Find the basis $\alpha$ and $\beta$ of $R^3$ and $P_3(R)$ respectibly such that $[T]^\beta _\alpha$ is a diagonal matrix.

Question

Let $T:\mathbf{R}^3 -> P_3(\mathbf{R})$ the linear transformation given by $T(a,b,c)=(a-c)x^2+bx$

Find the basis $\alpha$ and $\beta$ of $R^3$ and $P_3(\mathbf{R})$ respectibly such that $[T]^\beta _\alpha$ is a diagonal matrix.

Well I know that the matrix given by the basis of eigenvectors will be diagonal. But I want to find a shorter way. Any suggestions?

Answer 1

Hints:

1. Observe that the kernel of $T$ is the span of $(1,0,1)^T$.
2. The matrix of $T$ in a pair of bases $(\alpha_1,\alpha_2,\alpha_3),\ (\beta_1,\beta_2,\beta_3)$ is $\pmatrix{1&0&0\\0&1&0\\0&0&0}$ iff $$T(\alpha_1)=\beta_1,\ \ T(\alpha_2)=\beta_2,\ \ T(\alpha_3)=0\,.$$