Let $A$ be a real squared matrix. A scalar $\lambda\in\mathbb C$ is an eigenvalue for $A$ iff $$Av=\lambda v \tag A$$ for some complex vector $v$.

This condition can be equivalently written in terms of purely real quantities as the following system: $$\begin{cases} (A-\lambda_1 I)v_1 = - \lambda_2 v_2, \\ (A-\lambda_1 I)v_2 = \phantom{-}\lambda_2 v_1, \end{cases} \tag B$$ as can be seen by decomposing the quantities in (A) into real and imaginary parts: $\lambda=\lambda_1+i\lambda_2$ and $v=v_1+i v_2$. If we didn't know anything about complex numbers, we would be working directly on (B), asking for a pair of reals $\lambda_1,\lambda_2\in\mathbb R$ such that (B) is satisfied for some real vectors $v_1,v_2$.

This pair of conditions can be seen to imply to the following ones: $$\begin{cases} [(A-\lambda_1 I)^2 + \lambda_2^2 I ]v_1 = 0, \\ [(A-\lambda_1 I)^2 + \lambda_2^2 I ]v_2 = 0. \end{cases} \tag C$$ This follows from applying $(A-\lambda_1 I)$ twice to either $v_1$ or $v_2$, and using (B). This, on the other hand, is equivalent to the condition $$\det[(A-\lambda_1 I)^2 + \lambda_2^2 I] = 0. \tag D$$ See also this post about the equivalence of (A) and (D).

From complex analysis we know that, given an arbitrary real matrix $A$, there must be a pair of reals $\lambda_1,\lambda_2$ such that (D) is verified. Not knowing what complex numbers are, how would we go in finding such values for a given $A$? The determinant equation gives a polynomial of two variables which I'm not sure how to handle.

  • $\begingroup$ Not knowing what complex numbers are, we would consider pairs of real numbers, I suppose. But why should we not know complex numbers? $\endgroup$ Jan 20, 2020 at 19:56
  • $\begingroup$ cause it's fun to think about =)? Also, from complex analysis we know that these kinds of real systems are solvable, which suggests that there is also probably a way to solve them working only in $\mathbb R$. I'm just curious how one would go into handling the situation remaining in the reals $\endgroup$
    – glS
    Jan 20, 2020 at 20:00
  • $\begingroup$ When a real matrix has a complex conjugate eigenvalue pair, they correspond to an invariant plane. So, one approach is to search for invariant planes that aren’t generated by real eigenvectors. $\endgroup$
    – amd
    Jan 20, 2020 at 21:53

1 Answer 1


The $\lambda$-eigenspace of an operator is associated to the linear polynomial $p(x) = x - \lambda$:

  • The $\lambda$-eigenspace of $A$ is the set $\{v \mid p(A)v = 0\}$.
  • The presence of $(x - \lambda)$ in the characteristic polynomial of $A$ tells you that there is a $\lambda$-eigenspace.

For example, if the characteristic polynomial of $A$ is $x^2 - 2x - 3$, then I can factorise it to $(x-3)(x+1)$ and hence I know that the two eigenvalues are $3$ and $-1$.

However, over the real numbers not every real polynomial factorises into linear parts: the best we can do is linear parts and quadratic parts. Say for example that the characteristic polynomial of $A$ was $(x-1)(x^2 + 1)$. Then the $(x-1)$ term tells me that there is a one-dimensional subspace $L$ such that $(A-1)L = 0$ (i.e. a 1-eigenspace), while the $(x^2 + 1)$ part tells me that there is a two-dimensional subspace $P$ such that $(A^2 + 1)P = 0$.

For each irreducible quadratic you see, there is a pair of complex conjugates associated to it (the roots). This pair of complex conjugates are your $\lambda_1 \pm i\lambda_2$. For example, for the polynomial $x^2 + 1$, the complex conjugates would be $\pm i$.

If you for some reason absolutely didn't want to touch complex numbers, and you had an irreducible quadratic $ax^2 + bx + c = 0$ that you were trying to extract $\lambda_1$ and $\lambda_2$ from, you could just break up the quadratic formula into real and imaginary parts: $$\lambda_1 = \frac{-b}{2a}, \quad \lambda_2 = \pm \frac{\sqrt{4ac - b^2}}{2a}$$ This is not really a cheat, since we're just completing the square on $ax^2 + bx + c$ to make it look more like $(x-\lambda_1)^2 + \lambda_2^2$.


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