Question

Express the matrix $A = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}$ as a sum of symmetric and skew-symmetric matrices.

Hint:

Remember: $A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T)$ always works.

Answer 1

Concept: Any square matrix $A$ can be written as: \[ A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T) \] where:

• $\frac{1}{2}(A + A^T)$ is symmetric
• $\frac{1}{2}(A - A^T)$ is skew-symmetric

Step 1: {Find transpose of $A$.}
\[ A^T = \begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix} \]
Step 2: {Compute symmetric part.}
\[ A + A^T = \begin{bmatrix} 4 & -3 & -3 \\ -3 & 6 & 2 \\ -3 & 2 & -6 \end{bmatrix} \] \[ \frac{1}{2}(A + A^T) = \begin{bmatrix} 2 & -\frac{3}{2} & -\frac{3}{2} \\ -\frac{3}{2} & 3 & 1 \\ -\frac{3}{2} & 1 & -3 \end{bmatrix} \]
Step 3: {Compute skew-symmetric part.}
\[ A - A^T = \begin{bmatrix} 0 & -1 & -5 \\ 1 & 0 & 6 \\ 5 & -6 & 0 \end{bmatrix} \] \[ \frac{1}{2}(A - A^T) = \begin{bmatrix} 0 & -\frac{1}{2} & -\frac{5}{2 }\\ \frac{1}{2} & 0 & 3 \\ \frac{5}{2} & -3 & 0 \end{bmatrix} \]
Step 4: {Final expression.}
\[ A = \begin{bmatrix} 2 & -\frac{3}{2} & -\frac{3}{2}\\ -\frac{3}{2} & 3 & 1 \\ -\frac{3}{2} & 1 & -3 \end{bmatrix} + \begin{bmatrix} 0 & -\frac{1}{2} & -\frac{5}{2 }\\ \frac{1}{2} & 0 & 3 \\ \frac{5}{2} & -3 & 0 \end{bmatrix} \]
Step 5: }
Conclusion.}
Thus, $A$ is expressed as the sum of a symmetric and a skew-symmetric matrix.