# An operator $T:\mathbb{R}^4\to \mathbb{R}^4$ such that $T$ has no (real) eigenvalues.

Give an example of an operator $T:\mathbb{R}^4\to \mathbb{R}^4$ such that $T$ has no (real) eigenvalues.

How can I find this operator?

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First, find a polynomial of degree 4 with no real roots. Then form the companion matrix of this polynomial (look up "companion matrix" if you are not familiar with this term - you'll be glad you did).

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Consider a rotation about the origin.

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... or two rotations, on ${\mathbb R}^2$ and ${\mathbb R}^2$. –  Robert Israel Aug 27 '12 at 4:53
think about block multiplication. –  James S. Cook Aug 27 '12 at 4:54
@Potato, What is the definition of rotation for $\mathbb{R}^4$? Thanks. –  Hiperion Aug 27 '12 at 4:55
An orthogonal matrix with determinant 1. For the purposes of constructing your example, use @RobertIsrael's hint. –  Potato Aug 27 '12 at 4:58

Another angle to look at this problem is to take the differential equation $(D^2+1)^2[y]=0$ (or $y''''+2y''+y=0$ if you prefer) and reduce it to a system of four first order ODEs by reduction of order $x_1=y,x_2=y',x_3=y'',x_4=y'''$. The matrix of this system of ODEs will have complex eigenvalues $i,-i$ corresponding to a pair of complex e-vectors and a pair of generalized complex e-vectors. This is the complementary matrix to the ODE.

$$\left[ \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & 0 & -2 & 0 \end{array} \right]$$

This matrix corresponds to $T: \mathbb{R}^4 \rightarrow \mathbb{R}^4$ with no real e-values.

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This is essentially the companion matrix for the quartic $(x^2+1)^2$. –  Gerry Myerson Aug 27 '12 at 7:13

A simple example is the linear operator corresponding to $\left( \begin{array}{ccc} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1\\ 0 & 0 & 1 & 0\\ \end{array} \right)$

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This does not correspond to a linear transformation from $\mathbb{R}^4$ to itself! –  Michael Joyce Aug 27 '12 at 5:36
Quite right. Very late. It has been fixed. –  Euler....IS_ALIVE Aug 27 '12 at 5:45

You can construct examples by doing the following:

1. Choose any diagonal matrix with non-real values on it's diagonal:

$A = \begin{pmatrix} a_{1,1} & 0 & 0 & 0 \\ 0 & a_{2,2} & 0 & 0 \\ 0 & 0 & a_{3,3} & 0 \\ 0 & 0 & 0 & a_{4,4} \end{pmatrix}$

By choosing correctly $M \in GL(4,\mathbb C)$, you can make the operator $T=M^{-1}AM$ real: $T : \mathbb R^4 \longrightarrow \mathbb R^4$.

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These examples are incorrect, for this is not a $T:\mathbb{R}^4\to\mathbb{R}^4$. –  akkkk Aug 27 '12 at 8:56
Not necessarily you can solve linear equations and make your operator real by choosing correctly the entries of $M$. –  Ilies Zidane Aug 27 '12 at 9:23
sure, but you said to choose "any". –  akkkk Aug 27 '12 at 9:33
Okay, you can. But then you need to prove this. More specifically, you would have to give an example or otherwise prove the existence of such $M$. Your answer is not very helpful like this. You translated the problem of finding $T$ to a much harder question of finding the right $M$. –  akkkk Aug 27 '12 at 9:50
I don't think I believe this. It seems to imply that given any 4 complex numbers there's a quartic polynomial with real coefficients (the characteristic polynomial of $T$) with those 4 numbers as roots. –  Gerry Myerson Aug 27 '12 at 12:43