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Find a basis for $U=\{A\in \mathbb{R}^{2\times 2}: A=A^t\}$

Now It easy to see that becuase $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}=\begin{pmatrix} a & c \\ b & d \end{pmatrix}$$ We have $$\begin{pmatrix} a & c \\ c & d \end{pmatrix}\in Span\{\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\}$$

But how do I slove it using system of linear equations? Do I need to look at

$$\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x & y \\ z & t \end{pmatrix}=\begin{pmatrix} a & c \\ b & d \end{pmatrix}?$$

gbox
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    Your matrix is not equal to the span of your proposed set, but is rather an element of it. I'm also not sure what you mean by "how do I show it". Show what exactly? – user23793 Jan 01 '18 at 22:22
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    See https://math.stackexchange.com/questions/620757/find-a-basis-for-the-vector-space-of-symmetric-matrices-with-an-order-of-n-tim and https://math.stackexchange.com/questions/138944/questions-on-symmetric-matrices and https://math.stackexchange.com/questions/2194415/how-is-this-a-basis-for-the-vector-space-of-symmetric-2x2-matrices and https://math.stackexchange.com/questions/870519/prove-basis-for-symmetric-matrix and probably quite a few others. – Gerry Myerson Jan 01 '18 at 22:27
  • @user23793 edited, I mean how do I solve the qeustion using a system of linear equations – gbox Jan 01 '18 at 22:32

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For $\Bbb{R}^{2\times2}$ you have the standard basis $$b_1:=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},\quad b_2:=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},\qquad b_3:=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix},\qquad b_4:=\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}.$$ With respect to this basis, the subspace $U$ is the zero locus of the linear equation $x_2=x_3$. This means $\{b_1,b_2+b_3,b_4\}$ is a basis for $U$, as you already found.

Servaes
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