# Degenerate eigenvalues and finding normalized eigenvectors

I am trying to find the eigenvalues and the normalized eigenvectors of the matrix: $$\begin{bmatrix}0 & i & 0\\0 & 0& i\\i & 0 & 0 \end{bmatrix}$$

It is stated that the eigenvalues are : $$\lambda_1 = i\omega$$, $$\lambda_2 = i$$ and $$\lambda_3 = i\omega^2$$ and two of the normalised eigenvectors are: $$\frac{1}{\sqrt 3}\begin{bmatrix}1\\1\\1\end{bmatrix}$$ and $$\frac{1}{\sqrt 3}\begin{bmatrix}1\\\omega^2\\\omega\end{bmatrix}$$.

I'm not sure what $$\omega$$ means in this case and how to arrive to this result, given that $$i$$ is a degenerate eigenvalue.

• I suspect that $\omega$ has been used in this way previously in your study materials. – amd Apr 23 at 17:38

The matrix is equal to $$iM$$, where$$M=\begin{bmatrix}0&1&0\\0&0&1\\1&0&0\end{bmatrix}.$$Therefore, its eigenvalues are the eigenvalues of $$M$$ times $$i$$. But the characteristic polynomial of $$M$$ is $$1-x^3$$ and therefore its eigenvalues are $$\{1,\omega,\omega^2\}$$, where$$\omega=-\frac12\pm\frac{\sqrt3}2i$$

$$\omega$$ is used for a lot of different things in math, but in this context it's generally used to represent a non-principal cube root of $$1$$. That is, $$\omega$$ is a number such that $$\omega^3=1$$ but $$\omega \neq 1$$. While there are two numbers that satisfy that relation, they are conjugates of each other and thus indistinguishable by polynomials over the reals. (Taking different values to be $$\omega$$ will result in the expressions you have evaluating to the same set of eigenvalues, albeit in a different order.) If a particular number is needed, it's generally taken to be the first one, going counter-clockwise, from $$1$$ in the complex plane, i.e. -$$\frac12+\frac{\sqrt3}2i$$.

$$i$$ is not degenerate; $$\frac{1}{\sqrt 3}\begin{bmatrix}1\\1\\1\end{bmatrix}$$ is the only (up to scaling) eigenvalue for that eigenvalue. Both its arithmetic and geometric multiplicities are $$1$$.

• Another usual notation is $j$. – Bernard Apr 23 at 17:40

Here a direct calculation.

The characteristic polynomial of our matrix $$A=\left[\begin{array}{rrr} 0 & i & 0 \\ 0 & 0 & i \\ i & 0 & 0 \end{array}\right]$$ is given by \begin{align*} \chi_A(t) &= \det(t\cdot I_3-A) \\ &= \left|\begin{array}{rrr} t & -i & 0 \\ 0 & t & -i \\ -i & 0 & t \end{array}\right| \\ &\overset{\ast}{=}\left|\begin{array}{rrr} t - i & t - i & t - i \\ 0 & t & -i \\ -i & 0 & t \end{array}\right| \\ &= (t-i)\left|\begin{array}{rrr} 1 & 1 & 1 \\ 0 & t & -i \\ -i & 0 & t \end{array}\right| \\ &\overset{\circledast}{=}(t-i)\left|\begin{array}{rrr} 1 & 1 & 1 \\ 0 & t & -i \\ 0 & i & t + i \end{array}\right| \\ &= (t-i)\left|\begin{array}{rr} t & -i \\ i & t + i \end{array}\right| \\ &= (t-i)\cdot\{t^2+i\,t+i^2\} \\ &=(t-i)\cdot\{t^2+i\,t-1\} \\ &\overset{q}{=} (t-i)\cdot\left(t-\frac{-i+\sqrt{3}}{2}\right)\cdot \left(t-\frac{-i-\sqrt{3}}{2}\right) \end{align*} The equality marked $$\ast$$ is obtained by replacing row one with the sum of all three rows of the previous matrix. The equality marked $$\circledast$$ is obtained by adding $$i$$ times row one to row three. The equality marked $$q$$ is obtained from the quadratic formula.

Our factorization of $$\chi_A(t)$$ shows that the eigenvalues of $$A$$ are \begin{align*} \lambda_1 &= i & \lambda_2 &= \frac{-i+\sqrt{3}}{2} & \lambda_3 &= \frac{-i-\sqrt{3}}{2} \end{align*}