The linear operation L(x) is defined by the cross product L(x) = b × X, where b = [0 1 0]T and X = [x1x2x3]T are three dimensional vectors. The 3 × 3 matrix M of this operation satisfies \[{\text{L}}\left( {\text{x}} \right) = {\text{M}}\left[ {\begin{array}{*{20}{c}} {{{\text{x}}_1}} \\ {{{\text{x}}_2}} \\ {{{\text{x}}_3}} \end{array}} \right].\] Then the eigen values of M are A. 0, +1, -1 B. 1, -1, 1 C. i, -i, 1 D. i, -i, 0

The correct answer is $\boxed{\text{(C)}}$.

The eigenvalues of a matrix are the roots of its characteristic polynomial. The characteristic polynomial of a matrix $M$ is given by

$$p(x) = \det(xI – M).$$

In this case, we have

$$p(x) = \det \begin{bmatrix} x & -x_2 & -x_3 \\ x_1 & x & -x_3 \\ x_2 & x_3 & x \end{bmatrix} = x^3 – x.$$

The roots of this polynomial are $x = 0$ and $x = \pm 1$. Therefore, the eigenvalues of $M$ are $0$, $1$, and $-1$.

To see this, we can also note that the matrix $M$ can be written as

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

This matrix is diagonalizable, with eigenvalues $0$, $1$, and $-1$. Therefore, the eigenvalues of $M$ are also $0$, $1$, and $-1$.

Here is a brief explanation of each option:

• Option (A): The eigenvalues of $M$ cannot be $0$, $+1$, and $-1$, because these three numbers are not the roots of the characteristic polynomial of $M$.
• Option (B): The eigenvalues of $M$ cannot be $1$, $-1$, and $1$, because these three numbers are not the roots of the characteristic polynomial of $M$.
• Option (C): The eigenvalues of $M$ are $0$, $1$, and $-1$, because these are the roots of the characteristic polynomial of $M$.
• Option (D): The eigenvalues of $M$ cannot be $i$, $-i$, and $1$, because these three numbers are not the roots of the characteristic polynomial of $M$.