$3$-by-$3$ Real Matrices with Repeated Eigenvalues

Question

The questions I have are as follows.

1. Prove that for $3 \times 3$ matrices with repeated eigenvalues, all eigenvalues are real.

2. Prove that if two eigenvalues of $3 \times 3$ are complex conjugate, then in some real basis, it takes the form $\begin{bmatrix} a & b & 0 \\ -b & a & 0 \\ 0 & 0 & \lambda \end{bmatrix}$.

I have already proven that if the 3x3 matrix has distinct eigenvalues, then there are either 3 real eigenvalues or 1 real eigenvalue and 2 complex conjugate eigenvalues. How can I use this fact to prove 1? Can I just make an eigenvalue equal to $a+bi$ with $b=0$ and prove it?

As for second question, I have no clue how to do this. Any help as to how should I approach this?

Thank you.

Answer 1

I guess you are under the assumption that your matrices have real coefficients. Therefore if $z$ is a root of the characteristic polynomial, then also $\bar{z}$ is a root.

If $z\in\mathbb{C}$ and $z\notin\mathbb{R}$, then $z\ne\bar{z}$. If the double root is complex not real, you would have three complex non real roots: this is a contradiction, because a degree $3$ polynomial (with real coefficients) has at least a real root.

In a different wording: if the roots are $\lambda$ and $a\pm bi$, with $b\ne0$, then these roots are distinct.

Assume now that the eigenvalues are $\lambda\in\mathbb{R}$ and $a\pm bi\in\mathbb{C}$, with $b\ne0$. If $f\colon\mathbb{R}^3\to\mathbb{R}^3$ is defined by $f(v)=Av$ (where $A$ is the given matrix), then you want to find a basis such that \begin{align} f(v_1)&=av_1-bv_2\\ f(v_2)&=bv_1+av_2\\ f(v_3)&=\lambda v_3 \end{align} The choice of $v_3$ is obvious: it must be an eigenvector relative to $\lambda$.

Take an eigenvector $w$ relative to $a+bi$ and split it as $w=v_1+iv_2$, where $v_1,v_2\in\mathbb{R}^3$. Then $Aw=(a+bi)w$ translates into $$ Av_1+iAv_2=(av_1-bv_2)+i(bv_1+av_2) $$ Can you finish?

Answer 2

1) You're on the right track. Since the two cases are 3 real or 1 real and 2 complex conjugates, the only case where they're not all real is if the repeated roots are complex conjugates. Thus, we would need a number whose complex conjugate is itself, which must be a real number ($a+bi = a-bi \Rightarrow b = 0$). Thus, the 1 real and 2 complex conjugates case collapses to the 3 real case.

2) This is an application of the fact that two nonsingular matrices with the same eigenvalues are similar, and the fact that the matrix in (2) has the eigenvalues $\lambda, a\pm bi$.

Answer 3

HINT

Regarding $1)$ take the characteristic polynomial of the matrix, say $q(x)$.

The eigenvalues are its roots. Also $\deg q(x)\le3$

If it has repeated eigenvalues, it means its roots are $r_1=r_2, r_3$.

But if $r$ is a root, $\bar{r}$ must also be a root..