Matrix $A=B+C$ with $B$ symmetric and $C$ antisymmetric

Question

I am stumped on a question and am looking for some guidance on how to get it done.

The problem gives you:

$x_1 = \begin{bmatrix}9&-4&-2 \\-9&6&-3 \\10&-3&9\end{bmatrix}$

$x_1 = x_2 + x_3$.

$x_2$ is a symmetric matrix and $x_3$ is an antisymmetric matrix.

That is that part that confuses me. The only thing Ive gotten is that the $x_2$ has the $9,6,9$ on the diagonal and $x_3$ has $0$ for its diagonal. Can anyone help me down the right path.

Thanks

Answer 1

For any square matrix $A$ over a field $\Bbb F$ with $\text{char} \; \Bbb F \ne 2$, note that $A_+ = \frac{1}{2}(A + A^T)$ is symmetric, and $A_- = \frac{1}{2}(A - A^T)$ is skew-symmetric, where $A^T$ is the transpose of the matrix $A$. Furthermore, $A = A_+ + A_-$, so we can simply apply these formulas to the given matrix $x_1$ and we see that we can take

$x_2 = \begin{bmatrix} 9 & -\frac{13}{2} & 4 \\ -\frac{13}{2} & 6 & -3 \\ 4 & -3 & 9 \end{bmatrix} \tag{1}$

and

$x_3 = \begin{bmatrix} 0 & \frac{5}{2} & -6 \\ -\frac{5}{2} & 0 & 0 \\ 6 & 0 & 0 \end{bmatrix} \tag{2}$.

Incidentally, the decomposition $A = A_+ + A_-$ is unique; to see this just write $A = B_+ + B_-$ as another symmetric-plus-skew-symmetric representation of $A$. Then $A_+ - B_+ = A_- - B_-$, and since the left-hand side is symmetric while the right-hand side is skew, they are both zero. (Note that $A^T = A \; \text{and} \ A^T = -A \Rightarrow A = -A \Rightarrow A = 0$.)

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

Answer 2

To continue, we have $$x_1=x_2+x_3$$ and $$x_1 = \begin{bmatrix}9&-4&-2 \\-9&6&-3 \\10&-3&9\end{bmatrix} \quad x_2 = \begin{bmatrix}9&a&b \\a&6&c \\b&c&9\end{bmatrix} \quad x_3 = \begin{bmatrix}0&d&e \\-d&0&f \\-e&-f&0\end{bmatrix}$$ for $a,b,c,d,e,f \in \mathbb{R}$.

So we have simultaneous equations such as $$-4=a+d \quad \text{and} \quad -9=a-d$$ which, when we solve, imply $a=-\frac{13}{2}$ and $d=\frac{5}{2}$.

Now consider the simultaneous equations $$-3=c+f \quad \text{and} \quad -3=c-f$$ and $$-2=b+e \quad \text{and} \quad 10=b-e.$$

Answer 3

Note that for any square matrix $A$, we have $$A=\frac{1}{2}(A+A^T)+\frac{1}{2}(A-A^T)$$ where $A^T$ is the transpose of $A$. Note that $\frac{1}{2}(A+A^T)$ is symmetric because $$\frac{1}{2}(A+A^T)^T=\frac{1}{2}(A^T+A).$$ Similarly, $\frac{1}{2}(A-A^T)$ is antisymmetric because $\frac{1}{2}(A-A^T)^T=-\frac{1}{2}(A-A^T)$.

Therefore, in your case when $x_1=\begin{bmatrix}9&-4&-2 \\-9&6&-3 \\10&-3&9\end{bmatrix}$, we have $$x_2=\frac{1}{2}(x_1+x_1^T)\mbox{ and }x_3=\frac{1}{2}(x_1-x_1^T).$$