How do you find the minimal polynomial?

[Artusartos]

Homework Statement

If we have a transformation matrix $\begin{bmatrix} 1 & 2 & 4 \\0 & 0 & 0 \\0 & 0 & 0 \end{bmatrix}$

Homework Equations

The Attempt at a Solution

I found the characteristic polynomial of this matrix: $x^3 - x^2$ = $x^2(x-1)$...can anybody please help me?

Thanks in advance

 

[Simon Bridge]

Are you doing some sort of course?
There should be something in your course-notes that tell you how the minimal polynomial is related to the characteristic polynomial.

Also see:
http://uk.answers.yahoo.com/question/index?qid=20081204110100AAo5kS8

 

[Artusartos]

Are you doing some sort of course?
There should be something in your course-notes that tell you how the minimal polynomial is related to the characteristic polynomial.

Also see:
http://uk.answers.yahoo.com/question/index?qid=20081204110100AAo5kS8

I do know that the minimal polynomial divides the characteristic polynomial. But there is nothing else about how to find it in my notes. But thanks, I will look at the link that you gave me...

 

[HallsofIvy]

Surely there is a definition? The "minimal polynomial" of a linear transformation is the polynomial of lowest degree that is made equal to 0 by the linear transformation. Yes, it must divide the "characteristic polynomial". And the only divisors of $x^2(x- 1)$ are $x$, $x^2$, $x- 1$, $x(x- 1)$, and $x^2(x- 1)$. Which of does A make 0?

 

[Ray Vickson]

Homework Statement

If we have a transformation matrix $\begin{bmatrix} 1 & 2 & 4 \\0 & 0 & 0 \\0 & 0 & 0 \end{bmatrix}$

Homework Equations

The Attempt at a Solution

I found the characteristic polynomial of this matrix: $x^3 - x^2$ = $x^2(x-1)$...can anybody please help me?

Thanks in advance

You could do what Maple does, which is to regard the identity matrix I and the matrices A, A^2, A^3,... as being spread out into n^2-dimensional vectors, then use row operations to find the least m such that I,A,A^2, ... , A^m are linearly dependent, and to find the coefficients involved.

However, in your case that is overkill. Just follow the previous suggestions, and test the polynomials x^2 and x(1-x) to see if either of them annihilate A. If not, minimal = characteristic.

RGV

 

[Artusartos]

You could do what Maple does, which is to regard the identity matrix I and the matrices A, A^2, A^3,... as being spread out into n^2-dimensional vectors, then use row operations to find the least m such that I,A,A^2, ... , A^m are linearly dependent, and to find the coefficients involved.

However, in your case that is overkill. Just follow the previous suggestions, and test the polynomials x^2 and x(1-x) to see if either of them annihilate A. If not, minimal = characteristic.

RGV

By, "A", you mean the transformation matrix, right?

 

[Artusartos]

Surely there is a definition? The "minimal polynomial" of a linear transformation is the polynomial of lowest degree that is made equal to 0 by the linear transformation. Yes, it must divide the "characteristic polynomial". And the only divisors of $x^2(x- 1)$ are $x$, $x^2$, $x- 1$, $x(x- 1)$, and $x^2(x- 1)$. Which of does A make 0?

Thanks a lot. I think I was able to do it for this matrix...but I have another matrix that I'm a bit confused about...

$\begin{bmatrix} 1 & cos( \theta) \\0 & sin( \theta) \end{bmatrix}$ where theta is between zero and 2pi.


This is what the professor told us:

If theta is not equal to pi, then the minimal polynomial is equal to the characteristic polynomial. If theta equals pi, then the minimal polynomial is the negative of the identity. I found the characteristic polynomial to be...$f(x)= x^2 - x - x sin( \theta) + sin( \theta)$...but now, I have theta in the equation. So how will that work? I won't be able to substitute my matrix for x and get zero when theta is in there...

 

[Ray Vickson]

By, "A", you mean the transformation matrix, right?

Yes, of course. You have a matrix, and you need sometimes to give it a name.

RGV

 

[Artusartos]

You could do what Maple does, which is to regard the identity matrix I and the matrices A, A^2, A^3,... as being spread out into n^2-dimensional vectors, then use row operations to find the least m such that I,A,A^2, ... , A^m are linearly dependent, and to find the coefficients involved.

However, in your case that is overkill. Just follow the previous suggestions, and test the polynomials x^2 and x(1-x) to see if either of them annihilate A. If not, minimal = characteristic.

RGV

Can you expand on what you have said about that maple thing? Because our professor said something similar...but I'm not sure I understood him correctly...

He told us to use this matrix:

$\begin{bmatrix} 1 & 2 & 4 & 8\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{bmatrix}$

to make it linearly dependent (this question actually has many parts, so the matrix I gave is related to this one...except that this one is linearly dependent of course...this is the least one that is linearly dependent, by the way). So can you explain the maple method too? If you don't mind...

 

[Ray Vickson]

Can you expand on what you have said about that maple thing? Because our professor said something similar...but I'm not sure I understood him correctly...

He told us to use this matrix:

$\begin{bmatrix} 1 & 2 & 4 & 8\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{bmatrix}$

to make it linearly dependent (this question actually has many parts, so the matrix I gave is related to this one...except that this one is linearly dependent of course...this is the least one that is linearly dependent, by the way). So can you explain the maple method too? If you don't mind...

This won't work: is is a 3×4 matrix, so you cannot even square it. If we drop column 4 we obtain a 3×3 matrix A, but it is too easy to show what is happening, since it satisfies A² = A, so has minimum polynomial x²- x. Let's try instead the matrix
$$B = \begin{bmatrix}1&2&3\\1&0&0\\0&1&1 \end{bmatrix},\\ \text{ giving }\\ B^2 = \begin{bmatrix}3&5&6\\1&2&3\\1&1&1\end{bmatrix} \\ B^3 = \begin{bmatrix} 8&12&15\\3&5&6\\2&3&4\end{bmatrix}$$
Form the vectors
$$I \leftrightarrow v_0 = \begin{bmatrix}1&0&0&0&1&0&0&0&1\end{bmatrix}\\ A \leftrightarrow v_1 = \begin{bmatrix}1&2&3&1&0&0&0&1&1\end{bmatrix}\\ A^2 \leftrightarrow v_2 = \begin{bmatrix}3&5&6&1&2&3&1&1&1\end{bmatrix}\\ A^3 \leftrightarrow v_3 = \begin{bmatrix}8&12&15&3&5&6&2&3&4\end{bmatrix}$$
Now start doing row-operations on v0,v1,v2,v3:
$$v_1' = v_1-v_0 = \begin{bmatrix}0&2&3&1&-1&0&9&1&0\end{bmatrix}\\ v_2' = v_2 -3v_0 = \begin{bmatrix}0&5&6&1&-1&3&1&1&-2\end{bmatrix}\\ v_3' = v_3 - 8v_0 = \begin{bmatrix}0&12&15&3&-3&6&2&3&-4\end{bmatrix}\\$$
Since v_1' ≠ 0 and v_2' ≠ 0 we are not yet done with row-reduction. Now "zero out" the second components of v_2' and v_3', by subtracting multiples of v_1':
$$v_2'' = v_2' - (5/2)v_1' = \begin{bmatrix}0&0&-3/2&-3/2&3/2&3&1&-3/2&2\end{bmatrix}\\ v_3'' = v_3' - 6v_1' = \begin{bmatrix}0&0&-3&-3&3&6&2&-3&-4\end{bmatrix}$$
Since v_2'' ≠ 0 the minimal polynomial is of degree 3 (so is the characteristic polynomial). If we want to, we can even determine characteristic polynomial by continuing the process:
$$v_3''' = v_3'' - [(-3)/(-3/2)]v_2'' = v_3'' -2 v_2'' = \begin{bmatrix}0&0&0&0&0&0&0&0&0\end{bmatrix}$$
Thus, we have
$$0 = v_3''' = v_3 - 2v_2 - v_1 - v_0 \leftrightarrow A^3 - 2A^2 - A - I,$$
so the characteristic polynomial is
$$p(x) = x^3 - 2x^2 - x - 1.$$

RGV

 

[Artusartos]

This won't work: is is a 3×4 matrix, so you cannot even square it. If we drop column 4 we obtain a 3×3 matrix A, but it is too easy to show what is happening, since it satisfies A² = A, so has minimum polynomial x²- x. Let's try instead the matrix
$$B = \begin{bmatrix}1&2&3\\1&0&0\\0&1&1 \end{bmatrix},\\ \text{ giving }\\ B^2 = \begin{bmatrix}3&5&6\\1&2&3\\1&1&1\end{bmatrix} \\ B^3 = \begin{bmatrix} 8&12&15\\3&5&6\\2&3&4\end{bmatrix}$$
Form the vectors
$$I \leftrightarrow v_0 = \begin{bmatrix}1&0&0&0&1&0&0&0&1\end{bmatrix}\\ A \leftrightarrow v_1 = \begin{bmatrix}1&2&3&1&0&0&0&1&1\end{bmatrix}\\ A^2 \leftrightarrow v_2 = \begin{bmatrix}3&5&6&1&2&3&1&1&1\end{bmatrix}\\ A^3 \leftrightarrow v_3 = \begin{bmatrix}8&12&15&3&5&6&2&3&4\end{bmatrix}$$
Now start doing row-operations on v0,v1,v2,v3:
$$v_1' = v_1-v_0 = \begin{bmatrix}0&2&3&1&-1&0&9&1&0\end{bmatrix}\\ v_2' = v_2 -3v_0 = \begin{bmatrix}0&5&6&1&-1&3&1&1&-2\end{bmatrix}\\ v_3' = v_3 - 8v_0 = \begin{bmatrix}0&12&15&3&-3&6&2&3&-4\end{bmatrix}\\$$
Since v_1' ≠ 0 and v_2' ≠ 0 we are not yet done with row-reduction. Now "zero out" the second components of v_2' and v_3', by subtracting multiples of v_1':
$$v_2'' = v_2' - (5/2)v_1' = \begin{bmatrix}0&0&-3/2&-3/2&3/2&3&1&-3/2&2\end{bmatrix}\\ v_3'' = v_3' - 6v_1' = \begin{bmatrix}0&0&-3&-3&3&6&2&-3&-4\end{bmatrix}$$
Since v_2'' ≠ 0 the minimal polynomial is of degree 3 (so is the characteristic polynomial). If we want to, we can even determine characteristic polynomial by continuing the process:
$$v_3''' = v_3'' - [(-3)/(-3/2)]v_2'' = v_3'' -2 v_2'' = \begin{bmatrix}0&0&0&0&0&0&0&0&0\end{bmatrix}$$
Thus, we have
$$0 = v_3''' = v_3 - 2v_2 - v_1 - v_0 \leftrightarrow A^3 - 2A^2 - A - I,$$
so the characteristic polynomial is
$$p(x) = x^3 - 2x^2 - x - 1.$$

RGV

Thanks a lot :)