Proving that the given set is not linearly independent.

Question

Suppose that $p_0,p_1,p_2,p_3....p_m$ are polynomials in $\mathbb{P}_m(F)$ such that $p_j(2)=0$ for each $j$.

Prove that $(p_0,p_1,p_2,p_3....p_m)$ is not linearly independent in $\mathbb{P}_m(F)$

Attempt.

I am unable to figure out where I am doing wrong.

consider the list $p_0=x^2-2x,p_1=x^3-2x^2$.I will prove that these two vectors are linearly independent.

let $\alpha (x^2-2x)+\beta (x^3-2x^2)=0$

$(x-2)\left(\alpha (x^2)+\beta (x^3)\right)=0$

which implies $\alpha=0$ and $\beta=0$

Hence when

$\alpha (x^2-2x)+\beta (x^3-2x^2)=0$

and

$p_0(2)=0,p_1(2)=0$

it implies(strictly)

$\alpha=0,\beta=0$

$p_0=x^2-2x,p_1=x^3-2x^2$

They are linearly independent where we were supposed to show that they are not.

Tell me where I am wrong?

Answer 1

I imagine the notation $\mathbb P_m$ refers to polynomials of degree at most $m$. If this is the case, your counterexample is not valid because you use $m=1$, but you use a polynomial of degree $3$ which is not in $\mathbb P_1$.

Answer 2

The polynomials would all be divisible by $x-2$. Then it would amount to having $m+1$ (quotient) polynomials of degrees at most $m-1$ , which implies they're linearly dependent. That's because $P_{m-1}(F)$ has dimension $m$.