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The linear operation $L(x)$ is defined by the cross product $L(x)=b \times x$, where $b=\begin{bmatrix} 0 &1 & 0 \end{bmatrix}^T$ and $x=\begin{bmatrix} x_1 &x_2 & x_3 \end{bmatrix}^T$ 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 

(A) $0,+1,-1$
(B) $1,-1,1$
(C) $i,-i,1$
(D) $i,-i,0$

how to solve this..??

in Linear Algebra recategorized by
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1 Answer

4 votes
4 votes
$b=\begin{bmatrix} 0 & 1 & 0 \end{bmatrix}^{T}$

$b=\begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}$

$x=\begin{bmatrix} x_{1} & x_{2} &x _{3} \end{bmatrix}^{T}$

$x=\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix}$

 

$L(x)=b \times x$

$L(x)=\begin{vmatrix} i & j & k\\ 0 & 1 & 0\\ x_{1} & x_{2} & x_{3} \end{vmatrix}=x_{3} \ j -x_{1} \ k$

$L(x)=\begin{bmatrix} x_{3}\\ 0\\ -x_{1} \end{bmatrix}$

 

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

$\begin{bmatrix} x_{3}\\ 0\\ x_{1} \end{bmatrix}= M \begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix}$

Here,

$M=\begin{bmatrix} 0&0&1\\ 0&0&0\\ -1&0&0 \end{bmatrix}$

$M=\begin{vmatrix} 0- \lambda &0&1\\ 0&0-\lambda &0\\ -1&0&0-\lambda \end{vmatrix}$

$|A-\lambda I|=0$

$\begin{vmatrix} 0- \lambda &0&1\\ 0&0-\lambda &0\\ -1&0&0-\lambda \end{vmatrix}=0$

$-\lambda ^{3}-\lambda =0$

 

$\lambda=0,\iota ,-\iota$

 

Hence,Option(D)$\lambda=\iota ,-\iota ,0 .$

2 Comments

it is very helpful...ty
0
0
Small correction:

$L(x)=x_3 \hspace{1mm}i-x_1\hspace{1mm}k$
0
0
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