Can a Matrix Be Expressed as a Sum of Diagonalizable and Nilpotent Matrices?

  • Thread starter MathIdiot
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In summary: Yes, this is the theorem that states that a matrix can be decomposed into a product of a linear map and a non-zero diagonal matrix.
  • #1
MathIdiot
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How do you write a matrix as a sum of a diagonalizable matrix and a nilpotent matrix?

It would be great if you could describe the steps in Layman's terms because I am not so hot in Linear Algebra.

Thanks
 
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  • #2
What sort of normal forms do you know for matrices?
 
  • #3
I'm not sure what you mean. The only ideas about "normal" I know is normalizing a vector. Normalizing a basis of vectors as well as normalizing a basis of orthogonal vectors to get an orthonormal basis. And orthogonal matrices. Is this at all what your implying?
 
  • #4
No; I'm using "normal form" to describe the category of things like row echelon form or diagonal form -- methods for representing matrices in some sort of specialized format which is easy to manipulate.
 
  • #5
Oh, ok (sorry).
They are square, they don't have any other specialized format.
 
  • #6
it is not true in general,so first assume the field is algebraically closed. then assume you know that every linear map satisfies some minimal polynomial

then use the euclidean algorithm to decompose the space into a direct sum on each factor of which the polynomial is of form (X-c)^r.

then note that if T satisfies (X-c)^r, then it is the sum of T-cId and cId, where cID is diagonalizable, and T-cId satisfies X^r, hence is nilpotent.

done.
 
  • #7
Ok, I hope you can clarify a few things from your last statement.
Those assumptions sound fine.
However what is the euclidean algorithm?
Also, what is meant by "T satisfies (X-c)^r"?
Lastly how do you get cID? What I mean really is, how does one achieve this diagonalizable matrix?
 
  • #8
Euclidean algorithm, I believe, goes like this:

let f, g be polynomials in F[x] such that g does not equal 0. Then there exists uniquely determined polynomials q and r such that

f = qg + r

and r = 0 or deg r < deg gMathwonk, is this what is known as the primary decomposition theorem?
 
Last edited:

FAQ: Can a Matrix Be Expressed as a Sum of Diagonalizable and Nilpotent Matrices?

What does it mean for a matrix to be diagonalizable?

A matrix is diagonalizable if it can be written in the form of A = PDP-1, where D is a diagonal matrix and P is an invertible matrix. This means that the matrix can be simplified to have non-zero values only on the diagonal, making it easier to work with in calculations.

How do you determine if a matrix is diagonalizable?

A matrix is diagonalizable if it has n linearly independent eigenvectors, where n is the size of the matrix. This means that the eigenvectors can be used to form the invertible matrix P in the equation A = PDP-1. If a matrix does not have n linearly independent eigenvectors, it is not diagonalizable.

What is the significance of a matrix being diagonalizable?

A diagonalizable matrix has many useful properties and makes solving systems of linear equations and finding powers of a matrix easier. It also allows for a better understanding of the behavior of a system described by the matrix. Additionally, diagonalizable matrices have simpler forms that can be used to calculate things like determinants and eigenvalues.

What is a nilpotent matrix?

A nilpotent matrix is a square matrix where Ak = 0 for some positive integer k. This means that when the matrix is raised to the power of k, it becomes a zero matrix. Nilpotent matrices are always singular and have eigenvalues of 0.

How do you determine if a matrix is nilpotent?

A matrix is nilpotent if its eigenvalues are all equal to 0. This means that the trace and determinant of the matrix are also equal to 0. Additionally, the rank of a nilpotent matrix is always less than its size, and the matrix must have at least one zero eigenvalue. These conditions can be used to determine if a matrix is nilpotent.

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