# Proving/verifying dimension and basis

I'm coming from a computer science background and am currently trying to formalize my linear algebra knowledge by going through Linear Algebra Done Right. I have an intuitive grasp on most of the basic concepts but I struggle with trying to prove things rigorously. Here's an example:

Let $U=\{p\in \mathcal{P}_4(F): p(2)=p(5)\}$. Find a basis of U. Extend the basis to a basis of $\mathcal{P}_4(F)$. Find a subspace $W$ of $\mathcal{P}_4(F)$ such that $\mathcal{P}_4(F)=U\oplus W$.

It was pretty straightforward for me to find a basis $\{x^4-203x,x^3-39x,x^2-7x,1\}$, but I have no idea how to really prove that it's in fact a basis. One technique I know is to show that a set is (1) linearly independent, and (2) of the same size as the dimension of the vector space. Condition 1 can be proved by stating that each vector contains a polynomial of a different degree, and so none can be constructed as a linear combination of the others. However, how would I prove condition 2? Specifically, how do I prove that the vector space is of dimension 4? Intuitively it makes sense that

1. $\mathcal{P}_4(F)$ has dimension 5
2. $U$ excludes some members from that space so it has to have fewer dimensions
3. The condition $p(2)=p(5)$ places a single constraint on U, so it makes sense that it has one less dimension.

However, I have no idea how to formalize this.

Help with this specific question would be great, but what I'm really curious about is if there are standard ways to find the dimension of spaces with definitions like this, or if there are ways to verify a basis without relying on the dimension of the vector space.

• Well, first of all, Axler's book might be a bit tough for you given your background. It also has a fairly unorthodox approach that underplays matricies in favor of linear transformations and delibrately avoids determinants,both of which I question. Friedberg/Insel/Spence or Curtis might be better choices-both are careful and rigorous with many examples and give more standard approaches to the subject. Jul 18, 2016 at 21:53
• Well I have taken an intro course in undergrad. However, I'm focusing on machine learning now and find myself trying to understand concepts like eigenvalues and determinants at a deeper level. Would you still recommend that I switch textbooks? Jul 18, 2016 at 22:09

Let's start by finding a basis of $$U=\{p\in P_4:p(2)=p(5)\}$$ To do so, suppose $p\in P_4$ is given by $$p(t)=a_0+a_1t+a_2t^2+a_3t^3+a_4t^4$$ Then $p\in U$ if and only if $$a_{0} + 2 \, a_{1} + 4 \, a_{2} + 8 \, a_{3} + 16 \, a_{4} = a_{0} + 5 \, a_{1} + 25 \, a_{2} + 125 \, a_{3} + 625 \, a_{4}$$ That is, $p\in U$ if and only if $$3 \, a_{1} + 21 \, a_{2} + 117 \, a_{3} + 609 \, a_{4}=0$$ Dividing by $3$ gives $p\in U$ if and only if $$a_{1} + 7 \, a_{2} + 39 \, a_{3} + 203 \, a_{4}=0\tag{1}$$ Now, (1) implies that $p\in U$ if and only if \begin{align*} p(t) &= a_0-(7 \, a_{2} + 39 \, a_{3} + 203 \, a_{4})t+a_2t^2+a_3t^3+a_4t^4 \\ &= a_0+a_2(-7\,t+t^2)+a_3(-39\,t+t^3)+a_4(-203\,t+t^4) \end{align*} This proves that $$U=\DeclareMathOperator{Span}{Span}\Span\{ 1,-7\,t+t^2,-39\,t+t^3,-203\,t+t^4 \}\tag{2}$$ To prove that the polynomials listed in (2) are linearly independent, suppose $$\lambda_1\cdot 1+\lambda_2\cdot(-7\,t+t^2)+\lambda_3\cdot(-39\,t+t^3)+\lambda_4\cdot(-203\,t+t^4)=0$$ Then $$\lambda_{4} t^{4} + \lambda_{3} t^{3} + \lambda_{2} t^{2} - 7 \, \lambda_{2} t - 39 \, \lambda_{3} t - 203 \, \lambda_{4} t + \lambda_{1}=0\tag{3}$$ In particular, the coefficients of $t^0$, $t^2$, $t^3$, and $t^4$ in (3) must vanish. Hence $\lambda_1=\lambda_2=\lambda_3=\lambda_4=0$ so our polynomials are linearly independent.

This shows that the polynomials listed in (2) is a basis of $U$.

Finally, to find a subspace $W$ of $P_4$ satisfying $P_4=U\oplus W$ we need only find $\dim P_4-\dim U=5-4=1$ linearly independent polynomials in $P_4$ not in $U$. Of course $p(t)=t$ does the trick so we can take $W=\Span\{t\}$.

Another way to verify that the polynomials in (2) are linearly independent is to check their Wronskian. In this case the Wronskian is $$W=\det \left[\begin{array}{rrrr} 1 & t^{2} - 7 \, t & t^{3} - 39 \, t & t^{4} - 203 \, t \\ 0 & 2 \, t - 7 & 3 \, t^{2} - 39 & 4 \, t^{3} - 203 \\ 0 & 2 & 6 \, t & 12 \, t^{2} \\ 0 & 0 & 6 & 24 \, t \end{array}\right] = 48 \, t^{3} - 504 \, t^{2} + 1872 \, t - 2436 \not\equiv 0$$ This also proves that the polynomials are linearly independent (assuming we are working over $\Bbb R$).

• +1 Excellent computational answer! Great for a beginner! Jul 18, 2016 at 21:45
• Just to confirm - a subspace being "proper" means that it has fewer dimensions than the original vector space? If so, I understand that. I think the key step that I was missing was establishing that the basis vectors I proposed spanned $U$, and that $4 \leq$ dim$U$. Jul 18, 2016 at 22:05
• @user1340033 "Proper" means "proper subset" which is equivalent to being strictly smaller in dimension. Jul 18, 2016 at 22:42
• Reviewing this again, I'm failing to see why dim $U \geq 4$. Since $U$ is spanned by 4 vectors, isn't 4 actually an upper bound on the number of dimensions it has? Granted, we could then go on to say that they're obviously linearly independent vectors and so dim $U=4$. I'm just trying to understand what I'm missing in the current proof. Aug 1, 2016 at 9:28
• @user1340033 You are correct. My answer has been edited. Aug 4, 2016 at 19:50