# Find the dimension of a vector subspace

I'm doing a problem on finding the dimension of a linear subspace, more specifically

if $\:$ {$f \in \mathcal P_n(\mathbf F): f(1)=0, f'(2)=0$} is a subspace of $P_n$, what is this dimension of this subspace? Here $\mathcal P_n(\mathbf F)$ denotes a vector space of Polynomials of degree $n$ over the real number field.

At first glance, I thought the dimension is infinity, but I think perhaps since the degree is restricted, the dimension should be finite. Yet I find it hard to specify the number of dimensions. Being a beginner of linear algebra, I would like to hear some detailed explanation on how to solve this type of problems.

• two things: you need to add some extra info about what you've tried, any thoughts you have, etc. Second, what is $P_n(F)$? Jul 5, 2015 at 14:04
• I don't know what $\mathcal P_n(\mathbf F)$ is, but your set seems like a kernel of dimension $2$. Jul 5, 2015 at 14:07
• Hint. If you know why the dimension of the space of polynomials of degree at most $n$ is $n+1$ then you can look at what those conditions imply about the coefficients of $f$ in your subspace. (I suspect someone will post this as an answer.) Jul 5, 2015 at 14:08
• @xhimi I don't think $2$ has to be a root. Just the derivative is zero there. Jul 5, 2015 at 14:11
• @xhimi Note that the set definition says nothing about $f(2)=0$. Jul 5, 2015 at 14:11

Intuitively, dimension is the number of degrees of freedom. The elements of $\mathcal{P}_n(\mathbb R)$ are polynomials of degree $n$ (more precisely, at most $n$), so they look like $$a_0 + a_1 x + a_2 x^2 + \dotsb + a_{n-1} x^{n-1} + a_n x^n$$ To specify such a polynomial, you have to specify $n+1$ numbers, the coefficients $a_0,a_1,\dotsc,a_n$. So there are $n+1$ degrees of freedom in this "space" of polynomials.

To prove that formally, you'd want to think of polynomials $a_0+\dotsb+a_nx^n$ as being linear combinations of the polynomials $1,x,x^2,\dotsc,x^n$, and show that these latter polynomials form a basis. This is done in chapter 2 of Axler.

Again intuitively, a constraint that specifies a single number reduces the number of degrees of freedom by 1. Thus imposing the constraint that we will only work with polynomials $f(x)$ satisfying $f(1)=0$ should, we expect, reduce the dimension from $n+1$ to $n$.

The formal version of this is the rank-nullity theorem (Axler's theorem 3.4), which is why everybody's giving answers involving it. I see Axler doesn't do that until chapter 3, though.

So I think the only thing you can do at this point is to produce an explicit basis for the subspace in question. Exercise 8 in chapter 2 is similar; have you tried that? (And for playing with polynomials, exercises 9 and 12 in the same chapter are good.)

(I have the 2nd edition of Axler's text; hopefully it matches yours.)

• +1 for pointing the OP to the relevant places in his text. Jul 5, 2015 at 14:39

Hint

For any linear function $T:V\to W$ we have that $\dim\ker T+\dim\text{Im} T=\dim V$.

Now consider $T(f)=(f(1),f'(2))$.

• Another sorry I do not understand ker and Im. I appreciate your help though. Jul 5, 2015 at 14:18

If you want to do this from scratch, it's a slog but you can actually display a basis for the given subspace:

$p(x)=(-a_{1}-a_{2}-\cdots -a_{n})+(-4a_{2}-12a_{3}-\cdots -n2^{n-1}a_{n})x+a_{2}x^{2}+\cdots +a_{n}x^{n}$.

Substituting again for $a_{1}$, we have

$(-(-4a_{2}-12a_{3}-\cdots -n2^{n-1})-a_{2}-\cdots -a_{n})+(-4a_{2}-12a_{3}-\cdots -n2^{n-1})a_{n}x+a_{2}x^{2}+\cdots +a_{n}x^{n}$.

This is, finally,

$(4a_{2}+12a_{3}+\cdots +n2^{n-1}a_{n}-a_{2}-\cdots -a_{n})+(-4a_{2}-12a_{3}-\cdots -n2^{n-1}a_{n})x+a_{2}x^{2}+\cdots +a_{n}x^{n}$

$=(3a_{2}+11a_{3}+\cdots +(n2^{n-1}-1)a_{n})+(-4a_{2}-12a_{3}-\cdots -n2^{n-1}a_{n})x+a_{2}x^{2}+\cdots +a_{n}x^{n}$

Factoring out the $a_{i}$, we have

$p(x)=(x^{2}-4x+3)a_{2}+(x^{3}-12x+11)a_{3}+\cdots +(x^{n}-n2^{n-1}x+(n2^{n-1}-1))a_{n}$

To finish, just observe that the set of $n-1$ vectors

$\left \{ x^{2}-4x+3,x^{3}-12x+11,\cdots ,x^{n}-n2^{n-1}x+(n2^{n-1}-1) \right \}$ spans the subspace and is linearly independent since all the vectors have distinct degrees. Therefore the dimension of the subspace is $n-1$.

Hint

Denote $\varphi$ the linear form $P \to P(1)$ and $\psi$ the one $P \to P^\prime(2)$. You can prove that $(\varphi, \psi)$ is an independent family using carrefuly selected polynomials.

Then the subspace $H=\{f \in \mathcal P_n(\mathbf F): f(1)=0, f'(2)=0\}$ is the orthogonal of the subspace of linear forms generated by $\{\varphi, \psi\}$.

• I'm sorry I'm a beginner of linear algebra, and do not understand what is orthogonal. But thanks for your answer. Jul 5, 2015 at 14:17

I suppose $\mathscr P_n$ denotes the vector space of polynomials with degree at most $n$. If that is the case, in the canonical basis $\{1, x,\dots,x^n\}$, the matrix of the linear map: \begin{aligned} \varphi\colon\mathscr P_n&\longrightarrow \mathbf R^2\\ f&\longmapsto (f(1),f'(2)) \end{aligned}\quad\text{is}\quad\begin{bmatrix} 1&1&1&1&\dots&1\\ 0&1&4&12&\dots&n 2^{n-1} \end{bmatrix} and it has clearly has rank $2$. Hence by the rank-nullity theorem, the subspace, which is $\ker\varphi$, has dimension $n-1$.

• I have not learned the rank-nullity theorem. I'm currently at Chapter 2 of Axler's Linear Algebra Done Right. Is there some other methods of reaching the correct answer(I think n-1 is the answer)? Jul 5, 2015 at 14:24
• If you know row-reduction, it is as clear: you have a suystem of linear equations of rank $2$ in a vector space of dimension $n+1$. The solutions are a subspace of dimension $n+1-2$. Jul 5, 2015 at 14:28