Question

Let \(X = (\mathbb{R}^3, \|\cdot\|_1)\), where \(\left\|\begin{pmatrix} x \\ y \\ z \end{pmatrix}\right\|_1 = |x| + |y| + |z|\), and let \(T: X \to X\) be the linear transformation defined by \[ T\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 & 1 & 3 \\ 2 & 2 & -2 \\ 1 & 3 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix}. \] The operator norm of \(T\) is equal to ______. (Answer in integer)

Hint:

For a linear map on \((\mathbb{R}^3,\|\cdot\|_1)\), the operator norm equals the largest sum of absolute values found in any single column of its matrix.

Answer 1

Correct Answer: 8

Step 1: Use the shape of the unit ball in $\|\cdot\|_1$.
The closed unit ball of $(\mathbb{R}^3, \|\cdot\|_1)$ is the octahedron with corners at $\pm e_1, \pm e_2, \pm e_3$, that is, the six points $(\pm1,0,0)$, $(0,\pm1,0)$, $(0,0,\pm1)$. Every point $v$ with $\|v\|_1 \le 1$ is a convex combination of these six corners.
Step 2: Use convexity to reduce the search to the corners.
The function $v \mapsto \|Tv\|_1$ is convex, since it is the composition of the linear map $T$ with the convex norm $\|\cdot\|_1$. A convex function on a convex set attains its maximum at an extreme point, so the maximum of $\|Tv\|_1$ over the unit ball occurs at one of the six corners. Also $\|T(-e_i)\|_1 = \|Te_i\|_1$, so it is enough to check just $e_1, e_2, e_3$.
Step 3: Evaluate $T$ at each of the three basis vectors.
The matrix of $T$ is $A = \begin{pmatrix} 1 & 1 & 3 \\ 2 & 2 & -2 \\ 1 & 3 & -3 \end{pmatrix}$, and $Te_j$ is the $j$-th column of $A$.
$Te_1 = (1,2,1)$, so $\|Te_1\|_1 = 1+2+1 = 4$.
$Te_2 = (1,2,3)$, so $\|Te_2\|_1 = 1+2+3 = 6$.
$Te_3 = (3,-2,-3)$, so $\|Te_3\|_1 = 3+2+3 = 8$.

Step 4: Pick the largest value.
Among $4, 6, 8$, the largest is $8$, attained at $v=e_3$. Since the maximum of $\|Tv\|_1$ over the whole unit ball must be one of these three numbers by Step 2, the operator norm of $T$ is exactly $8$. \[ \boxed{\|T\| = 8} \]