1. **Stating the problem:** We have a vector space $M$ of matrices of the form $\begin{pmatrix} a & b \\ c & 0 \end{pmatrix}$ with the constraint $a - 2b + c = 0$. We are given an inner product defined by $(A,B) = \operatorname{trace}(B^T A)$ for all $A,B \in M$. Also, a linear operator $L$ acts on quadratic polynomials $ax^2 + bx + c$ as: $$L(ax^2 + bx + c) = (a + 3b + 6c)x^2 + \left(-\frac{4}{3}a - 3b - 4c\right)x + \left(\frac{2}{3}a + b + c\right).$$ 2. **Understanding the problem:** We want to analyze the space $M$ and the inner product, and understand the linear operator $L$ acting on polynomials. 3. **Step 1: Express the constraint $a - 2b + c = 0$** This means $a = 2b - c$. 4. **Step 2: Substitute $a$ in the polynomial and in $L$:** Let $a = 2b - c$, then the polynomial is: $$p(x) = (2b - c)x^2 + bx + c.$$ 5. **Step 3: Apply $L$ to $p(x)$:** Calculate each coefficient: - Coefficient of $x^2$: $$a + 3b + 6c = (2b - c) + 3b + 6c = 5b + 5c.$$ - Coefficient of $x$: $$-\frac{4}{3}a - 3b - 4c = -\frac{4}{3}(2b - c) - 3b - 4c = -\frac{8}{3}b + \frac{4}{3}c - 3b - 4c = -\frac{17}{3}b - \frac{8}{3}c.$$ - Constant term: $$\frac{2}{3}a + b + c = \frac{2}{3}(2b - c) + b + c = \frac{4}{3}b - \frac{2}{3}c + b + c = \frac{7}{3}b + \frac{1}{3}c.$$ 6. **Step 4: Write $L(p(x))$ in terms of $b$ and $c$ only:** $$L(p(x)) = (5b + 5c)x^2 + \left(-\frac{17}{3}b - \frac{8}{3}c\right)x + \left(\frac{7}{3}b + \frac{1}{3}c\right).$$ 7. **Summary:** - The space $M$ is characterized by matrices with $a = 2b - c$. - The inner product is $(A,B) = \operatorname{trace}(B^T A)$. - The linear operator $L$ acts on polynomials with coefficients $a,b,c$ satisfying $a=2b-c$ and transforms them as above. This completes the analysis of the first problem.