Show that a operator is self-adjoint in $\mathbf{R}^{3}$ Let $A:\mathbf{R}^{3}\rightarrow \mathbf{R}^{3}$ be given by $A(x,y,z)=(3x-z,2y,-x+3z)$. Show that $A$ is self-adjoint.

My work, (I consider the inner product like dot product in $\mathbf{R}^{3}$):
Let $w,z\in \mathbf{R}^{3}$ where $w=(x_{1},y_{1},z_{1})$ and $z=(x_{2},y_{2},z_{2})$ and $x_{i},y_{i},z_{i} \in \mathbf{R}$ for $i=1,2$, so:
\begin{eqnarray*}
\langle Aw,z \rangle &=& \langle (3x_{1}-z_{1},2y_{1},-x_{1}+3z_{1}),(x_{2},y_{2},z_{2}) \rangle \\
&=&3x_{1}x_{2}-z_{1}x_{2}+2y_{1}y_{2}-x_{1}z_{2}+3z_{1}z_{2}
\end{eqnarray*}
By other way:
\begin{eqnarray*}
\langle w,Az \rangle &=& \langle (x_{1},y_{1},z_{1}),(3x_{2}-z_{2},2y_{2},-x_{2}+3z_{2}) \rangle \\
&=&3x_{1}x_{2}-z_{2}x_{1}+2y_{1}y_{2}-x_{2}z_{1}+3z_{1}z_{2}
\end{eqnarray*}
But if $A$ is self-adjoint then $\langle Aw,z\rangle=\langle w,Az \rangle$ and I can't see this.
 A: Your calculations are correct, you just need to reorder the terms. Does this help?
\begin{eqnarray*}
\langle Aw,z \rangle &=& \langle (3x_{1}-z_{1},2y_{1},-x_{1}+3z_{1}),(x_{2},y_{2},z_{2}) \rangle \\
&=&3x_{1}x_{2}-\color{blue}{z_{1}x_{2}}+2y_{1}y_{2}-\color{red}{x_{1}z_{2}}+3z_{1}z_{2}
\end{eqnarray*}
\begin{eqnarray*}
\langle w,Az \rangle &=& \langle (x_{1},y_{1},z_{1}),(3x_{2}-z_{2},2y_{2},-x_{2}+3z_{2}) \rangle \\
&=&3x_{1}x_{2}-\color{red}{z_{2}x_{1}}+2y_{1}y_{2}-\color{blue}{x_{2}z_{1}}+3z_{1}z_{2}
\end{eqnarray*}
A: You can do this problem "abstractly".
Observe that the linear map $A$ is the following matrix:
$$
A=\begin{pmatrix}
3&0&-1\\
0&2&0\\
-1&0&3
\end{pmatrix}
$$
and it is symmetric.
For any (row) vector $v\in V:=\mathbb{R}^3$, $A(v)=vA$ where the right-hand side is matrix multiplication.
So now you have
$$
\begin{align}
\langle A(w),z\rangle
&=(wA)z^T\\
&=((wA)z^T)^T&(\textrm{since you have a 1-by-1 matrix})\\
&=z(wA)^T\\
&=z(A^Tw^T)=(zA^T)w^T\\
&=(zA)w^T=\langle w,A(z)\rangle
=\langle A(z),w\rangle
\end{align}
$$
