Answer:
[tex]\left[\begin{array}{ccc}0&1&0\\1&0&0\\0&0&0\end{array}\right],\left[\begin{array}{ccc}0&1&0\\1&0&0\\0&0&0\end{array}\right],\left[\begin{array}{ccc}0&1&0\\1&0&1\\0&1&0\end{array}\right][/tex]
Step-by-step explanation:
We are given that vector space of all symmetric matrix
M(R)={[tex]\left[\begin{array}{ccc}a&b&c\\d&e&f\\g&h&i\end{array}\right][/tex]
where a,b,c,d,e,f,g,h,i[tex]\in R[/tex]}
Let A=[tex]\left[\begin{array}{ccc}a&b&c\\d&e&f\\g&h&i\end{array}\right][/tex]
A'=[tex]\left[\begin{array}{ccc}a&d&g\\b&e&h\\c&f&i\end{array}\right][/tex]
A is symmetric
Then A'=A
[tex]\left[\begin{array}{ccc}a&b&c\\d&e&f\\g&h&i\end{array}\right][/tex]=[tex]\left[\begin{array}{ccc}a&d&g\\b&e&h\\c&f&i\end{array}\right][/tex]
a=a , e=e, i=i
a-a=0 , e-e=0, i-i=0
b=d, c=g ,f=h
Hence, the matrix
[tex]\left[\begin{array}{ccc}0&b&c\\b&0&f\\c&f&0\end{array}\right][/tex]
Therefore, the basis are
[tex]\left[\begin{array}{ccc}0&1&0\\1&0&0\\0&0&0\end{array}\right],\left[\begin{array}{ccc}0&1&0\\1&0&0\\0&0&0\end{array}\right],\left[\begin{array}{ccc}0&1&0\\1&0&1\\0&1&0\end{array}\right][/tex]
There are three elements to generate an element of vectors of all symmetric matrix [tex]3\times 3[/tex] matrix
[tex]\left[\begin{array}{ccc}0&b&c\\b&0&f\\c&f&0\end{array}\right][/tex]=[tex]b\left[\begin{array}{ccc}0&1&0\\1&0&0\\0&0&0\end{array}\right]+c\left[\begin{array}{ccc}0&1&0\\1&0&0\\0&0&0\end{array}\right]+f\left[\begin{array}{ccc}0&1&0\\1&0&1\\0&1&0\end{array}\right][/tex]
