Linearly dependent or not I have a set and I would like to know if it is linearly dependent or not:
$(x^2+x,2x^2+1,x^2-x+1)$
Usually I use matrices to show that sets are linearly dependent or not.
In this case I wanted to use specific values like:
$x=0: (0,1,1)$
but the other values I could have used are complex numbers.
What is then another way to prove its linearly dependence?
 A: Well, for three functions (in this case) $f_1,f_2,f_3$, to be linearly independent, we need to check that the only solution to the equation $\alpha f_1+ \beta f_2+\gamma f_3=0$ is $\alpha=\beta=\gamma=0$.
So, solving $\alpha f_1+ \beta f_2+\gamma f_3=0$, we get:
$$\alpha(x^2+x)+\beta(2x^2+1)+\gamma(x^2-x+1)=0$$.
Collecting together constants, $x-$terms and $x^2$-terms, we get:
$$(\alpha+2\beta+\gamma)x^2+(\alpha-\gamma)x+(\beta+\gamma)=0.$$
Now compare $x^2$-terms, $x$-terms, and constant terms on the RHS to give:
$(1) \alpha+2\beta+\gamma=0$
$(2) \alpha-\gamma=0$
$(3) \beta+\gamma=0$.
Solving these (try it yourself, by Gaussian elimination!), we get: $(\alpha,\beta,\gamma)=(t,-t,t))\neq(0,0,0)$ (where $t$ is any real number), so, by the definition 
 of linear dependence, these functions are, indeed, linearly dependent.
Note that the concept is true for whether vectors $\vec v_1, \vec v_2, ..., \vec v_n$ are linearly independent. Just solve $\alpha_1 \vec v_1+\alpha_2 \vec v_2+...+\alpha_n \vec v_n=\vec 0,$ and see if the only solution is $\alpha_1=\alpha_2=...=\alpha_n=0$ (in which case they're independent).
The same holds for linear dependence of matrices.
A: Notice that $(2x^2 + 1) - (x^2 + x) = (x^2 - x +1)$. So the three vectors are linearly dependent.
In this case, looking at the three vectors for long enough will show that they're linearly dependent.
A: Hint:
Apart from Crockett's comment, you may want do the following:


*

*Let $\mathbf{u}_i$ every one of the components of your vector $\mathbf{u} =( (x^2 + x), (2 x^2 + 1), (x^2 - x + 1))$.

*Expand the vector as a linear combination as follows: $\alpha \mathbf{u}_1 + \beta \mathbf{u}_2 + \gamma \mathbf{u}_3$.

*Note that this yields to the following re-arrangement of the situation:
$$(\alpha + 2 \beta + \gamma) \, x ^2 + (\alpha - \gamma) \, x + (\beta + \gamma) \, x^0. $$

*Note that $(x^2,x,1)$ is the base of the vector space of polynomials in $x$.

*Does the matrix (or its determinant): 
$$\left(\begin{array}{ccc} 1 & 2 & 1 \\ 1 & 0 & - 1 \\ 0 & 1 & 1  
\end{array}\right)$$ tell you something? 


Cheers!
