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Worked Examples · Example 17

Q.Express the matrix A=[123−4−10351]A = \begin{bmatrix} 1 & 2 & 3 \\ -4 & -1 & 0 \\ 3 & 5 & 1 \end{bmatrix} as the sum of a symmetric and a skew-symmetric matrix.

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Any square matrix splits as A=12(A+A′)+12(A−A′)A = \tfrac12(A+A') + \tfrac12(A-A'), where the first part is symmetric and the second is skew-symmetric.

A=P+Q,P=12(A+A′),Q=12(A−A′)A = P + Q,\qquad P = \tfrac{1}{2}(A + A'),\qquad Q = \tfrac{1}{2}(A - A')

PP is symmetric (P′=PP'=P) and QQ is skew-symmetric (Q′=−QQ'=-Q); A′A' is the transpose of AA.

  1. Write the transpose of A=[123−4−10351]A = \begin{bmatrix} 1 & 2 & 3 \\ -4 & -1 & 0 \\ 3 & 5 & 1 \end{bmatrix}:

A′=[1−432−15301].A' = \begin{bmatrix} 1 & -4 & 3 \\ 2 & -1 & 5 \\ 3 & 0 & 1 \end{bmatrix}.

  1. Add to get A+A′A + A':

A+A′=[2−26−2−25652].A + A' = \begin{bmatrix} 2 & -2 & 6 \\ -2 & -2 & 5 \\ 6 & 5 & 2 \end{bmatrix}.

  1. The symmetric part P=12(A+A′)P = \tfrac12(A+A'):

P=[1−13−1−1523521].P = \begin{bmatrix} 1 & -1 & 3 \\ -1 & -1 & \tfrac{5}{2} \\ 3 & \tfrac{5}{2} & 1 \end{bmatrix}.

  1. Subtract to get A−A′A - A':

A−A′=[060−60−5050].A - A' = \begin{bmatrix} 0 & 6 & 0 \\ -6 & 0 & -5 \\ 0 & 5 & 0 \end{bmatrix}.

  1. The skew-symmetric part Q=12(A−A′)Q = \tfrac12(A-A'): …

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