Minimal Polynomial A nxn Matrix

• wurth_skidder_23
In summary, the conversation discusses proving the minimal polynomial for an n x n matrix with distinct eigenvalues is equal to (s-a_1)^(d_1)*...*(s-a_k)^(d_k). The index of each eigenvalue is defined as the nullspace of (A-a*I)^m, and the smallest index for which the nullspaces are equal is called the index of the eigenvalue. The conversation also mentions that the characteristic polynomial can be divided by the minimal polynomial, and discusses using good choices of f(A)v to prove the minimal polynomial.
wurth_skidder_23
Let A be an n x n matrix; denote its distinct eigenvalues by a_1,...,a_k and denote the index of a_i by d_i. How do I prove that the minimal polynomial is then:

m_A(s) = (s-a_1)^d_1*...*(s-a_k)^d_k

?

The characterstic polynomial is defined as:

p_A(s) = (s-a_1)*...*(s-a_n);

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the index of a_i by d_i
What do you mean by index? If you mean "multiplicity", then you can't prove it. For example, consider the minimum polynomials of the zero and identity matrices. (1 and x-1, respectively)

The book defines the index as:

"N_m = N_m(a) the nullspace of (A-a*I)^m. The subspaces N_m consist of generalized eigenvectors; they are indexed increasingly, that is N_1 is included in N_2 is included in...

Since these are subspaces of a finite-dimensional space, they must be equal from a certain index on. We denote by d = d(a) the smallest such index, that is, N_d = N_(d+1) = ... but N_(d-1) /= N_d_j

d(a) is called the index of the eigenvalue a."

This is from Peter D. Lax's book on linear algebra.

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If f is the minimal polynomial of A, then f(A) is the zero matrix. So f(A)v = 0 for any vector. Does that help?

That is also true for the characteristic polynomial. I know that the characteristic polynomial can be divided by a minimal polynomial and I want to show that the minimal polynomial is equal to (s-a_1)^(d_1)*...*(s-a_k)^(d_k)

That is also true for the characteristic polynomial.
Right. But you want the smallest polynomial that satisfies what I wrote. So you first want to prove:

(s-a_1)^(d_1)*...*(s-a_k)^(d_k)

satisfies what I wrote, and then you want to prove that any proper factor of it does not. I think good choices of f(A)v will do that.

What is a minimal polynomial of a nxn matrix?

The minimal polynomial of a nxn matrix is the unique monic polynomial of least degree that has the matrix as a root. In other words, it is the polynomial that, when evaluated at the matrix, gives the zero matrix.

How is the minimal polynomial related to the characteristic polynomial?

The characteristic polynomial is the determinant of the matrix multiplied by the identity matrix. The minimal polynomial is a factor of the characteristic polynomial, and they have the same roots.

How can the minimal polynomial be computed?

The minimal polynomial can be computed using various methods, such as using the Cayley-Hamilton theorem or finding the eigenvalues and eigenvectors of the matrix. It can also be computed using the characteristic polynomial and the matrix's Jordan canonical form.

What is the significance of the minimal polynomial in linear algebra?

The minimal polynomial is important in understanding the properties and behavior of a matrix. It provides information about the matrix's eigenvalues, diagonalizability, and the structure of its Jordan canonical form. It is also useful in solving systems of linear equations and in various applications, such as in coding theory.

Can a matrix have more than one minimal polynomial?

No, a matrix can only have one minimal polynomial. This is because the minimal polynomial is unique and is determined by the matrix's characteristic polynomial and Jordan canonical form. However, different matrices can have the same minimal polynomial.

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