Finding a Basis for P2(R): [2 + 5x + 4x^2]a = [1,2,3], etc

[zodiacbrave]

Homework Statement

If possible, find a basis a = {a1, a2, a3} of P2(R) such that...

[2 + 5x + 4x^2]a = [1, 2, 3], [1 + x + x^2]a = [4,1,2] and [x + x^2]a = [3, -5, 1]
2. The attempt at a solution

Basically, we have something like Ax = b for each of these, right?

A* [2,5,4] = [1,2,3]... A * [1,1,1] = [4,1,2], etc...

I thought that since a basis for R^3 would be [1,0,0] [0,1,0], [0,0,1]. I could go from basis a to basis standard.


I can find A, uniquely for each equation, but to find one common basis for all three, is difficult, I'm lost.

Thank you

 

[vela]

I think you have that backwards. (If I'm misinterpreting your notation, never mind.) When you write [2+5x+4x²]_a, you're referring to the representation of 2+5x+4x² in the {a_i} basis. What that means is that

1 a₁ + 2 a₂ + 3 a₃ = [2, 5, 4]

where [2, 5, 4] is the representation of 2+5x+4x² in the usual {1, x, x²} basis. If the matrix A has the vectors {a_i} as its columns, you'd have

$$A\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 2 \\ 5 \\ 4 \end{bmatrix}$$

The matrix A would be a 3x3 matrix, so you'd have 9 unknowns and only three equations. You can't determine A completely with just one vector equation. With all three vectors, however, you'll have a total of 9 equations which let's you determine A completely.

 

[zodiacbrave]

okay, so by observation, I can determine the matrix [0, 1, 0; 0,1,1;1,0,1] multiplied by the vector [1;2;3] is equal to the vector [2;5;4]

I am not understanding how to solve this, I understand 1 a1 + 2 a2 + 3 a3 = [2, 5, 4]

Does that mean a1 = [2;0;0]... a2 = [0; 5/2;0] and a3 = [0;0;4/3]?

 

[vela]

No, you'd read off the columns of the matrix to see what the a_i's are.

The mistake you're making is that you're only finding a solution to the first vector equation, but there's actually an infinite number. There's only one solution that'll satisfy all three vector equations simultaneously. The other two equations tell you

$$A\begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$$

$$A\begin{bmatrix} 3 \\ -5 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}$$

You can express all three equations as one matrix equation:

$$A\begin{bmatrix} 1 & 4 & 3 \\ 2 & 1 & -5 \\ 3 & 2 & 1\end{bmatrix} = \begin{bmatrix} 2 & 1 & 0 \\ 5 & 1 & 1 \\ 4 & 1 & 1\end{bmatrix}$$

Do you understand why?