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Linear Algebra With Applications

Otto Bretscher

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 442 pages

ISBN: 9780321796974

Chapter 8: Q11E (page 413)

The singular values of any triangular matrix are the absolute values of its diagonal entries.

Short Answer

Expert verified

The given statement is FALSE.

Step by step solution

01

Step 1: Definition of singular values:

The diagonal entries of the matrix are singular values, which are arranged in ascending order.

The singular values are found from the non- zero eigenvalues of $A^{T} A$ .

Hence, singular value is $蟽 = \sqrt{位}$ .

02

Step 2: To Find TRUE or FALSE

Let

$A = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}]$

Then, $A$ is upper triangular.

Let's check if both the singular values of $A$ are l or not.

$A^{T} A = [\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}] [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] A^{T} A = [\begin{matrix} 1 & 1 \\ 1 & 2 \end{matrix}]$

Find the singular values:

$(A^{T} A - 位 I) = 0$

$[\begin{matrix} 1 & 1 \\ 1 & 2 \end{matrix}] - 位 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = 0 [\begin{matrix} 1 & 1 \\ 1 & 2 \end{matrix}] - [\begin{matrix} 位 & 0 \\ 0 & 位 \end{matrix}] = 0 [\begin{matrix} 1 - 位 & 1 \\ 1 & 2 - 位 \end{matrix}] = 0 (1 - 位) (2 - 位) - (1) (1) = 0$

The characteristic polynomial is

$A^{T} A = (1 - 位) (2 - 位) - 1 = 位^{2} - 3 位 + 1$

The roots of the polynomial are $\frac{3 卤 \sqrt{5}}{2}$ .

The singular values are found from the non- zero eigenvalues of $A^{T} A$ .

Hence, singular value $蟽 = \sqrt{位}$ .Thus, the singular values of $A$ cannot be equal to 1.

Thus, the given statement is FALSE.

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