# polyvalm

Matrix polynomial evaluation

## Description

example

Y = polyvalm(p,X) returns the evaluation of polynomial p in a matrix sense. This evaluation is the same as substituting matrix X in the polynomial, p.

## Examples

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Find the characteristic polynomial of a Pascal Matrix of order 4.

X =  pascal(4)
X = 4×4

1     1     1     1
1     2     3     4
1     3     6    10
1     4    10    20

p = poly(X)
p = 1×5

1.0000  -29.0000   72.0000  -29.0000    1.0000

The characteristic polynomial is

$p\left(x\right)={x}^{4}-29{x}^{3}+72{x}^{2}-29x+1$

Pascal matrices have the property that the vector of coefficients of the characteristic polynomial is the same forward and backward (palindromic).

Substitute the matrix, X, into the characteristic equation, p. The result is very close to being a zero matrix. This example is an instance of the Cayley-Hamilton theorem, where a matrix satisfies its own characteristic equation.

Y = polyvalm(p,X)
Y = 4×4
10-10 ×

-0.0003   -0.0036   -0.0052   -0.0143
-0.0021   -0.0136   -0.0179   -0.0464
-0.0059   -0.0330   -0.0400   -0.1047
-0.0130   -0.0639   -0.0750   -0.1962

## Input Arguments

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Polynomial coefficients, specified as a vector. For example, the vector [1 0 1] represents the polynomial ${x}^{2}+1$, and the vector [3.13 -2.21 5.99] represents the polynomial $3.13{x}^{2}-2.21x+5.99$.

Data Types: single | double
Complex Number Support: Yes

Input matrix, specified as a square matrix.

Data Types: single | double
Complex Number Support: Yes

## Output Arguments

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Output polynomial coefficients, returned as a row vector.