# find eigen values of given matrix

If $w$ is a non-real cube root of unity, then eigenvalues of the matrix:

$$\begin{pmatrix} 1&1&1 \\ 1&w&w^2 \\ 1&w^2&w^4 \end{pmatrix} \begin{pmatrix} 1&0&0 \\ 0&-1&0 \\ 0&0&0 \end{pmatrix} \begin{pmatrix} 1&1&1 \\ 1&1/w&1/w^2 \\ 1&1/w^2&1/w^4 \end{pmatrix}$$

are which of these:

(a) $1,-1,0$

(b) $\frac{1}{3},-\frac{1}{3},0$

(c) $1,w,w^2$

(d) $3,-3,0$

Please suggest some trick to find eigen value of this matrix given as product of 3 matrices, because to actually compute product of these matrix would not be a feasible idea in a exam.

• Actually, calculating the product may not take so much time as to trying to ifnd out a "trick". After all, the center matrix is almost an identity matrix and I see bunch of reciprocals in the third w.r.t the first. Feb 2, 2015 at 14:44
• Product becomes little complicated Feb 2, 2015 at 14:46
• The hint of Omnomnomnom is very good. Think about it, and remember that $\omega^3=1$, so $\omega^3-1=0 \iff (1+\omega+\omega^2)(\omega-1)=0$, but $\omega \neq 1$ so $1 + \omega + \omega^2 =0$. Also if $\frac{1}{3}PAP^{-1}$ has eigenvalues $1,-1$, then $PAP^{-1}$ has eigenvalues multiplied by $3$. Feb 2, 2015 at 15:29
• @Ale Ya alright thank you very very much Feb 2, 2015 at 17:14
• @singularity it is useful to note that in general, $$\overline{\omega} = |\omega|^2/\omega$$ so, in the case that $|\omega| = 1$, we have $$\overline{\omega} = \omega^{-1}$$ so, we indeed have that the two matrices at the end are the conjugate transpose of each other. Feb 2, 2015 at 21:56

$$\frac 1{\sqrt{3}}\pmatrix{ 1&1&1\\ 1&\omega & \omega^2\\ 1 & \omega^2 & \omega^4 }$$ is unitary (since its columns are orthonormal). From there, it's easy to see that $$\left[ \frac 1{\sqrt{3}}\pmatrix{ 1&1&1\\ 1&\omega & \omega^2\\ 1 & \omega^2 & \omega^4 } \right]^{-1}= \frac 1{\sqrt{3}}\pmatrix{ 1&1&1\\ 1&\omega^{-1} & \omega^{-2}\\ 1 & \omega^{-2} & \omega^{-4} }$$
• Orthonormal means that any two columns are mutually orthogonal, and every column has length (AKA magnitude, norm) $1$. Feb 2, 2015 at 16:24
hint: first and third matrix are 'nearly' inverse of each other . let the first matrix be $A$ second $B$ , the third if you notice is $3A^{-1}$. so now $3ABA^{-1}-kAIA^{-1}=0$ therefore the solution of $k$ is nothing but $|3B-kI|=0$ which is quite simple.note:$3B-kI=0$ does not have to hold true , as product together can be zero. the determinant of other two matrices is non zero so $|B|$ has to be zero. hope this helps