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Expert-verified Found in: Page 393 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # If A is any symmetric 3x3 matrix with eigenvalues -2,3 , and 4 , and $\stackrel{\mathbf{\to }}{\mathbf{u}}{\mathbf{}}$is a unit vector in ${{\mathbf{ℝ}}}^{{\mathbf{3}}}$ , what are the possible values of the dot product $\stackrel{\mathbf{\to }}{\mathbf{u}}{\mathbf{.}}{\mathbit{A}}\stackrel{\mathbf{\to }}{\mathbf{u}}$ ?

The possible values of the dot product $\stackrel{\to }{u}.A\stackrel{\to }{u}$ are $-2\le \stackrel{\to }{u}.A\stackrel{\to }{u}\le 4$ , where is a unit vector

See the step by step solution

## Step 1: Symmetric matrix:

In linear algebra, a symmetric matrix is a square matrix that stays unchanged when its transpose is calculated. In that instance, a symmetric matrix is one whose transpose equals the matrix itself.

## Step 2: Find the possible values of the dot product u→.Au→ :

Given that, A is any symmetric 3x3 matrix with eigenvalues -2,3 and 4 , and $\stackrel{\to }{u}$ is a unit vector in ${\mathrm{ℝ}}^{3}$. From the spectral theorem, we know that there exists an orthonormal eigenbasis ${\stackrel{\to }{v}}_{3}forT$ , with associated real eigenvalues ${\lambda }_{1}=4,{\lambda }_{2}=3$ and ${\lambda }_{3}=-2$ (Arrange things so that $\left|{\lambda }_{1}\right|\ge \left|{\lambda }_{2}\right|\ge \left|{\lambda }_{3}\right|$ ). Now consider unit vector as represented below:

$\stackrel{\to }{u}={c}_{1}{\stackrel{\to }{v}}_{1}+{c}_{2}{\stackrel{\to }{v}}_{2}+{c}_{3}{\stackrel{\to }{v}}_{3}\phantom{\rule{0ex}{0ex}}⇒A\stackrel{\to }{u}={\lambda }_{1}{c}_{1}{\stackrel{\to }{v}}_{1}+{\lambda }_{2}{c}_{2}{\stackrel{\to }{v}}_{2}+{\lambda }_{3}{c}_{3}{\stackrel{\to }{v}}_{3}\phantom{\rule{0ex}{0ex}}⇒A\stackrel{\to }{u}=4{c}_{1}{\stackrel{\to }{v}}_{1}+3{c}_{2}{\stackrel{\to }{v}}_{2}+2{c}_{3}{\stackrel{\to }{v}}_{3}$

Now evaluate as follows:

$\stackrel{\to }{u}.A\stackrel{\to }{u}=\left({c}_{1}{\stackrel{\to }{v}}_{1}+{c}_{2}{\stackrel{\to }{v}}_{2}+{c}_{3}{\stackrel{\to }{v}}_{3}\right).\left(4{c}_{1}{\stackrel{\to }{v}}_{1}+3{c}_{2}{\stackrel{\to }{v}}_{2}+2{c}_{3}{\stackrel{\to }{v}}_{3}\right)\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{u}.A\stackrel{\to }{u}=4{c}_{1}^{2}+3{c}_{2}^{2}-2{c}_{3}^{2}\left(1\right)$

Since

$4{c}_{1}^{2}+3{c}_{2}^{2}-2{c}_{3}^{2}\le \left(4{c}_{1}^{2}+4{c}_{2}^{2}-4{c}_{3}^{2}=4\right)\left(2\right)$

and

$\left(-2{c}_{1}^{2}-2{c}_{2}^{2}-2{c}_{3}^{2}=-2\right)\le 4{c}_{1}^{2}+3{c}_{2}^{2}-2{c}_{3}^{2}\left(3\right)$

From (1),(2) and (3) we can imply that the possible values of the dot product areas represented below:

$-2\le \stackrel{\to }{u}.A\stackrel{\to }{u}\le 4$ ### Want to see more solutions like these? 