 # Prove that if B is symmetric, then |B| is the largest eigenvalue of B. ankarskogC 2021-01-06 Answered
Prove that if B is symmetric, then $‖B‖$ is the largest eigenvalue of B.
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Step 1
Calculation:
To find the norm of the matrices:
a)$\left(\begin{array}{cc}4& 0\\ 1& 3\end{array}\right)$
The 1-norm is $‖A{‖}_{1}={\text{max}}_{1\le j\le n}\left(\sum _{i=1}^{n}|{a}_{ij}|\right)$
Using the above formula
$‖A{‖}_{1}=\text{max}\left(4+1,0+3\right)$
$=\text{max}\left(5,3\right)$
=5
Step 2
b)$\left(\begin{array}{cc}5& 3\\ -3& 3\end{array}\right)$
Using the formula , we get
$‖A{‖}_{1}=\text{max}\left(5+3,3+3\right)$
$=\text{max}\left(8,6\right)$
=8
Step 3
c)$\left(\begin{array}{ccc}1& \frac{-2}{\sqrt{3}}& 0\\ 0& \frac{-2}{\sqrt{3}}& 1\\ 0& \frac{2}{\sqrt{3}}& 1\end{array}\right)$
Using the formula , we get
$‖A{‖}_{1}=\text{max}\left(1+0+0,\frac{2}{\sqrt{3}}+\frac{2}{\sqrt{3}}+\frac{2}{\sqrt{3}},0+1+1\right)$
$=\text{max}\left(1,\frac{6}{\sqrt{3}},2\right)$
$=\frac{6}{\sqrt{3}}$