row and j-th column of the product matrix AB, denoted as \( (AB)_{ij} \), is calculated as the dot product of the i-th row of A and the j-th column of B. This procedure is done for all row-column pairs, resulting in a new matrix whose dimensions are determined by the number of rows in A and the number of columns in B. It's important to note that matrix multiplication is not commutative; changing the order of the matrices generally changes the result.

Dot Product in Matrices

The dot product, also known as the scalar product, in the context of matrices is the sum of the products of corresponding entries of two sequences of numbers. When two matrices are multiplied, the resultant matrix is constructed using dot products. Each element of matrix AB is computed as the dot product between a row vector in matrix A and a column vector in matrix B.

So, for \( (AB)_{ij} \), we multiply corresponding elements of the i-th row of A and the j-th column of B and then sum these products to get a single number. This single number constitutes the element at the i-th row and j-th column in the matrix AB. This dot product operation is fundamental in understanding many matrix transformations and properties, including the aforementioned property of the transpose of a product of matrices.

Proof in Linear Algebra

Proofs form the backbone of linear algebra, ensuring that the properties and theorems we rely on have a solid foundation. When we state that transposing a product of matrices results in the product of their transposes in reverse order — \( (AB)^T = B^T A^T \) — we're making a claim that needs a rigorous proof. Such proofs typically involve showing that a particular operation holds true for all elements of the matrix.

For our current property, the proof involves several steps which start with understanding the definition of a transpose and matrix multiplication, then proceed with the element-wise comparison of matrices \( (AB)^T \) and \( B^TA^T \) and conclude by demonstrating that these matrices have matching elements, thereby proving they are equal. This notion reinforces a fundamental aspect of linear algebra: a formula or property is only as good as its proof, which must be logical, thorough, and universally applicable.