Linear Algebra - Raising a matrix to a power

In summary, the conversation discusses raising a matrix to the 50th power in a more efficient way using the binomial theorem. It also mentions the use of a "Jordan Block" matrix and provides a link for further reference.
  • #1
snesnerd
26
0
| x 1 0 |
| 0 x 1 |
| 0 0 x |

I need to raise this matrix to the 50th power. Of course I can not solve this the extremely long way. Here is my attempt at this:

Let A represent the matrix above, and let N represent the following matrix:

| 0 1 0 |
| 0 0 1 |
| 0 0 0 |

Then A = XI + N. Since XI * N = N * XI, we can apply the binomial theorem.

A^50 = sum from k = 0 to 50 of (50 C k)(XI)^(50-k)N^k by the binomial theorem

= (XI)^50 + 50N(XI)^49 + 1225N^(2)(XI)^48 + N^3

Now if you take N^3 you will get 0, and if you take N^2, you get the following:

| 0 0 1 |
| 0 0 0 |
| 0 0 0 |

So we get:

I(X)^50 + 50N(X)^49 + 1225N^2(X)^48 which is,

| X^50 50X^49 1225X^48 |
| 0 X^50 50X^49 |
| 0 0 X^50 |

I think this is right. If my work is right could someone help me make this more thorough? I do not want to miss any details and want to show all my work. Thanks.
 
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  • #2
snesnerd said:
| x 1 0 |
| 0 x 1 |
| 0 0 x |

I need to raise this matrix to the 50th power. Of course I can not solve this the extremely long way. Here is my attempt at this:

Let A represent the matrix above, and let N represent the following matrix:

| 0 1 0 |
| 0 0 1 |
| 0 0 0 |

Then A = XI + N. Since XI * N = N * XI, we can apply the binomial theorem.

A^50 = sum from k = 0 to 50 of (50 C k)(XI)^(50-k)N^k by the binomial theorem

= (XI)^50 + 50N(XI)^49 + 1225N^(2)(XI)^48 + N^3

Now if you take N^3 you will get 0, and if you take N^2, you get the following:

| 0 0 1 |
| 0 0 0 |
| 0 0 0 |

So we get:

I(X)^50 + 50N(X)^49 + 1225N^2(X)^48 which is,

| X^50 50X^49 1225X^48 |
| 0 X^50 50X^49 |
| 0 0 X^50 |

I think this is right. If my work is right could someone help me make this more thorough? I do not want to miss any details and want to show all my work. Thanks.

Your matrix is a so-called "Jordan Block", and raising it to large powers is a well-solved problem; see, eg., http://en.wikipedia.org/wiki/Jordan_matrix

RGV
 
  • #3
Your solution looks just fine to me.
 

1. What does it mean to raise a matrix to a power?

Raising a matrix to a power involves multiplying the matrix by itself a specified number of times. This is similar to raising a number to a power, but instead of multiplying the number by itself, we multiply the matrix by itself.

2. How do you raise a matrix to a power?

To raise a matrix to a power, we first need to make sure that the matrix is a square matrix (same number of rows and columns). Then, we simply multiply the matrix by itself the specified number of times. For example, if we want to raise a matrix A to the power of 3, we would multiply A by itself 3 times: A x A x A.

3. What are the applications of raising a matrix to a power?

Raising a matrix to a power is commonly used in linear algebra for solving systems of linear equations, calculating eigenvalues and eigenvectors, and finding the inverse of a matrix. It is also used in many other fields such as computer science, physics, and economics for various applications.

4. Are there any special properties of raising a matrix to a power?

Yes, there are several properties that apply when raising a matrix to a power. One important property is that the order in which you raise a matrix to a power does not matter. In other words, (A x A) x A is equal to A x (A x A). Another property is that the inverse of a matrix raised to a power is equal to the inverse of the matrix raised to the same power.

5. Can any matrix be raised to any power?

No, not all matrices can be raised to any power. The matrix must be a square matrix in order to be raised to a power. Additionally, not all square matrices have an inverse, so they cannot be raised to negative powers. It is important to check for these conditions before attempting to raise a matrix to a power.

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