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

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

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.

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 $$3.13{x}^{2}-2.21x+5.99$$.

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