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.I am trying to prove the following statement (Linear Algebra Done Right, Section 2.A, #17):

Suppose $p_0,p_1,\ldots,p_m$ are polynomials in $P_m(F)$ such that $p_j(2)=0$ for each $j.$ Prove that $p_0,p_1,\ldots,p_m$ is not linearly independent in $P_m(F)$

Here is my proof:

Consider expression $(*)$:

$$ a_0 p_0(x) + a_1 p_1(x)+\cdots+a_m p_m(x) = 0 \tag{$*$} $$

If the set $\{ p_0,p_1,\ldots,p_m \}$ is linearly independent, then the only coefficients that make expression $(*)$ true are $a_i=0$ for all $i$ – regardless of what $x$ is.

If expression $(*)$ holds true for all $x,$ it holds true, in particular, for $x = 2.$ Just let $x = 2.$ Then exists a set of $a_i,$ not all $0,$ such that the expression $(*)$ is true.

I am not sure if my proof is correct. Since I am not far enough into the text to encounter terms such as 'basis', I have refrained from using them. I am simply trying to prove the original statement using nothing more than the definition of linear independence.

Thanks in advance for your help!

Michael Hardy
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  • your proof is correct. For all scalars $a_j$ the expression will be zero at $x = 2$. – Maaour Dec 22 '19 at 19:32
  • To prove: they are linearly dependent, right? – Berci Dec 22 '19 at 19:34
  • Berci: yes, that's the aim. @Paul: hmm i kind of understand what you're saying here. Could you elaborate? Isn't it enough for me to show that there are some set of coefficients, not all 0, in order to show linear dependence? –  Dec 22 '19 at 19:37
  • Is there a set of linear independent polynomials $p_0, \cdot, p_m$ in $P_{m+1}$ such that each $p_i$ equals zero at $x=2$? (There are.) What does your proof say about this? Does your proof hold up? – Doug M Dec 22 '19 at 19:56
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    This doesn't work. Consider the set ${x-2,(x-2)^2}.$ That is a linearly independent set, despite the fact that this same argument would appear to prove it is linearly dependent. – Michael Hardy Dec 22 '19 at 20:44
  • @MichaelHardy Wow yes, you're right! okay back to the drawing board for me. I'll trying thinking along the lines of what Doug M posted as an answer –  Dec 22 '19 at 20:49
  • @maa'our : I seems you need to be more careful. Look at my counterexample in my comment above. – Michael Hardy Dec 22 '19 at 21:02
  • yes, you're right, I wasn't being careful. thanks! – Maaour Dec 23 '19 at 14:13

2 Answers2

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The set $\{p_0,\cdot ,p_m\}$ has $m+1$ members. Each can be written as $(x-2)p$ with $p$ an element in $P_{m-1}.$ As $P_{m-1}$ is of dimension $m$ it is not possible to find $m+1$ independent polynomials.

Doug M
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This doesn't work. Consider the set $\{x-2,(x-2)^2\}.$ That is a linearly independent set, despite the fact that this same argument would appear to prove it is linearly dependent.

Linear independence means this:

  • For every sequence of coefficients $a_0,\ldots,a_m,$
  • if for every value of $x\in F,$ $a_0 p_0(x) + \cdots + a_mp_m(x)=0$
  • then $a_0=\cdots=a_m=0.$

That argument has this form: $$ \forall a \, \big( (\forall x \cdots ) \longrightarrow \cdots\big). $$ But the argument presented in the question seems to take this form: $$ \forall a \, \forall x \, \big( \cdots \longrightarrow \cdots\big). $$ Observe that $\Big((\forall x \,\alpha) \longrightarrow \beta \Big)$ is equivalent to $\Big(\beta \text{ or } \exists x\, (\text{not } \alpha\Big).$

Michael Hardy
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    I know this is quite late, but I just want to thank you for taking the time to explain the logical structure of the statement. It massively helped shape my approach to math statements in general. –  Feb 19 '20 at 04:29
  • @Adib0y360 : I'm glad it helped. – Michael Hardy Feb 19 '20 at 21:02