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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. $i, -i, 0$

edited | 276 views

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)

edited
How did you get b*X=[ x3  0  x2 ] ???

Thanks for pointing it out! It was a typo :D

It should have been [x3 0 -x1]T
Edited. :)

i could not understand how u got it actually i could not get how u multiplied this ???

As far as I understand,

b,X are 3-d vectors, i.e having 3 components. Following representation is of scalar part.

b=[0 1 0]T
X=[x1 x2 x3]T

b x X =

|i j k|

|0 1 0|

|x1  x2 x3|

(I'm not good with latex or any kind of formatting this site uses :( that's just determinant form of Cross Prod.)

You get: x3 along i, 0 along j, x1 along k

i,j,and k are just three unit vectors, representing the dimensions.

So, your resultant vector (scalar part) is : [x3 0 -x1]T

Its like a point in 3-d space (like we did in physics).

Does it help?

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

+1 vote