Choosing a basis for a subspace in $\mathbb{R}^{3\times 3}$ and finding a coordinate vector with respect to it We observe the subspace known as $U\subset \mathbb{R}^{3\times 3}$ that is spanned by the following vectors:
$
        \begin{bmatrix}
        1 & 0 & 0 \\
        0 & -2 & 0 \\
        0 & 0 & 3 \\
        \end{bmatrix}
$, $
        \begin{bmatrix}
        0 & -3 & 0 \\
        0 & 2 & 0 \\
        0 & -1 & 0 \\
        \end{bmatrix}
$, $
        \begin{bmatrix}
        0 & 0 & 1 \\
        0 & -2 & 0 \\
        3 & 0 & 0 \\
        \end{bmatrix}
$, $
        \begin{bmatrix}
        0 & 0 & 0 \\
        -1 & 2 & -3 \\
        0 & 0 & 0 \\
        \end{bmatrix}
$.
Choose a basis for $U$ and show that $
        \begin{bmatrix}
        2 & -3 & -2 \\
        -3 & 8 & -9 \\
        -6 & -1 & 6 \\
        \end{bmatrix}
\in U$. 
Determine the coordinate vector for this vector with respect to the chosen basis for $U$.

Hints to how I solve this? 

We have:
$$ (1) \qquad a  \begin{bmatrix}
        1 & 0 & 0 \\
        0 & -2 & 0 \\
        0 & 0 & 3 \\
        \end{bmatrix}
+b  \begin{bmatrix}
        0 & -3 & 0 \\
        0 & 2 & 0 \\
        0 & -1 & 0 \\
        \end{bmatrix}
+c \begin{bmatrix}
        0 & 0 & 1 \\
        0 & -2 & 0 \\
        3 & 0 & 0 \\
        \end{bmatrix}
+d \begin{bmatrix}
        0 & 0 & 0 \\
        -1 & 2 & -3 \\
        0 & 0 & 0 \\
        \end{bmatrix}
=\begin{bmatrix}
        a & -3b & c \\
        -d & -2(a-b+c-d) & -3d \\
        3c & -b & 3a \\
        \end{bmatrix} =
\begin{bmatrix}
        0 & 0 & 0 \\
        0 & 0 & 0 \\
        0 & 0 & 0 \\
        \end{bmatrix}
$$
iff $ a=b=c=d=0$. So these matrices are linearly independent and they are a basis $\mathcal{U}$ fo $U$.
And, for $(a,b,c,d)=(2,1,-2,3)$ the linear combination $(1)$ gives the matrix
$$M=
 \begin{bmatrix}
        2 & -3 & -2 \\
        -3 & 8 & -9 \\
        -6 & -1 & 6 \\
        \end{bmatrix}
$$
so $(2,1,-2,3)$  are the components of this matrix in the basis $\mathcal{U}$.
