Showing LI without determinant or Reduced row echelon form. Consider the $\mathbb{R}$-vector space $ P^3(\mathbb{R})$ of polynomials over $\mathbb{R}$ with degree at most 3. Define $$W= \{f \in P^3(\mathbb{R}): f^{'}(2)=f(1)=0\}$$
EDIT: Calculate $\dim(W)$. by finding a basis for W.
We need to find a basis for W the easiest way to do this is to look for as many linearly independent vectors as we can, we start by writing down the equations  we get $ ax^3 + bx^2 + cx +d =0$ and $ 3ax^2 + 2bx +c =0$ where $ x=1$ in the first equation and $x=2$ in the second this is from $f(1)=f^{'}(2)=0$ we get
$\begin{bmatrix}
    a       & b &c &d & &0   \\
    12a      & 4b & c & 0&  &0 
\end{bmatrix}=\begin{bmatrix}
    a       & b &c &d& &0   \\
    0      & -8b & -11c & -12d&  &0 
\end{bmatrix} $
$\begin{bmatrix}
   a       &  0 &c  -\frac{11c}{8} &d -\frac{3d}{2}&  &0   \\
    0      & -b & -\frac{11c}{8} & -\frac{3d}{2}&  &0 
\end{bmatrix}=\begin{bmatrix}
 a       &  0 &  -\frac{3c}{8} & -\frac{d}{2}&  &0   \\
    0      & b & \frac{11c}{8} & \frac{3d}{2}&  &0 
\end{bmatrix} $
Now here i am a bit lost we can solve for a and b in terms of c and d but we cant actually find a value for anything. I do have a theorem that tells me that this spans my subspace W but it doesn't tell me that my two vectors are linearly independent, in fact I don't even really have 2 vectors I got infinitely many of them in some sense, I tried writing out $$c_1(ax^3 -\frac{3c}{8} x -\frac{d}{2})+c_2 (bx^2 +\frac{11c}{8}x+\frac{3d}{2})=0$$ but i just have too many variables to say that $c_1=c_2=0$ is the only solution. Normally i would just guess some values but my prof said we apparently are not able to do that and assume what we found was spanning but I don't see anywhere to go from here. Can someone give me a hint how to proceed?
We are going to need a and b to be nonzero for this to make any sense so lets pick $a=b=1$ we know by theorem that what we have is spanning but must show linearly independent.
we have that $1= \frac{3c}{8} +\frac{d}{2}$ and $1= -\frac{11c}{8} -\frac{3d}{2}$ this tells us that $1+3 =  -\frac{11c}{8}+\frac{9c}{8} -\frac{3d}{2} + \frac{3d}{2}$ or $4=-\frac{2c}{8}$ or $c=-16$ and $d = (1+6)2=14$.\  We now wish to show that
$c_1 ( x^3 +6x -7) +c_2 (x^2 -22x +21) =0 $ if and only if $c_1=c_2=0$. $ c_1 x^3 +c_2 x^2 + x(6c_1-22c_2) -7c_1 +21c_2 =0$ for this to be true for all x this tells us that $ -7c_1 +21c_2 =0$ or $ c_1 =3c_2$ Now we have that  $ 3c_2 x^3 +c_2 x^2 + x(18c_2-22c_2) -21c_2 +21c_2 =0$  but this tells us that $-4c_2 =0$ so we have that $c_2=0 $ but from this we know that $c_1=0$. This tells us that the vector $( x^3 +6x -7)$ and the vector $(x^2 -22x +21) $ are linearly independent. However, we know that they span our subspace W as well so they form a basis for W.
 A: For a polynomial $f = ax^3 + bx^2 + cx +d$ in $P^3(\mathbb R)$  one has that $f(1) = a+b+c+d$ and $f'(2) = 12a+4b+c$, so
$$
f \in W \implies 
\begin{cases} 12a+4b+c = 0 \\[1mm] \qquad \quad \& \\[0.5mm] 
a+b+c+d = 0 
\end{cases} \implies 
\begin{cases} c = -12a-4b \\[1mm] \qquad \quad \& \\[0.5mm] 
d = -a-b-c = 11a+3b \end{cases}
$$
that is,
\begin{align}
f &= ax^3+bx^2+(-12a-4b)x+(11a+3b) \\ &= a(x^3-12x+11) + b(x^2-4x+3).
\end{align}
In other words, every polynomial in $W$ can be written as a linear combination of the polynomials $f_1 = x^3-12x+11$ and $f_2 = x^2-4x+3$. Finally, since $f_1$ and $f_2$ are not scalar multiples, they are linearly independent; hence they form a basis for $W$.
A: Find dimension is simple, indeed we have that the general polynomial is in the form
$$f(x)=ax^3+bx^2+cx+d$$
and since we have two independent conditions $\dim(W)=2$.

To find a basis, form here
$$\begin{bmatrix}
 a       &  0 &  -\frac{3c}{8} & -\frac{d}{2}    \\
    0      & b & \frac{11c}{8} & \frac{3d}{2}   
\end{bmatrix}=\begin{bmatrix}
  0   \\
   0 
\end{bmatrix}$$
we can set $a=s$ and $b=t$ as free parameters and then determine $c$ and $d$.
Finally, collecting the terms for the free parameters in the solution for $f(x;s,t)$ we can find the basis.
