Simple question regarding vector-space dimensions If I am looking at $\mathbb{R}^3$, I know that it will have a basis consisting of $3$ linearly independent vectors -- and therefore I know the space to be $3$ dimensional. If I am looking at $\mathbb{R}^{3\times 3}$ for some reason my brain is telling me that I could use the same basis as for $\mathbb{R}^3$ e.g. $\{e_1, e_2, e_3\}$ because I can use these three vectors to form any of the column vectors that make up $\mathbb{R}^{3\times 3}$ -- but this implies that $\mathbb{R}^{3\times 3}$ is $3$-dimensional (which is how I can be certain I'm wrong).
I think that I'm thinking about this as if I had three lego blocks with which I could span $\mathbb{R}^3$, and making $3$ different constructions with these same 3 lego blocks I can span $\mathbb{R}^{3\times 3}$ because I am using the same $3$ unit vectors as I used for $\mathbb{R}^3$. 
Sorry if this is too unclear, I know I'm wrong I just don't know why. Thanks.

The standard basis vectors $e_1$, $e_2$ and $e_3$ of $\Bbb{R}^3$ are not elements of $\Bbb{R}^{3\times3}$, so they certainly cannot form a basis for that vector space. The vector space $\Bbb{R}^{3\times3}$ is also not made up of 'column vectors', which you seem to suggest. It can be seen as being made up of $3\times3$-matrices, in which case the matrices
$$\begin{pmatrix}
1&0&0\\
0&0&0\\
0&0&0
\end{pmatrix},\qquad
\begin{pmatrix}
0&1&0\\
0&0&0\\
0&0&0
\end{pmatrix},\qquad
\begin{pmatrix}
0&0&1\\
0&0&0\\
0&0&0
\end{pmatrix},$$
$$
\begin{pmatrix}
0&0&0\\
1&0&0\\
0&0&0
\end{pmatrix},\qquad
\begin{pmatrix}
0&0&0\\
0&1&0\\
0&0&0
\end{pmatrix},\qquad
\begin{pmatrix}
0&0&0\\
0&0&1\\
0&0&0
\end{pmatrix},$$
$$
\begin{pmatrix}
0&0&0\\
0&0&0\\
1&0&0
\end{pmatrix},\qquad
\begin{pmatrix}
0&0&0\\
0&0&0\\
0&1&0
\end{pmatrix},\qquad
\begin{pmatrix}
0&0&0\\
0&0&0\\
0&0&1
\end{pmatrix},$$
form a basis. They are easily verified to be linearly independent, and to span the whole space, which shows that the dimension of $\Bbb{R}^{3\times3}$ is indeed $9$.
