Note: This page has been translated by MathWorks. Click here to see

To view all translated materials including this page, select Country from the country navigator on the bottom of this page.

To view all translated materials including this page, select Country from the country navigator on the bottom of this page.

Polynomial with specified roots or characteristic polynomial

`p = poly(r)`

`p = poly(A)`

For vectors,

`r = roots(p)`

and`p = poly(r)`

are inverse functions of each other, up to roundoff error, ordering, and scaling.

The algorithms employed for `poly`

and `roots`

illustrate
an interesting aspect of the modern approach to eigenvalue computation. `poly(A)`

generates
the characteristic polynomial of `A`

, and `roots(poly(A))`

finds
the roots of that polynomial, which are the eigenvalues of `A`

.
But both `poly`

and `roots`

use `eig`

,
which is based on similarity transformations. The classical approach,
which characterizes eigenvalues as roots of the characteristic polynomial,
is actually reversed.

If `A`

is an `n`

-by-`n`

matrix, `poly(A)`

produces
the coefficients `p(1)`

through `p(n+1)`

,
with `p(1)`

`=`

`1`

,
in

$$\mathrm{det}\left(\lambda I-A\right)={p}_{1}{\lambda}^{n}+\dots +{p}_{n}\lambda +{p}_{n+1}\text{\hspace{0.17em}}.$$

The algorithm is

z = eig(A); p = zeros(n+1,1); p(1) = 1; for j = 1:n p(2:j+1) = p(2:j+1)-z(j)*p(1:j); end

This recursion is derived by expanding the product,

$$(\lambda -{\lambda}_{1})(\lambda -{\lambda}_{2})\dots (\lambda -{\lambda}_{n})\text{\hspace{0.17em}}.$$

It is possible to prove that `poly(A)`

produces
the coefficients in the characteristic polynomial of a matrix within
roundoff error of `A`

. This is true even if the eigenvalues
of `A`

are badly conditioned. The traditional algorithms
for obtaining the characteristic polynomial do not use the eigenvalues,
and do not have such satisfactory numerical properties.

Was this topic helpful?