Question on Linear Combinations and Vectors This is a very simple question I just wanted to make sure I was doing correctly. 
Express the vector 
$$
\underline{v} = \left(\matrix{2\\-1\\5\\-3\\6}\right)
$$as a linear combination of $\mathbf{e_1}, \mathbf{e_2}, \mathbf{e_3}, \mathbf{e_4}$, and $\mathbf{e_5}$ in $\mathbb{R}^5$.
So would I just write it out as 
$$
2\mathbf{e_1}-\mathbf{e_2}+5\mathbf{e_3}-3\mathbf{e_4}+6\mathbf{e_5}
$$
or is there more involved here?
 A: An easy way to see this is to put the standard basis vectors into a matrix.  This may be overkill, but this question is currently unanswered.  
Note that taking each standard basis vector in $\mathbb{R}^5$ as columns into a matrix simply yields the identity matrix $I_5$ or some permutation of it.
$
\begin{align*}
I_5 = [e_1, e_2, e_3, e_4, e_5] =
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
\end{bmatrix}
\end{align*}
$
Now a linear combination can be found by simply multiplying:
$
\begin{align*}
\small
I_5\underline{v} = 
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
2 \\
-1 \\
5 \\
-3 \\
6 \\
\end{bmatrix}
&=
2\begin{bmatrix}
1 \\
0 \\
0 \\
0 \\
0 \\
\end{bmatrix}
+
(-1)\begin{bmatrix}
0 \\
1 \\
0 \\
0 \\
0 \\
\end{bmatrix}
+
5\begin{bmatrix}
0 \\
0 \\
1 \\
0 \\
0 \\
\end{bmatrix}
+
(-3)\begin{bmatrix}
0 \\
0 \\
0 \\
1 \\
0 \\
\end{bmatrix}
+
6\begin{bmatrix}
0 \\
0 \\
0 \\
0 \\
1 \\
\end{bmatrix} \\
\\
&=
2e_1 -e_2 +5e_3 - 3e_4 + 6e_5
\end{align*}
$
