Linear Algebra: Quadratic Forms and Matrix Norms

I'm posting the answer to the first part so that you can review it in the meantime. I will answer the second part as soon as you clarify the meaning of "matrix norm" here.

3a. To compute $A$, we pick $a_{ii}$ to be the coefficient of $x_i^2$, and $a_{ij} = a_{ji}$ to be half of the coefficient of $x_i x_j$; we get the matrix \[ A = \begin{pmatrix} 1 & 1 & -2 \\ 1 & 0 & 1 \\ -2 & 1 & 1 \end{pmatrix}. \]  In fact, the matrix is symmetric, and we have \[ \begin{pmatrix} x_1 & x_2 & x_3 \end{pmatrix} \begin{pmatrix} 1 & 1 & -2 \\ 1 & 0 & 1 \\ -2 & 1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} x_1 & x_2 & x_3 \end{pmatrix} \begin{pmatrix} x_1 +x_2 - 2x_3 \\ x_1 + x_3 \\ -2x_1 + x_2 + x_3 \end{pmatrix} \] which is equal to \[ x_1^2 + x_1x_2 - 2x_1x_3 + x_1x_2 + x_2x_3 - 2x_1x_3 + x_2x_3 + x_3^2 \] or, collecting the monomials, to \[ x_1^2 + _3^2 + 2x_1x_2 - 4x_1x_3 + 2x_2x_3 \] which is exactly our quadratic form.

3b. First we compute the eigenvalues. We compute the characteristic polynominal \[ p(x) = \det \begin{pmatrix} 1-x & 1 & -2 \\ 1 & -x & 1 \\ -2 & 1 & 1-x \end{pmatrix} = -x^3 +2x^2 +5x -6. \] We notice immediately that $p(1) = 0$, so we can collect $(1-x)$ and get \[ p(x) = (1-x)(x^2-x-6) = -(x-1)(x+2)(x-3) \] so the eigenvalues are $-2,1,3$. We already computed \[ \begin{pmatrix} 1 & 1 & -2 \\ 1 & 0 & 1 \\ -2 & 1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} x_1 +x_2 - 2x_3 \\ x_1 + x_3 \\ -2x_1 + x_2 + x_3 \end{pmatrix} \] so all we have to do is to set $(x_1,x_2,x_3)$ equals to $-2$, $1$, and $3$ times itself.

Let's start with $1$. We get \[ \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} x_1 +x_2 - 2x_3 \\ x_1 + x_3 \\ -2x_1 + x_2 + x_3 \end{pmatrix}. \] Substituting $x_2 = x_1 + x_3$, we get $x_1 = 2x_1 - x_3$ which gives $x_1 = x_3$. Then, $x_2 = x_1 + x_3 = 2x_1$, and the eigenvector relative to $1$ is $(1,2,1)^T$.

For $3$, we get \[ \begin{pmatrix} 3 x_1 \\ 3 x_2 \\ 3 x_3 \end{pmatrix} = \begin{pmatrix} x_1 +x_2 - 2x_3 \\ x_1 + x_3 \\ -2x_1 + x_2 + x_3 \end{pmatrix}. \] Multiplying the first by $3$ and substituting the second in, we get $9x_1 = 3x_1 + 3x_2 - 6x_3 = 4x_1 - 5x_3$ which gives $5 x_1 = -5 x_3$, so $x_1 = -x_3$. Then, $x_2 = x_1 + x_3 = x_1 =- x_1 = 0$, and the eigenvector relative to $3$ is then $(1, 0, -1)^T$.

For $-2$, we get \[ \begin{pmatrix} -2 x_1 \\ -2 x_2 \\ -2 x_3 \end{pmatrix} = \begin{pmatrix} x_1 +x_2 - 2x_3 \\ x_1 + x_3 \\ -2x_1 + x_2 + x_3 \end{pmatrix}. \] Multiplying the first by $-2$ and substituting the second in, we get $4x_1 = -2x_1 + -2x_2 +4x_3 = -x_1 + 5x_3$ which gives $5 x_1 = 5 x_3$, so $x_1 = x_3$. Then, $-2 x_2 = x_1 + x_3 = 2x_1$, so $x_2 = -x_1$. The eigenvector relative to $3$ is then $(1, -1, 1)^T$.

To write the orthonormal matrix we have to divide them by the norm, and get \[ P = \begin{pmatrix} \frac{1}{\sqrt{6}}& \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} \\ \frac{2}{\sqrt{6}}& \frac{-1}{\sqrt{3}} & 0 \\ \frac{1}{\sqrt{6}}& \frac{1}{\sqrt{3}} & \frac{-1}{\sqrt{2}} \end{pmatrix} \]

3c. The maximum and the minimum of $q$ correspond to the maximum and the minimum of the norms of the eigenvalues, so the maximum is $3$ (which is achieved by $x_1 = \frac{1}{\sqrt{2}}$, $x_2 = 0$, $x_3 = \frac{-1}{\sqrt{2}}$), and the minum is $1$ (which is achieved by $x_1 = \frac{1}{\sqrt{6}}$, $x_2 = \frac{2}{\sqrt{6}}$, $x_3 = \frac{1}{\sqrt{6}}$).

4. As per the addendum,  the matrix norm is the norm induced by the vector norm, i.e. the maximum of the norm of $Ax$ for $x$ unit vector. This is equal to the square root of the largest (in norm) eigenvalue of $A^T A$.

For the first matrix, we have \[ A^T A = \begin{pmatrix} 2 & -1 \\ 2 & 2 \end{pmatrix} \begin{pmatrix} 2 & 2 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 5 & 2 \\ 2 & 8 \end{pmatrix}.  \] The characteristic polynomial is \[ (5-x)(8-x) - 4 = 40 -13x + x^2 - 4 = x^2 - 13x + 36 = (x-4)(x-9) \] so the largest eigenvalue is $9$, and \[ \lVert A \rVert = \sqrt{9} = 3. \]

For the second matrix, we have \[ B^T B = \begin{pmatrix} 1 & 3 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 3 & -1 \end{pmatrix} = \begin{pmatrix} 10 & -3 \\ -3 & 1 \end{pmatrix}. \] The characteristic polynomial is \[ (10-x)(1-x) - 9 = 10 -11x + x^2 - 9 = x^2 - 11x + 1 \] whose roots are $\frac{1}{2}(11-3\sqrt{13})$ and $\frac{1}{2}(11+3\sqrt{13})$. The largest of the two is the second, so \[ \lVert B \rVert = \sqrt{\frac{1}{2}(11+3\sqrt{13})} \]