Question
Express the matrix $B=\left[\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right]$ as the sum of a symmetric and a skew-symmetric matrix.
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Answer
$B' = \left[ {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ { - 2}&3&{ - 2} \\ { - 4}&4&{ - 3} \end{array}} \right]$
Let $P = \frac{1}{2}(B + B') = \left[ {\begin{array}{*{20}{c}} 2&{\frac{{ - 3}}{2}}&{\frac{{ - 3}}{2}} \\ {\frac{{ - 3}}{2}}&3&1 \\ {\frac{{ - 3}}{2}}&1&{ - 3} \end{array}} \right]$
$P' = \left[ {\begin{array}{*{20}{c}} 2&{\frac{{ - 3}}{2}}&{\frac{{ - 3}}{2}} \\ {\frac{{ - 3}}{2}}&3&1 \\ {\frac{{ - 3}}{2}}&1&{ - 3} \end{array}} \right] = P$
Thus $P = \frac{1}{2}(B + B')$ is a symmetric matrix
Let $Q = \frac{1}{3}(B - B') = \left[ {\begin{array}{*{20}{c}} 0&{\frac{{ - 1}}{2}}&{\frac{{ - 5}}{2}} \\ {\frac{1}{2}}&0&3 \\ {\frac{5}{2}}&{ - 3}&0 \end{array}} \right]$
$Q' = \left[ {\begin{array}{*{20}{c}} 0&{\frac{{ - 1}}{2}}&{\frac{5}{2}} \\ {\frac{{ - 1}}{2}}&0&{ - 3} \\ {\frac{{ - 5}}{2}}&3&0 \end{array}} \right]$
$Q' = \left[ {\begin{array}{*{20}{c}} 0&{\frac{{ - 1}}{2}}&{\frac{{ - 5}}{2}} \\ {\frac{1}{2}}&0&3 \\ {\frac{5}{2}}&{ - 3}&0 \end{array}} \right]$
Q' = -Q
Thus $Q = \frac{1}{2}(B - B')$ is a skew symmetric matrix
$P + Q = \left[ {\begin{array}{*{20}{c}} 2&{\frac{{ - 3}}{2}}&{\frac{{ - 3}}{2}} \\ {\frac{{ - 3}}{2}}&3&1 \\ {\frac{{ - 3}}{2}}&1&{ - 3} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 0&{\frac{{ - 1}}{2}}&{\frac{{ - 5}}{2}} \\ {\frac{1}{2}}&0&3 \\ {\frac{5}{2}}&{ - 3}&0 \end{array}} \right]$
Let $P = \frac{1}{2}(B + B') = \left[ {\begin{array}{*{20}{c}} 2&{\frac{{ - 3}}{2}}&{\frac{{ - 3}}{2}} \\ {\frac{{ - 3}}{2}}&3&1 \\ {\frac{{ - 3}}{2}}&1&{ - 3} \end{array}} \right]$
$P' = \left[ {\begin{array}{*{20}{c}} 2&{\frac{{ - 3}}{2}}&{\frac{{ - 3}}{2}} \\ {\frac{{ - 3}}{2}}&3&1 \\ {\frac{{ - 3}}{2}}&1&{ - 3} \end{array}} \right] = P$
Thus $P = \frac{1}{2}(B + B')$ is a symmetric matrix
Let $Q = \frac{1}{3}(B - B') = \left[ {\begin{array}{*{20}{c}} 0&{\frac{{ - 1}}{2}}&{\frac{{ - 5}}{2}} \\ {\frac{1}{2}}&0&3 \\ {\frac{5}{2}}&{ - 3}&0 \end{array}} \right]$
$Q' = \left[ {\begin{array}{*{20}{c}} 0&{\frac{{ - 1}}{2}}&{\frac{5}{2}} \\ {\frac{{ - 1}}{2}}&0&{ - 3} \\ {\frac{{ - 5}}{2}}&3&0 \end{array}} \right]$
$Q' = \left[ {\begin{array}{*{20}{c}} 0&{\frac{{ - 1}}{2}}&{\frac{{ - 5}}{2}} \\ {\frac{1}{2}}&0&3 \\ {\frac{5}{2}}&{ - 3}&0 \end{array}} \right]$
Q' = -Q
Thus $Q = \frac{1}{2}(B - B')$ is a skew symmetric matrix
$P + Q = \left[ {\begin{array}{*{20}{c}} 2&{\frac{{ - 3}}{2}}&{\frac{{ - 3}}{2}} \\ {\frac{{ - 3}}{2}}&3&1 \\ {\frac{{ - 3}}{2}}&1&{ - 3} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 0&{\frac{{ - 1}}{2}}&{\frac{{ - 5}}{2}} \\ {\frac{1}{2}}&0&3 \\ {\frac{5}{2}}&{ - 3}&0 \end{array}} \right]$
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