[change of basis] Consider the family β = (1 + x + x 2 , x − x 2 , 2 + x 2 ) of the polynomial space of degree ≤ 2, R2[x].

1. It is enough to show that $1, x, x^2$ can be expressed as linear combination of vectors in the basis.

We have \[ 1 = - \frac{1}{3} \cdot (1+x+x^2) + \frac{1}{3} \cdot (x-x^2) + \frac{2}{3} \cdot (2+x^2), \] \[ x = \frac{2}{3} \cdot (1+x+x^2) +\frac{1}{3} \cdot (x-x^2) -\frac{1}{3} \cdot (2+x^2), \] \[ x^2 = \frac{2}{3} \cdot (1+x+x^2) -\frac{2}{3} \cdot (x-x^2) -\frac{1}{3} \cdot (2+x^2), \] so it forms a basis.

2. The matrix is implicitly given by the formulae above, that is, \[ \frac{1}{3} \begin{bmatrix} -1 & 2 & 2 \\ 1 & 1 & -2 \\ 2 & -1 & -1 \end{bmatrix}; \] it is worth noticing that it's the inverse of the matrix \[ \begin{bmatrix} 1 & 0 & 2 \\ 1 & 1 & 0 \\ 1 & -1 & 1 \end{bmatrix}, \] which is obviously the matrix of change of basis from the canonical basis to the given one.

3-4. All base-change matrices are invertible by definition, so the kernel is $\{0\}$ and the image is the whole space of polynomials $\mathbb{R}_2[x]$, regardless of the basis we use.