1. **Problem Statement:** Verify that the function $\langle x,y \rangle = x^T M y$ defines an inner product on $\mathbb{R}^4$ where $$M = \begin{pmatrix} 2 & -1 & 0 & 0 \\ -1 & 3 & 0 & 0 \\ 0 & 0 & 4 & 1 \\ 0 & 0 & 1 & 2 \end{pmatrix}$$ and $x,y \in \mathbb{R}^4$. Vectors given are: $$u = \begin{pmatrix}1 \\ 2 \\ 0 \\ 1\end{pmatrix}, v = \begin{pmatrix}0 \\ 1 \\ 1 \\ 1\end{pmatrix}, w = \begin{pmatrix}1 \\ 0 \\ 2 \\ 0\end{pmatrix}$$ 2. **Recall the definition of an inner product:** An inner product $\langle \cdot, \cdot \rangle$ on $\mathbb{R}^n$ must satisfy for all $x,y,z \in \mathbb{R}^n$ and scalar $c$: - Positivity: $\langle x,x \rangle \geq 0$ and $\langle x,x \rangle = 0$ iff $x=0$ - Symmetry: $\langle x,y \rangle = \langle y,x \rangle$ - Linearity in the first argument: $\langle cx + y, z \rangle = c\langle x,z \rangle + \langle y,z \rangle$ 3. **Check symmetry:** Since $\langle x,y \rangle = x^T M y$, symmetry requires $M$ to be symmetric. Check $M^T$: $$M^T = \begin{pmatrix} 2 & -1 & 0 & 0 \\ -1 & 3 & 0 & 0 \\ 0 & 0 & 4 & 1 \\ 0 & 0 & 1 & 2 \end{pmatrix} = M$$ So $M$ is symmetric. 4. **Check positivity:** We must show $x^T M x > 0$ for all $x \neq 0$. Since $M$ is block diagonal: $$M = \begin{pmatrix} A & 0 \\ 0 & B \end{pmatrix}$$ where $$A = \begin{pmatrix} 2 & -1 \\ -1 & 3 \end{pmatrix}, B = \begin{pmatrix} 4 & 1 \\ 1 & 2 \end{pmatrix}$$ Check if $A$ and $B$ are positive definite by leading principal minors: - For $A$: - $\det(2) = 2 > 0$ - $\det(A) = 2 \times 3 - (-1) \times (-1) = 6 - 1 = 5 > 0$ - For $B$: - $\det(4) = 4 > 0$ - $\det(B) = 4 \times 2 - 1 \times 1 = 8 - 1 = 7 > 0$ Both $A$ and $B$ are positive definite, so $M$ is positive definite. Therefore, $x^T M x > 0$ for all $x \neq 0$. 5. **Check linearity:** Linearity in the first argument follows from matrix multiplication properties: $$\langle cx + y, z \rangle = (cx + y)^T M z = c x^T M z + y^T M z = c \langle x,z \rangle + \langle y,z \rangle$$ 6. **Conclusion:** Since $M$ is symmetric and positive definite, and the function is linear in the first argument, $\langle x,y \rangle = x^T M y$ defines an inner product on $\mathbb{R}^4$. 7. **Verification with given vectors:** Calculate $\langle u,v \rangle$: $$u^T M v = \begin{pmatrix}1 & 2 & 0 & 1\end{pmatrix} \begin{pmatrix} 2 & -1 & 0 & 0 \\ -1 & 3 & 0 & 0 \\ 0 & 0 & 4 & 1 \\ 0 & 0 & 1 & 2 \end{pmatrix} \begin{pmatrix}0 \\ 1 \\ 1 \\ 1\end{pmatrix}$$ First compute $M v$: $$M v = \begin{pmatrix} 2 \times 0 + (-1) \times 1 + 0 + 0 = -1 \\ -1 \times 0 + 3 \times 1 + 0 + 0 = 3 \\ 0 + 0 + 4 \times 1 + 1 \times 1 = 5 \\ 0 + 0 + 1 \times 1 + 2 \times 1 = 3 \end{pmatrix} = \begin{pmatrix} -1 \\ 3 \\ 5 \\ 3 \end{pmatrix}$$ Then $$u^T (M v) = 1 \times (-1) + 2 \times 3 + 0 \times 5 + 1 \times 3 = -1 + 6 + 0 + 3 = 8$$ Similarly, $\langle v,u \rangle = v^T M u$ will also be 8, confirming symmetry. **Final answer:** The function $\langle x,y \rangle = x^T M y$ with the given $M$ defines an inner product on $\mathbb{R}^4$.