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I need help with the following problem

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

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  • $\begingroup$ What is your question? $\endgroup$ – Trevor Wilson Nov 26 '13 at 17:40
  • $\begingroup$ Can you explain what you have tried to do? $\endgroup$ – A.E Nov 26 '13 at 17:42
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In general,

$$\mathbf{M} = \dfrac{\mathbf{M} + \mathbf{M}^t}{2} + \dfrac{\mathbf{M} - \mathbf{M}^t}{2}.$$

The first term is a symmetric matrix; the second is skew-symmetric.

This is an instance of a much more general theorem. The "transpose" operation, done twice, is the identity. Such operations are called "involutions". And involutions all "split" their domains this way:

If you call that involution (think of "transpose") $T$, we have $T^2 - I = 0$. That means that $+1$ and $-1$ are the only eigenvalues for $T$,. Furthermore, the domain of $T$ splits into a direct sum of the $+1$ and $-1$ eigenspaces. That's the larger theorem I was referring to.

In fact, the solution above (for transpose) generalizes as well: for any vector in the domain of $T$, we have $$ v = \frac{1}{2} (v + T(v)) + \frac{1}{2} (v - T(v)) $$ which splits $v$ as a sum of a vector that's invariant under $T$ and one that negates when $T$ is applied, i.e., it splits $v$ as a linear combination of a multiple of a $+1$ eigenvector and a multiple of a $-1$ eigenvector.

Notice that the fraction $\frac{1}{2}$ appears here. That means that this trick, as written, only works over fields of characteristic not equal to two. You might ask whether it's still true in characteristic two. You could try some examples to find out.

The theorem can be applied to show that every function on the reals is the sum of an even and an odd function, that every complex number is the sum of a real and a pure imaginary (OK, that's pretty trivial), etc.

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  • $\begingroup$ So we have to give this answer in exam as well $\endgroup$ – user110715 Nov 26 '13 at 17:50

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