If a matrix $M$ has no real eigenvalues, and if I don't know how to prove that its characteristic polynomial has no real roots, can I always prove it using Gaussian elimination on $M-xI$ and succeeding in reaching a complete row echelon form with diagonal entries either of degree 0 or 2 that have no roots ?
——–Edit following one of the answers :
Say I want to prove that a real matrix $M$ has no real eigenvalues. Suppose its characteristic polynomial $P$ isn't easy to split into quadratic polynomials. If I could use Gaussian elimination on $M-xI$ to make it upper triangular in such a way that half of its diagonal entries are nonzero constants and the other half are rootless quadratic polynomials, the question would be resolved. My question is : can this always be done ?
(If so, does that mean that for any polynomial $P$ with no real roots, applying this to method to $C(P)-xI$, where $C(p)$ denotes the companion matrix of $P$, provides a method for factorising $P$ into quadratics ?)
Best Answer
By "no eigenvalues", I assume you mean "no real eigenvalues". If that's what you mean, then yes that works. For instance, the matrix $$ M = \pmatrix{0&-1\\1&0} $$ has no real eigenvalues. To verify that, we can compute $M - x I$ and row-reduce: $$ M - x I = \pmatrix{-x & -1\\1&-x} \leadsto \pmatrix{1&-x\\-x & -1} \leadsto \pmatrix{1&-x\\0&-x^2-1}. $$ Now, note that $-x^2 - 1$ is never zero (for real values $x$). So, we can divide the bottom row by $-x^2 - 1$. So, we have row-reduced $M - xI$ to $$ M - x I \leadsto \pmatrix{1&-x\\0&1} \leadsto \pmatrix{1&0\\0&1}. $$ So for any real $x$, $M - xI$ can be row-reduced to the identity matrix. So for any real $x$, $M - xI$ is invertible, which is to say that $x$ is not an eigenvalue of $M$.
Thus, $M$ indeed has no real eigenvalues.
Keep in mind that along this process, we did ultimately end up with an entry equal to $-(x^2 + 1)$; $x^2 + 1$ is in fact the characteristic polynomial of $M$. So a long story short: if you're computing (or checking for the existence of) eigenvalues by hand, calculating $\det(M - xI)$ in some way is still the way to go.