Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $B = ${$1-t^2, t-t^2, 2-2t+t^2$}. Check that $B$ is a basis for $P_2$ and find $[3+t-6t^2]_B$

In mathematical notation, what exactly does $[3+t-6t^2]_B$ mean? That $[3+t-6t^2]$ is in the subspace of $B$?

share|cite|improve this question
up vote 1 down vote accepted

You know that $\{1,t,t^2\}$ is a basis of $\mathbb P_2$ by construction, since all the vectors in $\mathbb P_2$ are assumed to be linear combinations of those $3$ (i.e. the set $\{1,t,t^2\}$ generates $\mathbb P_2$) and of course if $a_1 + a_2 t + a_3 t^2 = 0$ then all the coefficients vanish (i.e. $\{1,t,t^2\}$ is a linearly independent set). You don't need to do this work, but it is a useful thing to know that $\mathbb P_n$ has dimension $n+1$ in general (in this case, that $\mathbb P_2$ has dimension $3$).

Now since $\mathbb P_2$ has dimension $3$, all you have to prove to show that $\{1 - t^2, t - t^2, 2 - 2t + t^2\}$ is a basis of $\mathbb P_2$ is that it is linearly independent, OR that it is generating. Once you have shown one, the dimension of $\mathbb P_2$ will imply that this set is a basis of $\mathbb P_2$. To show that it is linearly independent, you can show that if $$ a_1(1-t^2) + a_2(t-t^2) + a_3(2-2t+t^2) = 0, $$ then $a_1 = a_2 = a_3 = 0$. If you do the math, $$ a_1(1-t^2) + a_2(t-t^2) + a_3(2-2t+t^2) = (a_1 + 2 a_3) 1 + (a_2 - 2a_3) t + (-a_1 - a_2 + a_3) t^2. $$ Showing that this polynomial is always zero amounts to solving the system of linear equations $$ -a_1 - a_2 + a_3 = 0, \quad a_2 - 2a_3 = 0, \quad a_1 + 2 a_3 = 0. $$ and proving that the unique solution is the trivial one. I leave this work to you, just switch to the matrix form of the system and find the row-echelon form.

If you wished to show that the set was generating instead, take an arbitrary polynomial $p(t) = b_0 + b_1 t + b_2 t^2 \in \mathbb P_2$. You need to show that there exists $x_1, x_2, x_3 \in \mathbb R$ such that $$ x_1(1-t^2) + x_2(t-t^2) + x_3(2-2t+t^2) = p(t) = b_0 + b_1 t + b_2 t^2, $$ i.e. that the system $$ x_1 + 2x_3 = b_0, \quad x_2 - 2x_3 = b_1, \quad -x_1 - x_2 +x_3 = b_2 $$ has a solution. Again, to show this, switch to matrix form and find the row-echelon form.

The last question you have been asked, i.e. what is $[3 + t - 6t^2]_B$, actually only demands that you solve the system $$ x_1 + 2x_3 = 3, \quad x_2 - 2x_3 = 1, \quad -x_1 - x_2 +x_3 = -6 $$ so that you can write $3+t -6t^2 = x_1(1-t^2) + x_2(t-t^2) + x_3(2-2t+t^2)$, i.e. $$ [3 + t - 6t^2]_B = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}. $$ All this notation says is "what are the coefficients of $p(t)$ when expressed as a linear combination of the elements of the basis $B$". Since the coefficients are uniquely determined when $B$ is a basis, you can just find the coefficients and put them in column vector form.

Hope that helps,

share|cite|improve this answer

To check that a set $B=\{v_1,...,v_n\}\subset V$ is a basis of $V$, you have to prove two things:
(1) The vectors of $B$ are lineraly independent, i.e. if $\sum_{i=1}^na_iv_i=0$ then $a_1=...a_n=0$.
(2) The vectors of $B$ span $V$, i.e. for all $v\in V$ there exist $a_1,...,a_n$ such that $\sum_{i=1}^na_iv_i=v$.
If you know that $\dim V=n$ then (1) and (2) are equivalent, so you can prove either one of them.
Here $V=P_2=\mathbb{R}_{\leq2}[t]$ (this is the more standart notation for the space of polynomial with real coefficients of degree at most 2). To check that $p_1,p_2,p_3$ are lineraly independent, just expand $\sum_{i=1}^3a_iv_i=0$ and check that this really implies that $a_1=a_2=a_3=0$.
The notation $[3+t-6t^2]_B$ means the coordinates of $3+t-6t^2$ w.r.t basis $B$. In other words, if $B=(p_1,p_2,p_3)$ then $[3+t-6t^2]_B=\left(\begin{array}{c}a_1\\a_2\\a_3\end{array}\right)$ such that $a_1p_1+a_2p_2+a_3p_3=3+t-6t^2$

share|cite|improve this answer
Okay. So, how would I find this from B? How would I prove that B is the basis for P_2? I'm not sure where to start. – Grace C Nov 7 '12 at 7:43
@Dennis : In standard linear algebra courses for undergrad students, $\mathbb P_2$ or $P_2$ is frequently used to denote this polynomial space. I've never seen $\mathbb R_{\le 2} [t]$ in my life though. Maybe we just had different books. – Patrick Da Silva Nov 7 '12 at 8:41
@Patrick Da Silva: How do you denote the space of polynomials with rational coefficients? complex coefficients? it is much more convenient to have the base field mentioned somewhere. – Dennis Gulko Nov 7 '12 at 8:45
@Dennis : As I said, in "standard linear algebra courses for undergrad students", $\mathbb P_2$ is frequently used. The reason for this notation is mainly because in those courses the field of scalars from the vector space is often assumed to be $\mathbb R$ for simplicity. Of course, if you do advanced linear algebra you might want to lose that restriction, but if you keep it, $\mathbb P_2$ is very clear in that context. – Patrick Da Silva Nov 7 '12 at 8:56
@PatrickDaSilva: When I did my undergrad linear algebra course (and when I teach it now for undergrads of all departments) we never assumed the field to be $\mathbb{R}$. The field was always $F$ and in examples $F$ was usually $\mathbb{Q},\mathbb{R}, \mathbb{C}$ or $\mathbb{Z}/p\mathbb{Z}$. But I guess it depends on the college/university you study/work in. – Dennis Gulko Nov 7 '12 at 20:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.