Question

For a given matrix \(A = \begin{bmatrix} 4 & 4 \\ 0 & 4 \\ -4 & 4 \end{bmatrix}\), and \(A^T\) representing its transpose, which of the following options is the CORRECT representation of \(A^T A\)?

• A. \(\begin{bmatrix} 4 & 0 \\ 0 & -4 \end{bmatrix}\)
• B. \(\begin{bmatrix} 32 & 0 \\ 0 & 48 \end{bmatrix}\)
• C. \(\begin{bmatrix} 32 & 0 \\ 0 & 0 \\ 0 & 48 \end{bmatrix}\)
• D. \(\begin{bmatrix} 4 & 0 \\ 0 & 0 \\ 0 & -4 \end{bmatrix}\)

Hint:

Multiply the transpose of A by A; the result is a 2 by 2 symmetric matrix formed from dot products of the columns of A.

Answer 1

The Correct Option is B

Step 1: View A in terms of its columns:
Write matrix A as two column vectors, $A = [\mathbf{a}_1 \ \mathbf{a}_2]$, where $\mathbf{a}_1 = (4, 0, -4)$ and $\mathbf{a}_2 = (4, 4, 4)$. The product $A^T A$ can then be understood as the matrix of dot products between these columns, since $(A^T A)_{ij} = \mathbf{a}_i \cdot \mathbf{a}_j$.
Step 2: Compute the norm squared of the first column:
The entry at position (1,1) is $\mathbf{a}_1 \cdot \mathbf{a}_1 = 4^2 + 0^2 + (-4)^2 = 16 + 0 + 16 = 32$, which is the squared length of the first column vector.
Step 3: Compute the norm squared of the second column:
The entry at position (2,2) is $\mathbf{a}_2 \cdot \mathbf{a}_2 = 4^2 + 4^2 + 4^2 = 16 + 16 + 16 = 48$, which is the squared length of the second column vector.
Step 4: Compute the cross term between the two columns:
The entries at position (1,2) and (2,1) are both equal to $\mathbf{a}_1 \cdot \mathbf{a}_2 = 4 \times 4 + 0 \times 4 + (-4) \times 4 = 16 + 0 - 16 = 0$, since $A^T A$ is always symmetric.
Step 5: Assemble the result and confirm:
Combining these values gives $A^T A = \begin{bmatrix} 32 & 0 \\ 0 & 48 \end{bmatrix}$, a symmetric 2 by 2 matrix, which matches option B and confirms the direct multiplication carried out by the row and column method.
Final Answer:
\[ \boxed{A^T A = \begin{bmatrix} 32 & 0 \\ 0 & 48 \end{bmatrix}} \]