<s>[INST] What is the dimension of the following function space? Construct a function space out of the following:
$v_1 = x^4+x^3+x^2+x+1$
$v_2 = x^4+x^2+1$
$v_3 = x^3+x$
$v_4 = x+1$
Where the space S is defined as linear product of the $v$'s: 
$S=lin(v_1,v_2,v_3,v_4)$
What is the dimension of S? Since $v_1$ is the linear product of $v_2$ and $v_3$ and the rest of the "vectors" appear to not be linearly dependent, is the correct answer 3? (Almost seems too easy to be the correct answer).
 [/INST] Pass from polynomials to coordinate vectors:
$$v_1\to(1,1,1,1,0)\;,\;\;v_2\to(1,0,1,0,1)\;,\;\;v_3\to(0,1,0,1,0)\;,\;\;v_4\to(0,0,0,1,1)$$
Now form the corresponding matrix with the above as rows, and reduce the matrix (Gauss elementary operations):
$$\begin{pmatrix}
1&1&1&1&0\\
1&0&1&0&1\\
0&1&0&1&0\\
0&0&0&1&1\end{pmatrix}\stackrel{R_2-R_1}\longrightarrow\begin{pmatrix}
1&1&1&1&0\\
0&\!-1&0&\!-1&1\\
0&1&0&1&0\\
0&0&0&1&1\end{pmatrix}\stackrel{R_3-R_2}\longrightarrow\begin{pmatrix}
1&1&1&1&0\\
0&\!-1&0&\!-1&1\\
0&0&0&0&1\\
0&0&0&1&1\end{pmatrix}\stackrel{R_3\leftrightarrow R_4}\longrightarrow$$$${}$$
$$\longrightarrow\begin{pmatrix}
1&1&1&1&0\\
0&\!-1&0&\!-1&1\\
0&0&0&1&1\\
0&0&0&0&1\end{pmatrix}$$$${}$$
We see we get a matrix of rank $\;4\;$ and then the four vectors above, and thus the four original ones (the polynomials) as well,  are linearly independent, and this means
$$\dim\text{ Span}\,\{v_1,v_2,v_3,v_4\}=4$$
Added after correction by the OP:
$$\begin{pmatrix}
1&1&1&1&1\\
1&0&1&0&1\\
0&1&0&1&0\\
0&0&0&1&1\end{pmatrix}\stackrel{R_2-R_1}\longrightarrow\begin{pmatrix}
1&1&1&1&0\\
0&\!-1&0&\!-1&0\\
0&1&0&1&0\\
0&0&0&1&1\end{pmatrix}\stackrel{R_3-R_2}\longrightarrow\begin{pmatrix}
1&1&1&1&0\\
0&\!-1&0&\!-1&0\\
0&0&0&0&0\\
0&0&0&1&1\end{pmatrix}\stackrel{R_3\leftrightarrow R_4}\longrightarrow$$$${}$$
and thus we get at once the dimension is three...
</s>