I have the following matrix $C$ whose elements $a$ and $b$ are real.
$\begin{bmatrix} a & b\\ -b&a \end{bmatrix}$
I am supposed to show that the set of such matrices is closed under addition. I am not really sure how I am supposed to do this. Could somebody give me a hint ?
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$\begingroup$$\begin{bmatrix} a & b\\ -b&a \end{bmatrix}+\begin{bmatrix} c & d\\ -d&c \end{bmatrix}=\begin{bmatrix} a+c & b+d\\ -(b+d)&a+c \end{bmatrix}$
$\endgroup$ 6 $\begingroup$I think it means that if you have two matrices with that structure and you add them, you get another matrix with the same structure. This is, let a,b,c and d be real numbers and consider the following matrices: $$ A= \left( \begin{array}{cc} a & b \\ -b & a \end{array} \right) $$ and $$ B= \left( \begin{array}{cc} c & d \\ -d & c \end{array} \right) $$ then $$ A+B= \left( \begin{array}{cc} a+c & b+d \\ -(b+d) & a+c \end{array} \right) $$
which has the same structure as the two matrices that you're adding together. These matrices are really useful to represent complex numbers, where A represents the complex number a+bi and B represents the complex number c+di. They're useful because the product of these numbers is $$(a+bi)(c+di)=(ac-bd) +(ad+bc)i$$ which is represented by the matrix $$ P= \left( \begin{array}{cc} ac-bd & ad+bc \\ -(ad+bc) & ac-bd \end{array} \right) $$ and you can check that $P=AB$, the product of both matrices.
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