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Exercise 7.1 · Q17

Q.Express the following matrices as the sum of a symmetric matrix and a skew-symmetric matrix:

(i) (4−23−5)\begin{pmatrix} 4 & -2 \\ 3 & -5\end{pmatrix}
(ii) (33−1−2−21−4−52)\begin{pmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{pmatrix}
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By Theorem 7.2, any square matrix MM splits as M=12(M+MT)+12(M−MT)M=\tfrac12(M+M^T)+\tfrac12(M-M^T), where the first term is symmetric and the second is skew-symmetric; we apply this to both given matrices.

Step 1. (i) Transpose M=(4−23−5)M=\begin{pmatrix}4&-2\\3&-5\end{pmatrix}. MT=(43−2−5)M^T=\begin{pmatrix}4&3\\-2&-5\end{pmatrix}.

Step 2. (i) Compute P=12(M+MT)P=\tfrac12(M+M^T).

M+MT=(811−10)⇒P=(41212−5)M+M^T=\begin{pmatrix}8&1\\1&-10\end{pmatrix}\Rightarrow P=\begin{pmatrix}4&\frac12\\\frac12&-5\end{pmatrix} (symmetric, since PT=PP^T=P).

Step 3. (i) Compute Q=12(M−MT)Q=\tfrac12(M-M^T).

M−MT=(0−550)⇒Q=(0−52520)M-M^T=\begin{pmatrix}0&-5\\5&0\end{pmatrix}\Rightarrow Q=\begin{pmatrix}0&-\frac52\\\frac52&0\end{pmatrix} (skew-symmetric, since QT=−QQ^T=-Q).

Step 4. (i) Check P+Q=MP+Q=M. (4+012−5212+52−5+0)=(4−23−5)=M\begin{pmatrix}4+0&\frac12-\frac52\\\frac12+\frac52&-5+0\end{pmatrix}=\begin{pmatrix}4&-2\\3&-5\end{pmatrix}=M ✓.

Step 5. (ii) Transpose M=(33−1−2−21−4−52)M=\begin{pmatrix}3&3&-1\\-2&-2&1\\-4&-5&2\end{pmatrix}. MT=(3−2−43−2−5−112)M^T=\begin{pmatrix}3&-2&-4\\3&-2&-5\\-1&1&2\end{pmatrix}.

Step 6. (ii) Compute P=12(M+MT)P=\tfrac12(M+M^T).

M+MT=(61−51−4−4−5−44)⇒P=(312−5212−2−2−52−22)M+M^T=\begin{pmatrix}6&1&-5\\1&-4&-4\\-5&-4&4\end{pmatrix}\Rightarrow P=\begin{pmatrix}3&\frac12&-\frac52\\\frac12&-2&-2\\-\frac52&-2&2\end{pmatrix} (symmetric).

Step 7. (ii) Compute Q=12(M−MT)Q=\tfrac12(M-M^T). …

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