Isomorphism between symmetric and upper triangular matrices Question: Determine if the vector spaces $V=S_{3}$, the 3x3 symmetric matrices, and $W=U_{3}$, the 3x3 upper triangular matrices, are isomorphic. If they are, give an explicit isomorphism $T: V \rightarrow W$.
Attempt: I've determined that they are isomorphic, since they both have dimension 6. I'm just not sure what to try as an isomorphism. Since the proof that all vector spaces of equal dimension are isomorphic uses coordinate vectors, I'd guess that those are the easiest ones to construct.
Could I map $S_{3}$ to $R^{3}$ using the coordinate vector map, then use the inverse of the coordinate vector map to bring $R^{3}$ to $U_{3}$?
$F : S_{3} \rightarrow R^{3}$ and $G : R^{3} \rightarrow U_{3}$, so if $T=F \circ G$, then $T : S_{3} \rightarrow U_{3}$. Seem legit?
 A: Hint: every symmetric matrix is a upper triangular matrix in which the entries have been reflected about the diagonal.
Full solution:
Your bases will be likely along the following lines:
For $S_3$: $$\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}\right], \left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix}\right],\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}\right], \left[ \begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}\right], \left[ \begin{matrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{matrix}\right], \left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}\right]$$
Define your isomorphism by sending the above to the following matrices:
$$\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}\right], 
\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix}\right],
\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}\right], 
\left[ \begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}\right], 
\left[ \begin{matrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}\right], 
\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}\right],$$
respectively.
This is more than likely your basis for $U_3$.
It should be rather clear that this is invertible.
A: A basis of the space of $W$ is
$$
\{E_{i,j}\}_{i<j}\cup \{E_{i,i}\}
$$
A basis of the space of $V$ is
$$
\{E_{i,j} + E_{j,i}\}_{i<j}\cup \{E_{i,i}\}
$$
Hence, let us define $T:V\to W$ such as $T$ is linear and
$$\begin{cases}
T(E_{i,j} + E_{j,i}) = E_{i,j} & (i<j)\\
T(E_{i,i}) = E_{i,i}
\end{cases}$$
It maps a basis to a basis, hence it is an isomorphism:


*

*if $T(x) = 0$, then with $x = \sum_{i<j} x_{i,j}(E_{i,j} + E_{j,i})
+\sum x_kE_{k,k},$
$$
0 = T(x) = \sum_{i<j} x_{i,j}E_{i,j}+\sum x_iE_{i,i}\\
\implies  x_{i,j}=0, x_k =0
$$for every $i<j, k$. Hence $T$ is into.

*If $y=\sum_{i<j} y_{i,j}E_{i,j}+\sum y_iE_{i,i}$, $y=T(x)$ with
$$
y= \sum_{i<j} y_{i,j}(E_{i,j} + E_{j,i})
+\sum y_kE_{k,k}
$$hence $T$ is onto.

