Finding the co-ordinate vector

Question

I can find the co-ordinate vectors for all $x$ in $R^n$ but I can't wrap my head around the ones for $x$ in $P_n$.

Here is a question:

• Let $V$ be the space $P_3$ of all polynomials of degree at most $3$ and take $\beta$ to be the basis $\{ v_1=1,v_2=1+t,v_3=t+t^2,v_4= t^2 +t^3\}$. Find the co-ordinate vector of $v = a + bt + ct^2 + dt^3 $.

This is the solution:

• If $v = x_1v_1 + x_2v_2 + x_3v_3+ x_4v_4$, then $x_1 + x_2 = a$, $x_2 + x_3 = b$, $x_3 + x_4 = c$ and $x_4 = d$. Hence $x_3 = c - d$, $x_2 = b - c + d$, and $x_1 = a - b + c - d$. Then the coordinate vector for $v$ is

  $$[a-b+c-d,b-c+d,c - d,d]^t.$$

I don't seem to know how $a$, $b$, $c$ and $d$ were found. Can someone please explain the solution to me?

Answer 1

Can you find the co-ordinate vectors of all $x\in\mathbb{R}^n$ with respect to any basis? For example, the co-ordinate vector of $(1,1,1)^t$ with respect to the standard basis is $(1,1,1)^t$, but with respect to the basis $(1,-1,0)^t,(0,1,-1)^t,(-1,0,1)^t$ it's something else.

If you know how to solve these problems in $\mathbb{R}^n$, then the principle is the same for any vector space. In this case, you have $v=a+bt+ct^2+dt^3$ (this isn't what you wrote, but I guess it's what you meant), and some basis $v_1,v_2,v_3,v_4$. As the solution says, you want to write $v$ as a linear combination of the basis vectors, so you need to find $x_1,x_2,x_3,x_4$ such that

$$v=x_1v_1+x_2v_2+x_3v_3+x_4v_4.$$

Plugging in the definitions of all the $v$s gives

$$a+bt+ct^2+dt^3=x_1+x_2(1+t)+x_3(t+t^2)+x_4(t^2+t^3)$$

and now all you have to do is expand out and compare coefficients. (This makes use of the fact that $1,t,t^2,t^3$ is another basis of $P_3$ - this is implicit in Alex's answer, which features a change of basis matrix between $v_1,v_2,v_3,v_4$ and $1,t,t^2,t^3$.)

Answer 2

The question can be made a lot clearer by given $$v = at^3 + bt^2 + ct + d$$ find the representation $$v = x_1v_1 + x_2v_2 + x_3v_3 + x_4v_4$$ See that $$v = x_1(1) + x_2(1+t) + x_3(t+t^2) + x_4(t^2+t^3) = x_4 t^3 + (x_3+x_4) t^2 + (x_2+x_3) t + (x_1+x_2)$$ so you need to solve $$\pmatrix{0&0&0&1\\0&0&1&1\\0&1&1&0\\1&1&0&0}\pmatrix{x_1\\x_2\\x_3\\x_4} = \pmatrix{a\\b\\c\\d}$$ Whose solution you already found to be $x = (a-b+c-d, b-c+d, c-d, d)^T$. Note that I used the form $at^3+bt^2+ct+d$ and you wrote $a+bt+ct^2+dt^3$ so $a,d$ and $b,c$ are swapped.