How can we prove that a nxn real matrix A is a root of a given polynomial?

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In summary: The adjugate matrix is usually denoted by \text{A}. It is a matrix that has the same rank as the original matrix (nxn), but whose determinant is the inverse of the determinant of the original matrix. So, if you have \text{A} and \text{A}_n, then \text{A}_n=\text{A}_n-1. So, if you want to find the inverse of a matrix, you can use the adjugate matrix.
  • #1
universedrill
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Homework Statement


1) Prove that: nxn real matrix A is a root of f(X)= a[n].X^n+...+a[0].I, where a[n],...,a[0] are coefficients of the polynomial P(t)= det [A-t.I]
2) Let 5x5 real matrix A be satisfied: A^2008 = 0. Prove that: A^5=0.

2. The attempt at a solution
I tried to solve problem 2 with an general idea: nxn matrix A: A^m=0 (m>n). Prove: A^n=0.
Let P(t)=det [A-t.I]. So, deg P(t)=n, t is a real number.
Let t is a root of P(t), we get:
det[A-tI]=0 -> the equation: (A-tI)X=0 has a root X which is different from 0
-> AX = tIX=tX -> A(AX)=A(tX)
->A^2.X=t(AX)=t(tX)=t^2.X ->... -> A^m.X=t^m.X
Because X differ from 0 and A^m =0, we find out t^m =0 -> t=0
Thus, P(t)= t^n.
Now, the important thing is proving problem 1. I remember that the problem 1 seem to be a theorem? Can you help me prove that, or find meterials saying that? Thanks
 
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  • #2
1) In other words, they are asking for a proof of the Cayley-Hamilton theorem? That I believe should be rather difficult, since proof of this theorem was omitted when I took my intermediate linear algebra course this semester.

2) Use result 1 to prove it. Instead of evaluating det(A-tI), what is det(A^2008 -tI) ?

P.S. Use of square brackets [ ] can be confusing. Use the normal parantheses instead.
 
  • #3
Thanks, but
Defennder said:
2) Instead of evaluating det(A-tI), what is det(A^2008 -tI) ?
I don't understand clearly what you mean.
And, is there any solution where theorem 1 isn't used for problem 2?
 
  • #4
Well, it appears that the problem has been set up in such a way such that you can use the result of theorem 1 (even if you do not know how to prove it) to do 2). And I don't know which part of what I wrote you do not understand. What don't you understand about finding det(A^2008 -tI) ?
 
  • #5
There are a few different methods to prove (1). Have you studied adjugate matrices yet? If so, can you reason that [itex]\text{adj} (A-tI_n)[/itex] exists? If so, what can you say about [itex](A-tI_n) \cdot \text{adj} (A-tI_n)[/itex]?
 

1. What is a nxn matrix?

A nxn matrix is a matrix with equal number of rows and columns. This means that the number of rows is equal to the number of columns.

2. How is a nxn matrix represented?

A nxn matrix is represented by writing the values in rows and columns, separated by commas. For example, a 3x3 matrix would look like this:

[1, 2, 3]
[4, 5, 6]
[7, 8, 9]

Each row is enclosed in square brackets and the rows are separated by a new line.

3. What are the uses of a nxn matrix?

Nxn matrices have various uses in mathematics, computer science, and engineering. They are used for solving systems of linear equations, representing transformations in geometry, and storing data in computer algorithms. They are also used in applications such as image processing, data analysis, and cryptography.

4. How is a nxn matrix multiplied?

To multiply two nxn matrices, the number of columns in the first matrix must be equal to the number of rows in the second matrix. The product matrix will have the same number of rows as the first matrix and the same number of columns as the second matrix. The multiplication is done by multiplying the corresponding elements in each row of the first matrix with the corresponding elements in each column of the second matrix, and then adding the products.

5. How is the determinant of a nxn matrix calculated?

The determinant of a nxn matrix is calculated by expanding the matrix along any row or column and finding the sum of the products of the elements and their corresponding cofactors. The cofactor of an element is calculated by finding the determinant of the submatrix formed by removing the row and column containing that element. This process is repeated until a 2x2 matrix is obtained, for which the determinant can be easily calculated. The final result is the determinant of the original nxn matrix.

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