Question

If \( \omega \) is a complex cube root of unity, show that:

\[ \left( \begin{bmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{bmatrix} + \begin{bmatrix} \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \\ \omega & \omega^2 & 1 \end{bmatrix} \right) \begin{bmatrix} 1 \\ \omega \\ \omega^2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

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Solution

We use the identity of cube roots of unity:

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Step 1: Add the matrices

\[ = \begin{bmatrix} 1+\omega & \omega+\omega^2 & \omega^2+1 \\ \omega+\omega^2 & \omega^2+1 & 1+\omega \\ \omega^2+\omega & 1+\omega^2 & \omega+1 \end{bmatrix} \begin{bmatrix} 1 \\ \omega \\ \omega^2 \end{bmatrix} \] —

Step 2: Multiply (Single Column Result)

\[ = \begin{bmatrix} (1+\omega)\cdot1 + (\omega+\omega^2)\cdot\omega + (\omega^2+1)\cdot\omega^2 \\ (\omega+\omega^2)\cdot1 + (\omega^2+1)\cdot\omega + (1+\omega)\cdot\omega^2 \\ (\omega^2+\omega)\cdot1 + (1+\omega^2)\cdot\omega + (\omega+1)\cdot\omega^2 \end{bmatrix} \] —

Step 3: Simplify

Using \( \omega^3 = 1 \) and \( 1+\omega+\omega^2=0 \), Each row becomes: \[ (1+\omega+\omega^2) + (1+\omega+\omega^2) = 0 \] —

Final Result

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Conclusion

Hence proved.

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