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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}

(i) Let $(v_1, v_2)$ be a basis for $text{Im}(p)$ and $(v_3)$ be a basis for $ker(p)$. Prove that $B' = (v_1, v_2, v_3)$
is a basis for $V$ .

I proved that for all $v$ ∈ V we have $v =big(v-p(v)big)+p(v)$.

If $yinker(p) ∩ text{Im}(p)$, then $y = p(v)$ for some $v ∈ V$.

Then, $y = p(y) = p^2(v) = 0$. That is, $(p ◦ p)(y) = p(p(y)) = p(v) = 0$ Hence, $ker(p) ∩ text{Im}(p)= {0}$.

Also, $ker(p) + text{Im}(p) ∈ V$, since each is a subspace of $V$ and their sum is a subspace as well.

(ii) Also, I am asked to compute $M = M_{B'}(p)$.
I know that $A^2 = A$, so $p(v) = pbig(p(w)big) = (p ◦ p)(w) = p (w) = v$, but I struggling to continue further.

How should I better approach this?

1) I think you got the general idea but you should write it out more clearly.

Notice that $A^2 = A$

Let $vin Im ; p$ then $exists w in V : v = Aw.$ And $$Av = A^2w = A w = v.$$

Therefore the restriction of $p$ to the subspace $Im ; p$ is the identity.

Let $v in ker p in Im ; p$ then

$$v = underbrace{v - p(v)}_{in ker;p} + underbrace{p(v)}_{in Im ;p}.$$

Now show that $ Im ; p cap ker p = {0}$. Let $y in Im ; p cap ker p Rightarrow exists v in V: y = p(v).$

Since $y in Im ; p : p(y) = y$ but $y in ker p$ so we also have $p(y) = 0.$

So $ Im ; p cap ker p = {0}$ and we have $V = Im ; p oplus ker p$.

Therefore
$$ B = {v_1,v_2,v_3}$$
is a basis of $V$.

2) Let
$$A =
begin{bmatrix}
frac{2}{3} & -frac{1}{3} & -frac{1}{3} \
-frac{1}{3} & frac{2}{3} & -frac{1}{3} \
-frac{1}{3} & -frac{1}{3} &frac{2}{3} \
end{bmatrix}
$$

Since $B' = {v_1,v_2,v_3 }$ is a basis of $V$ we have $ forall v in V , exists lambda_1,lambda_2,lambda_3 : v = sum_i lambda_i v_i$.

Finally since $v_1,v_2 in Im ; p$ and $v_3 in ker p$ and $p$ is linear:

begin{align*}
p(v) &= p (sum_i lambda_i v_i) \
&= sum_i lambda_ip(v_i) \
&=lambda_1 v_1 + lambda_2 v_2 + lambda_3p(v_3) \
&= lambda_1 v_1 + lambda_2 v_2 + 0.
end{align*}

The matrix in the basis $B'$ is

$$M =
begin{bmatrix}
1 & 0 & 0 \
0& 1 & 0 \
0& 0 & 0\
end{bmatrix}.
$$

You can see $p$ as a projection of $R_2[X]$ onto the subspace $ Im ; p$.