Matrix with complex eigenvalues but real entries

I'm doing construction problems and two of the problems ask me to make matrix that has complex eigenvalues.

The first one needs to be a 3 by 3 upper triangular matrix whose entries are real but have complex eigenvalues

The second one is 2x2 singular matrix whose eigenvalue is 3i.

In general how do i force a matrix to have complex eigenvalues

• Do you know about rational canonical form? – saulspatz Mar 9 at 22:00
• no that was not taught – Samurai Bale Mar 9 at 22:01
• Neither of these matrices can exist. Can you see why? – Joppy Mar 9 at 22:05
• Oops. I didn't look at the specs closely. You don't need to know about canonical forms. – saulspatz Mar 9 at 22:06
• for the first one since it is triangular the eigenvalues are on the diagonal and those need 2 be real, for the second determinant is 0 and the product of the eigenvalues = the determinant of the matrix in this case its 3i which can't be the case – Samurai Bale Mar 9 at 22:11

How about real Skew-symmetric matrix with full rank & even number of dimension: $$A\in \mathbb{R}^{n\times n}, A^\top= -A,~rank(A)=n, n\geq 2, n:\text{even number}$$

To understand it, need to prove for $$A\in \mathbb{R}^{n\times n}, A^\top= -A$$, (1) if $$\lambda$$ is a real eigenvalue of $$A$$, then $$\lambda=0$$. (2) if $$\lambda$$ is a complex eigenvalue, then it is pure imaginary.

Let $$v\neq 0$$ be the corresponding eigenvector; i.e., $$Av=\lambda v$$. Then,

(1) if $$\lambda$$ is real:

$$\lambda (v^\top v)=v^\top A v = (v^\top A v)^\top = v^\top A^\top v = v^\top (-A) v = -\lambda (v^\top v)$$

Because $$v^\top v\neq 0$$, we arrive at $$\lambda=0$$.

(2) if $$\lambda$$ is complex, let $$^*$$ denote the conjugate transpose:

$$\lambda (v^* v)=v^* A v= (v^* A v)^\top = \overline{ (v^* A v)^* }= \overline{v^* A^* v} = \overline{v^* (-\overline{A}) v} =\overline{v^* (-A) v}= \overline{-\lambda} (v^* v)$$

Because $$v^* v\neq 0$$, $$\lambda$$ is pure imaginary.

To construct a matrix with complex eigenvalue, it suffices to avoid zero eigenvalue; i.e., require $$rank(A)=n$$. However, if $$n$$ is an odd number, the characteristic polynomial of $$A$$ must has at least one real solution, which is exactly $$0$$ by property (1). Thus, we should also force $$n$$ to be an even number.