Question

Show that the product \(\mathrm{AA}^{\mathrm{T}}\) is a symmetric matrix.

Short answer

The product \( \boldsymbol{A} \boldsymbol{A}^{\text{T}} \) is symmetric because \( c_{ij} = c_{ji} \).

Step-by-step solution

- Recall matrix multiplication

Matrix multiplication involves taking the dot product of rows from the first matrix and columns from the second matrix. For two matrices \(\boldsymbol{A}\) and \(\boldsymbol{B}\), the element \((i, j)\) of the product matrix is the dot product of the \(\boldsymbol{i}\)-th row of \(\boldsymbol{A}\) and the \(\boldsymbol{j}\)-th column of \(\boldsymbol{B}\).

- Understand the transpose of a matrix

The transpose of a matrix \(\boldsymbol{A}\), denoted \(\boldsymbol{A}^{\mathrm{T}}\), is formed by swapping rows with columns. Explicitly, if \(a_{ij}\) are the elements of \(\boldsymbol{A}\), then the elements \((j, i)\) in \(\boldsymbol{A}^{\mathrm{T}}\) are \(a_{ji}\).

- Define the product \( \boldsymbol{A} \boldsymbol{A}^{\text{T}} \)

The product \(\boldsymbol{A} \boldsymbol{A}^{\text{T}}\) is a matrix where each element is the dot product of the \(i\)-th row of \(\boldsymbol{A}\) and the \(j\)-th row (which is now a column) of \(\boldsymbol{A}\). Mathematically, the element at position \((i,j)\) in the product matrix is \[ c_{ij} = \boldsymbol{a}_i \boldsymbol{a}_j^{\text{T}} \] where \( \boldsymbol{a}_i \) and \( \boldsymbol{a}_j \) are the \(i\)-th and \(j\)-th rows of \(\boldsymbol{A}\).

- Prove symmetry

A matrix \( \boldsymbol{M} \) is symmetric if it equals its transpose, i.e., \( \boldsymbol{M} = \boldsymbol{M}^{\text{T}} \). For \(\boldsymbol{C} = \boldsymbol{A} \boldsymbol{A}^{\text{T}}\), \( c_{ij} = \boldsymbol{a}_i \boldsymbol{a}_j^{\text{T}} \) and \( c_{ji} = \boldsymbol{a}_j \boldsymbol{a}_i^{\text{T}} \). Because the dot product is commutative, \( \boldsymbol{a}_i \boldsymbol{a}_j^{\text{T}} = \boldsymbol{a}_j \boldsymbol{a}_i^{\text{T}} \). Hence, \( c_{ij} = c_{ji} \) for all \( i \) and \( j \), and \( \boldsymbol{A} \boldsymbol{A}^{\text{T}} \) is symmetric.

Key concepts

matrix multiplication

Matrix multiplication is a fundamental concept in linear algebra. When we multiply two matrices \(\boldsymbol{A}\) and \(\boldsymbol{B}\), we compute the dot product of rows from \(\boldsymbol{A}\) with columns from \(\boldsymbol{B}\). This means for the element in the \(i,j\)-th position of the result, we multiply corresponding elements from the \(i\)-th row of \(\boldsymbol{A}\) and the \(j\)-th column of \(\boldsymbol{B}\), then sum them up.

For two matrices \(\boldsymbol{A} = [a_{ij}]\) and \(\boldsymbol{B} = [b_{ij}]\), the product \(\boldsymbol{C}\) is given by: \[ c_{ij} = \boldsymbol{a}_i \boldsymbol{b}_j \text{, where } \boldsymbol{a}_i \text{ is the row vector and } \boldsymbol{b}_j \text{ is the column vector.} \]

Key points to remember:

• The number of columns in \(\boldsymbol{A}\) must be equal to the number of rows in \(\boldsymbol{B}\).
• Matrix multiplication is not commutative, \(\boldsymbol{A} \boldsymbol{B} e \boldsymbol{B} \boldsymbol{A}\).
• Order matters in matrix multiplication.

matrix transpose

A transpose of a matrix transforms its rows into columns and vice versa. If \(\boldsymbol{A}\) is a matrix, its transpose is denoted by \(\boldsymbol{A}^{\text{T}}\).

To get the transpose, simply flip the elements over its diagonal.
For example:
If \(\boldsymbol{A}\) = \(\begin{bmatrix} 1 & 2 \ 3 & 4 \ \end{bmatrix}\)
Then, \(\boldsymbol{A}^{\text{T}}\) = \(\begin{bmatrix} 1 & 3 \ 2 & 4 \ \end{bmatrix}\)

Key points to remember:

• Transposing twice gives the original matrix back: \(\boldsymbol{(A}^{\text{T}}\boldsymbol{)}^{\text{T}} = \boldsymbol{A}\).
• The transpose of a sum is the sum of the transposes: \(\boldsymbol{(A + B)}^{\text{T}} = \boldsymbol{A}^{\text{T}} + \boldsymbol{B}^{\text{T}}\).
• The transpose of a product reverses the order: \(\boldsymbol{(A B)}^{\text{T}} = \boldsymbol{B}^{\text{T}} \boldsymbol{A}^{\text{T}}\).

dot product

The dot product is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number.

For vectors \(\boldsymbol{u} = [u_1, u_2, ..., u_n]\) and \(\boldsymbol{v} = [v_1, v_2, ..., v_n]\), the dot product \(\boldsymbol{u} \boldsymbol{\bullet} \boldsymbol{v}\) is defined as:
\[ \boldsymbol{u} \boldsymbol{\bullet} \boldsymbol{v} = \text{Σ} (u_i v_i) \]
If \( n = 3 \), then the dot product becomes:
\[ \boldsymbol{u} \boldsymbol{\bullet} \boldsymbol{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 \]
Key points to remember:

• The dot product is commutative: \(\boldsymbol{u} \boldsymbol{\bullet} \boldsymbol{v} = \boldsymbol{v} \boldsymbol{\bullet} \boldsymbol{u}\).
• The dot product follows the distributive property: \(\boldsymbol{u} \boldsymbol{\bullet} (\boldsymbol{v} + \boldsymbol{w}) = \boldsymbol{u} \boldsymbol{\bullet} \boldsymbol{v} + \boldsymbol{u} \boldsymbol{\bullet} \boldsymbol{w}\).
• The dot product of orthogonal vectors (90-degree angle) is zero.

symmetric property

A matrix is called symmetric if it is equal to its transpose. In other words, if \(\boldsymbol{A}\) is symmetric, then \(\boldsymbol{A} = \boldsymbol{(A}^{\text{T}}\boldsymbol{)}\).

For elements \(a_{ij}\) of \(\boldsymbol{A}\), symmetry implies: \[ a_{ij} = a_{ji} \text{ for all } i \text{ and } j \]
To prove that \(\boldsymbol{A} \boldsymbol{A}^{\text{T}} \) is symmetric, consider: \[ c_{ij} = \boldsymbol{a_i} \boldsymbol{\bullet} \boldsymbol{a_j}^{\text{T}} \] And since the dot product is commutative: \[ c_{ji} = \boldsymbol{a_j} \boldsymbol{\bullet} \boldsymbol{a_i}^{\text{T}} = \boldsymbol{a_i} \boldsymbol{\bullet} \boldsymbol{a_j}^{\text{T}} \] Which means, \(\boldsymbol{A} \boldsymbol{A}^{\text{T}} \) is indeed symmetric because \(c_{ij} = c_{ji}\).

Key points for symmetric matrices:

• All eigenvalues of a real symmetric matrix are real.
• Symmetric matrices are necessarily square matrices.
• The entries along the diagonal of a symmetric matrix do not change when transposing.