The question, pure curiosity, is whether you can solve a quadratic with the use of matrices?

And if yes, does that method also work for higher polynomials?

Say for example I have a quadratic such as written below: \begin{equation} x^2+3x+2=0 \end{equation} By simple other known methods, one ends up with $x = -1$ or $-2$.

(I already know it is possible to find the solution of a system of simultaneous equations... Maybe does that help?)

Can you get to those results with matrices? Thanks a lot for your contribution and answers!

  • $\begingroup$ I don't think so, at least not in a more or less "natural" way, as far as I can see. What you can do with matrices is to classify two-dimensional or three-dimensional quadratics, and to find out whether some given quadratic by means of the symmetric matrix determined by its coefficients, and the matrix's determinant, signature, etc. $\endgroup$ – Timbuc Oct 28 '14 at 18:33
  • $\begingroup$ matrices are usually used to solve linear equations. $\endgroup$ – Mustafa Said Oct 28 '14 at 18:57

Here's one way to do use matrices here: we are given the problem $$ u^2 + 3u + 2 = 0 $$ substituting $u = x/y$, we have $$ x^2/y^2 + 3x/y + 2 = 0 \implies\\ x^2 + 3xy + 2y^2 = 0 \implies\\ \pmatrix{x&y} \pmatrix{1&3/2\\3/2&2} \pmatrix{x\\y} = 0 $$ this is now a problem about quadratic forms, which can be solved using knowledge about symmetric/Hermitian matrices.

This particular method doesn't extend to higher degree polynomials.

Another approach: the zeros of a polynomial are precisely the eigenvalues of the associated companion matrix. This works for polynomials of any degree.

In your particular case, we can find the zeros of $x^2 + 3x + 2$ by finding the eigenvalues of the matrix $$ \pmatrix{0&-2\\1&-3\\} $$

  • 3
    $\begingroup$ Matlab, for example, finds the roots of a polynomial (numerically) by constructing the companion matrix and computing its eigenvalues. $\endgroup$ – Robert Israel Oct 28 '14 at 20:00
  • $\begingroup$ Wow! Thanks! That is extremly useful to know! Thanks a lot! $\endgroup$ – user3604362 Oct 31 '14 at 18:32

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