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The linear operation $L(x)$ is defined by the cross product $L(x)= b \times x$, where $b=\left[0 1 0\right]^{T}$ and $x=\left[x_{1} x_{2} x_{3}\right]^{T}$ are three dimensional vectors. The $3 \times 3$ matrix $M$ of this operation satisfies

$L(x)=M\begin{bmatrix}
x_{1}& \\
x_{2}&  \\
x_{3}& \\
\end{bmatrix}$

Then the eigenvalues of $M$ are

  1. $0, +1, -1$
  2. $1, -1, 1$
  3. $i, -i, 1$
  4. $i, -i, 0$

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1 Answer

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L(x) = b x X = [x3 0 -x1]T (Using definition of Cross Prod. from wikipedia ;) See comment below for details )

L(x) = M(3x3) x [x1 x2 x3]T

Now, what should M do with X? :D :D

Exchange the top and bottom rows, place sign, nullify middle row.

That  can be achieved with M:

[0 0 1
0 0 0
-1 0 0]

Its Eigenvalues turn out to be: 0, +i. -i (A)

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4 Comments

yes its very helpful thank u
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how 0, +i, -i? isn't it 0, +1, -1?
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Solving, [0 0 1] [0 0 0] [-1 0 0] Don't we get, x(x^2+1)=0? ;)
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