Eigenvalues and characteristic polynomials

[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?

 

[HallsofIvy]

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.

 

[dobedobedo]

Jesus christ, I feel stupid. Thanks.

 

[mathwonk]

thats how people make up exercises and test problems.

 

[dobedobedo]

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

 

[Hurkyl]

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.

 

[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?

 

[HallsofIvy]

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

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.

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]?

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.

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

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.

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

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.

 

[DonAntonio]

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?

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

 

[dobedobedo]

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

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?

 

[divergentgrad]

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?

\left( \begin{array}{cc} <br /> 0 &amp; 1\\ -3 &amp; 4\\ \end{array} \right) ?<br />