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).
Best Answer
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...