LA2.1.4

From Exampleproblems

If w is cube root of unity,show that ${\begin{bmatrix} 1 & w & w^2 \\ w & w^2 & 1 \\ w^2 & 1 & w \end{bmatrix}+\begin{bmatrix} w & w^2 & 1 \\ w^2 & 1 & w \\ w & w^2 & 1 \end{bmatrix}}\begin{bmatrix} 1 \\ w \\ w^2 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\,$

w is a cube root of unity implies $1+w+w^2=0,w^3=1\,$

Now ${\begin{bmatrix} 1 & w & w^2 \\ w & w^2 & 1 \\ w^2 & 1 & w \end{bmatrix}+\begin{bmatrix} w & w^2 & 1 \\ w^2 & 1 & w \\ w & w^2 & 1 \end{bmatrix}}\begin{bmatrix} 1 \\ w \\ w^2 \end{bmatrix}= \begin{bmatrix} 1+w & w+w^2 & w^2+1 \\ w+w^2 & w^2+1 & 1+w \\ w^2+w & 1+w^2 & w+1 \end{bmatrix} \begin{bmatrix} 1 \\ w \\ w^2 \end{bmatrix}\,$

$\begin{bmatrix} -w^2 & -1 & -w \\ -1 & -w & -w^2 \\ -1 & -w & -w^2 \end{bmatrix} \begin{bmatrix} 1 \\ w \\ w^2 \end{bmatrix}\,$

$\begin{bmatrix} -w^2-w-w^3 \\ -1-w^2-w^4 \\ -1-w^2-w^4 \end{bmatrix}=\begin{bmatrix} -w^2-w-1 \\ -1-w^2-w \\ -1-w^2-w \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\,$

Main Page:Linear Algebra

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