# Eigenvalues and characteristic polynomials

1. Oct 26, 2012

### dobedobedo

Hello guise.

I am familiar to a method of diagonalizing an nxn-matrix which fulfills the following condition:
the sum of the dimensions of the eigenspaces is equal to n.

As to the algorithm itself, it says:

1. Find the characteristic polynomial.
2. Find the roots of the characteristic polynomial.
3. Let the eigenvectors $v_{i}$ be the column vectors of some matrix S.
4. Let the eigenvalues $\lambda_{i}$ be the elements of some diagonal matrix, ordered to CORRESPOND the order of the eigenvectors in S.
5. Our Diagonalization of A should be:
$A = S \cdot A \cdot S^{-1} = (v_{1}...v_ {i}...v_ {n}) \cdot (\lambda_{1}...\lambda_{i}...\lambda_{n}) \cdot S^{-1}$

My question is: how do I find at least one such matrix A Corresponding to some randomly created polynomial of degree $m$ with integer roots? If it is too difficult to solve this for an arbitrary $m$, that's okay. But let's say for $m = 5$? Or for the much simpler case of $m = 2$?
Is this somehow related to the quadratic form?

2. Oct 26, 2012

### HallsofIvy

Staff Emeritus
Given the m roots, construct the m by m diagonal matrix, D, having those roots on the main diagonal. For any invertible matrix, P, $A= PDP^{-1}$ will have characteristic polynomial with those roots.

3. Oct 26, 2012

### dobedobedo

Jesus christ, I feel stupid. Thanks.

4. Oct 26, 2012

### mathwonk

thats how people make up exercises and test problems.

5. Oct 27, 2012

### dobedobedo

A follow-up question: how do I find a matrix with integer elements whose inverse also has integer elements? ^^

6. Oct 27, 2012

### Hurkyl

Staff Emeritus
You need its determinant to be 1 or -1.

Any integer matrix is a product of integer elementary matrices corresponding to the row operations:
• Adding an integer multiple of one row to another
• Swapping two rows
• Multiplying a row by a constant

As long as you only use 1 or -1 for that last one, all three sorts are invertible.

7. Oct 27, 2012

### dobedobedo

Hahahaha! How brilliant. Okay. Now I can finally randomly create integer matrices A such that $A = SDS^{-1}$ are integer matrices as well! Other questions:

-How do I find a matrix A with integer elements, which has an inverse with rational elements? I get the feeling that it always is that a "rational" matrix always has a "rational" inverse [but I do not know how to show it]?

-How do I find a matrix A with rational elements which has an inverse with integer elements?

-How do I prove that a real matrix never has an inverse with complex elements?

Last edited by a moderator: Oct 27, 2012
8. Oct 27, 2012

### HallsofIvy

Staff Emeritus
You can't create a matrix A such that this will work for all S. You need both A and S having determinants 1 or -1.

The inverse of matrix A is the matrix of cofactors divided by the determinant of A so, yes, since the rational numbers is closed under addition, subtraction, multiplication, and division by non-zero numbers, the inverse of a matrix with rational entries (including integers) will always have rational entries.

An obvious way is to start with an invertible matrix with integer coefficients and let A be its inverse! More generally if the determinant of A is a common multiple of its denominators (and so an integer) then its inverse will have integer entries.

Again, every element of A-1 is its cofactor divided by the matrix of A. Since the real numbers is closed under the four basic operations, that will always give real numbers.

9. Oct 27, 2012

### DonAntonio

If you don't have the eigenvalues you use the companion matrix of a monic polynomial:
$${}$$
$$\left(\begin{array} {}0&1&0&0&...&0\\0&0&1&0&...&0\\...&...&...&...&...&...\\0&0&0&...&0&1\\\!\!-c_0&\!\!-c_1&\!\!-c_2&\!\!-c_3&...&\!\!-c_{n-1}\end{array}\right)$$

The coolest feature of the above matrix is that $\,f(x)=c_0+c_1x+...+c_{n-1}x^{n-1}+x^n\,$ is not only its

characteristic but also its minimal polynomial...!

DonAntonio

10. Oct 27, 2012

### dobedobedo

Very interesting! But I am somewhat confused. How do I actually use it? I was wondering how to find my matrix A out of some given polynomial - but that would require me to know it's roots, due to the design of the algorithm. For instance, if I have my matrix {{2,1},{1,2}} then it's characteristic polynomial is x^2-4x+3. If I use the formula you gave me, it's corresponding matrix would be {{0,1},{0,0},{-3,-2}}, which is non-square and therefore not a matrix A which fulfills the criteria of the algorithm? Or am I completely mistaken?

Last edited: Oct 27, 2012
11. Nov 4, 2012

$\left( \begin{array}{cc} 0 & 1\\ -3 & 4\\ \end{array} \right) ?$