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I'm trying to construct a normal matrix $A\in \mathbb R^{n\times n}$ such that all it's eigenvalues are complex but at least one of then also has a positive real part (well, at least two then, since the entries are real they'd come in pairs). Is that possible?

If the desired matrix exists, we'd find it among normal matrices which are neither hermitian nor skew-hermitian. E.g.

$$ \begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \\ \end{pmatrix} $$

which is orthogonal up to scaling .

My attempt: obviously, $n$ needs to be even, so I've tried a companion matrix of $t^2+2t+2$

$$ \begin{pmatrix} 0 & -2 \\ 1 & -2 \\ \end{pmatrix} $$

(it's eigenvalues are $1\pm i$) which turned out not to be normal.

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Just add a positive multiple of the identity to a rotation matrix, to make the eigenvalues have positive real part. $$ \begin{pmatrix}3&-1\\1&3\end{pmatrix}. $$

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In order to get a real matrix with no real eigenvalue, $n$ must be even. Then you can construct any normal matrix with nonreal eigenvalues having positive real parts as follows:

  1. Take a block diagonal matrix $D:=D_1\oplus\cdots\oplus D_k$, $k=n/2$, with the diagonal blocks $D_i$ of the form $$\tag{1} D_i=\begin{bmatrix}\alpha&\beta\\-\beta&\alpha\end{bmatrix}, $$ where $\alpha$ and $\beta$ are real, $\alpha>0$ and $\beta\neq 0$. It is easy to see that all blocks $D_i$ (and hence $D$) are normal with eigenvalues $\alpha\pm i\beta$.
  2. Take any $n\times n$ orthogonal matrix $Q$.
  3. Set $A:=QDQ^T$.

With a slight modification of Item 1, you can construct any real normal matrix with any real (by $1\times 1$ blocks $D_i$) and complex (by $2\times 2$ blocks $D_i$ of the form (1)) eigenvalues (where the complex eigenvalues of course appear in conjugate pairs) by this process. In fact, Items 1-3 define $A$ by constructing its real Schur form.

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