# polyvalm

Matrix polynomial evaluation

## Syntax

``Y = polyvalm(p,X)``

## 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.

## Version History

Introduced before R2006a