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The following is a homework problem:

For a fixed $a \in \mathbb{R}$, determine the dimension of the subspace of $P_n(\mathbb{R})$ (the vector space of polynomials of at most degree $n$) defined by: $$W = \{f \in P_n(\mathbb{R})\,|\, f(a) = 0\}.$$

I can't imagine what the basis would look like for $W$ though. Or do I not even need the basis? Can I figure out the dimension without knowing the basis?

Could you give some helpful hints or a framework for solving this and similar problems?

Note: I'm not looking for the answer itself since this is homework.

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Are you sure $P_n$ stands for the space of polynomials ? It is not the space of polynomials which degree is smaller than $n$ ? –  Student Feb 6 '12 at 0:21
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Use Tailor expansion at point $x=a$, and note that polyomials $(x-a)^k$ are linearly independent –  no identity Feb 6 '12 at 0:21
    
Student: Good call! I fixed the question. –  Casey Patton Feb 6 '12 at 0:40
    
It can be noted that by division algorithm, any $f\in W$ can be written as $f=(x-a)g$ where $g\in P_{n-1}(\mathbb{R})$. This gives the dimension of $W$ easily. I hope I am correct. –  Debashish Jun 13 at 8:34

3 Answers 3

up vote 2 down vote accepted

There is a linear map on $\rm \mathbb{P}_n(\mathbb{R})$ given by $\rm L_a :f(x)\to f(x+a)$ for any $\rm a\in\mathbb{R}$. It is invertible because we can see $\rm L_a L_{-a}=L_{-a}L_a=Id$ as linear maps. Hence $\rm L_a$ preserves dimensions of subspaces. We may then write $\rm L_{-a} W=V=\{ p(x)\in \mathbb{P}_n(\mathbb{R}):p(0)=0\}$. It is easy to see that $\rm B=\{x,x^2,\dots,x^n\}$ is a basis for this space by showing that $\rm p(0)=0\,$ is equivalent to $\rm p_0=0$ (the constant term vanishes).

One may use this to show $\rm L_{-a}B=\{x-a,(x-a)^2,\dots,(x-a)^n\}$ is a vector space basis for $\rm W$.

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For $f\in W$ define $T(f) = f(a)$. This is a linear mapping from $P_n(R)$ onto its field. Since $\dim(P_n(R)) = n+1$, $\dim(W) = \dim(\ker(T)) = n$.

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This is the rank-nullity theorem: $\dim\,\operatorname{ker} T+ \dim\,\operatorname{img} T = \dim \,(\text{domain})$. –  anon Feb 6 '12 at 1:02

Suppose $f(x)=c_0x^n+c_1x^{n-1}+\cdots+c_n$. You want $f(a)=0$, which means you want $$a^nc_0+a^{n-1}c_1+\cdots+c_n=0$$ This is a system of one homogeneous linear equation in the $n+1$ unknowns $c_0,c_1,\dots,c_n$. Do you know how to find the dimension of, and a basis for, the vector space of solutions of a system of homogeneous linear equations?

The other answers are fine, but this one works even if you haven't yet studied linear transformations.

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