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)
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)
Show 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.
Updated On: Jul 21, 2026
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Correct Answer: 8
Solution and Explanation
Step 1: Recall what the operator norm means here.
For a linear map \(T: X \to X\) where \(X = (\mathbb{R}^3, \|\cdot\|_1)\), the operator norm is defined as \[ \|T\| = \sup_{\|v\|_1 = 1} \|Tv\|_1. \] The matrix of \(T\) is \(A = \begin{pmatrix} 1 & 1 & 3 \\ 2 & 2 & -2 \\ 1 & 3 & -3 \end{pmatrix}\).
Step 2: Get an upper bound for \(\|Tv\|_1\) in terms of the columns of \(A\).
Write \(v = (x,y,z)\) with \(\|v\|_1 = |x|+|y|+|z| = 1\). Then \(Tv = xA_1 + yA_2 + zA_3\), where \(A_1, A_2, A_3\) are the three columns of \(A\). By the triangle inequality, \[ \|Tv\|_1 \leq |x|\,\|A_1\|_1 + |y|\,\|A_2\|_1 + |z|\,\|A_3\|_1. \] If \(M\) is the largest of \(\|A_1\|_1, \|A_2\|_1, \|A_3\|_1\), the right side is at most \(M(|x|+|y|+|z|) = M\). So \(\|T\| \leq M\), where \(M\) is the largest column sum of absolute values.
Step 3: Compute the three column sums.
Column 1: \(|1|+|2|+|1| = 4\).
Column 2: \(|1|+|2|+|3| = 6\).
Column 3: \(|3|+|-2|+|-3| = 8\).
The largest of these is \(M = 8\), from column 3.
Step 4: Show this bound is actually attained.
Take \(v = e_3 = (0,0,1)\), which has \(\|v\|_1 = 1\). Then \(Tv = A_3 = (3,-2,-3)\), and \[ \|Tv\|_1 = |3|+|-2|+|-3| = 8. \] So the bound \(M=8\) from Step 2 is reached at \(v=e_3\), which means the supremum defining \(\|T\|\) equals \(8\).
Final Answer:
The operator norm of \(T\) is 8. \[ \boxed{\|T\| = 8} \]
For a linear map \(T: X \to X\) where \(X = (\mathbb{R}^3, \|\cdot\|_1)\), the operator norm is defined as \[ \|T\| = \sup_{\|v\|_1 = 1} \|Tv\|_1. \] The matrix of \(T\) is \(A = \begin{pmatrix} 1 & 1 & 3 \\ 2 & 2 & -2 \\ 1 & 3 & -3 \end{pmatrix}\).
Step 2: Get an upper bound for \(\|Tv\|_1\) in terms of the columns of \(A\).
Write \(v = (x,y,z)\) with \(\|v\|_1 = |x|+|y|+|z| = 1\). Then \(Tv = xA_1 + yA_2 + zA_3\), where \(A_1, A_2, A_3\) are the three columns of \(A\). By the triangle inequality, \[ \|Tv\|_1 \leq |x|\,\|A_1\|_1 + |y|\,\|A_2\|_1 + |z|\,\|A_3\|_1. \] If \(M\) is the largest of \(\|A_1\|_1, \|A_2\|_1, \|A_3\|_1\), the right side is at most \(M(|x|+|y|+|z|) = M\). So \(\|T\| \leq M\), where \(M\) is the largest column sum of absolute values.
Step 3: Compute the three column sums.
Column 1: \(|1|+|2|+|1| = 4\).
Column 2: \(|1|+|2|+|3| = 6\).
Column 3: \(|3|+|-2|+|-3| = 8\).
The largest of these is \(M = 8\), from column 3.
Step 4: Show this bound is actually attained.
Take \(v = e_3 = (0,0,1)\), which has \(\|v\|_1 = 1\). Then \(Tv = A_3 = (3,-2,-3)\), and \[ \|Tv\|_1 = |3|+|-2|+|-3| = 8. \] So the bound \(M=8\) from Step 2 is reached at \(v=e_3\), which means the supremum defining \(\|T\|\) equals \(8\).
Final Answer:
The operator norm of \(T\) is 8. \[ \boxed{\|T\| = 8} \]
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