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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?

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    That is correct. – 2017-02-08
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    Whilst we can infer what you mean by $\mathbf{e_i}$ it would be great to explain what you mean exactly i.e. definition. – 2017-02-08
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    I believe it's just referring to the basis vectors. – 2017-02-08
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    Yes. This is the usual notation for the standard basis vectors – 2017-02-08

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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*} $