Compute $\text{rk}(p)$ using Gauss reduction on $A$. Compute $\dim\big(\ker(p)\big)$.

Question

Let $B = (1, X, X^2)$ be a basis for $\mathbb{R}_2[X]$ and $p ∈ \mathcal{L}\big(\mathbb{R}_2[X]\big)$ be the linear map defined by $p(1) = \frac{1}{3}(2 − X − X^2)$, $p(X) = \frac{1}{3}(−1 + 2X − X^2)$ and $p(X^2) = \frac{1}{3} (−1 − X + 2X^2)$.

By definition of the task, I found that $A=M_B(p)$, which is the matrix of $p$ with respect to the basis $B$.

\begin{bmatrix} \frac{2}{3} & -\frac{1}{3} & -\frac{1}{3} \\ -\frac{1}{3} & \frac{2}{3} & -\frac{1}{3} \\ -\frac{1}{3} & -\frac{1}{3} &\frac{2}{3} \\ \end{bmatrix}

Compute $\text{rk}(p)$ using Gauss reduction on $A$. Compute $\dim\big(\ker(p)\big)$.

I found $\text{rk}(p)$ which is $2$. I looked at the definition of $\ker$, and $\dim(\ker)$ but apparently I'm quite struggling to find it. How should I go about it?

Answer 1

There is a theorem called the rank-nullity theorem, which states that for a linear transformation $p:V\to W, \text{rk}(p)+\dim(\ker(p))=\dim(V)$. In your case, $\dim(\ker(p))=3-\text{rk}(p)=3-2=1$.