# Eigendecomposing real matrices without knowing complex numbers

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.

• Not knowing what complex numbers are, we would consider pairs of real numbers, I suppose. But why should we not know complex numbers? Jan 20, 2020 at 19:56
• 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
– glS
Jan 20, 2020 at 20:00
• 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.
– amd
Jan 20, 2020 at 21:53

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 $$(x-\lambda_1)^2 + \lambda_2^2$$.