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Consider the following linear space $U=\{A\in \mathcal{M}_{3\times 3}\,|\,A=A^T\}$ (the space of symetric matices of order 3). I wonder what is the basis of this space. I was thinking about the following candidates: $A_1=\begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix}$, $A_2=\begin{bmatrix}0&0&0\\0&1&0\\0&0&1\end{bmatrix}$, $A_3=\begin{bmatrix}0&0&1\\0&0&0\\1&0&0\end{bmatrix}$, $A_4=\begin{bmatrix}0&1&0\\1&0&0\\0&0&0\end{bmatrix}$, $A_5=\begin{bmatrix}0&0&0\\0&0&1\\0&1&0\end{bmatrix}$. My idea is to obtain a matrix $B$ such that $b_{ij}=1$, i.e., $$A_1+…+A_5=B.$$ The set $B(U)=\{A_i, \,i=1…5\}$ is linearly independent, but the question is, if it is a basis? If I consider any linear combination $$c_1A_1+…+c_5A_5,$$ then in the result matrix I will have coefficients $c_1,…c_5$, while a matrix from $U$ has nine coefficients, which means that some of them will be duplicated.

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Well, a vector space usually has infinitely many basis. But your set isn't one of them, because the matrix$$\begin{bmatrix}0&0&0\\0&1&0\\0&0&0\end{bmatrix}$$cannot be expressed as a linear combination of its elements.

Forget your matrix $A_2$ and use$$\begin{bmatrix}0&0&0\\0&1&0\\0&0&0\end{bmatrix}\text{ and }\begin{bmatrix}0&0&0\\0&0&0\\0&0&1\end{bmatrix}$$instead. Then you will have a basis.

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The space of all $n\times n $ matrices over $\mathbb{R}$ that are symmetric has dimension $\dfrac{n(n+1)}{2}$.

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    $\begingroup$ But upper triangular or lower triangular matrices are not symmetric. $\endgroup$ – Giuseppe Negro Feb 14 '18 at 12:04
  • $\begingroup$ The condition that i have mentioned makes them symmetric $\endgroup$ – Upstart Feb 14 '18 at 12:07

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