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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

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  • $\begingroup$ Do you know about rational canonical form? $\endgroup$ – saulspatz Mar 9 at 22:00
  • $\begingroup$ no that was not taught $\endgroup$ – Samurai Bale Mar 9 at 22:01
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    $\begingroup$ Neither of these matrices can exist. Can you see why? $\endgroup$ – Joppy Mar 9 at 22:05
  • $\begingroup$ Oops. I didn't look at the specs closely. You don't need to know about canonical forms. $\endgroup$ – saulspatz Mar 9 at 22:06
  • $\begingroup$ 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 $\endgroup$ – Samurai Bale Mar 9 at 22:11
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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.

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