Answer to Question #281192 in Statistics and Probability for Torjan

Question

[LA] Check that {1,(x+1),(x+1)^2} is a basis of the vector space of polynomial over R of degree at most 2. Find the coordinate of 3+x+2x^2 with respect to the basis.

Answer 1

1. Consider the standard basis $B=\{1, x, x^2\}$ on $P_2.$ Using this basis, we can write the elements using coordinate vectors as

$$[1]_B=\begin{bmatrix} 1\\ 0 \\ 0 \end{bmatrix},[x+1]_B=\begin{bmatrix} 1\\ 1 \\ 0 \end{bmatrix},[(x+1)^2]_B=\begin{bmatrix} 1\\ 2 \\ 1 \end{bmatrix}$$

$(x+1)^2=1+2x+x^2$

Because $\dim P_2=3,$ this set is a basis if and only if these three vectors are linearly independent. To verify this, consider the matrix

$$\begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{pmatrix}$$

This matrix is upper-triangular, and the diagonal entries are all non-zero. This implies the matrix is non-singular, and so the columns are linearly independent.

Thus, the set $\{1, (x+1), (x+1)^2\}$ is a basis of $P_2.$

2.

Using the technique of completing the square, we can factor the polynomial $f(x)=3+x+2x^2$ as we like. Specifically,

$$f(x)=3+x+2x^2$$

$$=2(x^2+2x+1)-4x-2+3+x$$

$$=2(x+1)^2-3(x+1)+4(1)$$

Hence, we have the linear combination

$$f(x)=3+x+2x^2=4(1)-3(x+1)+2(x+1)^2.$$