Diagonalization of a matrix ${}$

Question

When we do Diagonalization of a matrix, what does it mean when 1 of the condition is that the characteristic polynomial has to 'split'? Let's say i calculate the determinant with lambda, what do i have to look for so the characteristic polynomial split? does it have to do with the degree of the found polynomial?

Answer 1

A polynomial splits if it can written as a product of linear factors. That is, $p(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$ splits iff there exists $\lambda_1,\ldots,\lambda_n$ such that

$$ p(x) = a_n(x - \lambda_1) (x-\lambda_2) \cdots (x - \lambda_n). $$

The polynomial $x^2 - 1$ splits since $x^2 - 1 = (x-1)(x+1)$. Does the polynomial $x^2 + 1$ split? It depends. If we're in the real numbers, then no it doesn't. However, if we allow ourselves to use complex numbers, then it does: $x^2 +1 = (x+i)(x-i)$.

So the question of whether a polynomial splits depends over which field you take your numbers to be in. The complex numbers have a special property called algebraic closure which means that every complex-valued polynomial splits over $\Bbb C$. (This is the so-called Fundamental theorem of Algebra.) The real numbers do not have this property, as the example $x^2 + 1$ demonstates.

A real polynomial can be factored down into a product of quadratic and linear polynomials, however: $p(x) = q_1(x) \cdots q_n(x)$. Once you have done this, if any of the quadratic polynomial, $q_k$ has no real roots, then you can conclude $p$ does not split over $\Bbb R$. If it factors down to linear terms, then definitionally it splits.