The discussion revolves around finding the minimal polynomial of a transformation matrix, specifically a 3x3 matrix with a row of zeros. Participants explore the relationship between the minimal polynomial and the characteristic polynomial, as well as methods for determining the minimal polynomial through various approaches.
There is an ongoing exploration of different methods to find the minimal polynomial, including a reference to using row operations and linear dependence. Participants are sharing insights and clarifying definitions, but no consensus has been reached on a specific method or solution.
Some participants mention constraints related to the matrices being discussed, such as dimensions and the implications of having rows of zeros. There is also mention of specific cases depending on the value of theta in one of the matrices, which adds complexity to the discussion.
Simon Bridge said: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 said: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
Ray Vickson said: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
HallsofIvy said: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?
Artusartos said:By, "A", you mean the transformation matrix, right?
Ray Vickson said: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 said: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 said: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},\\<br /> \text{ giving }\\<br /> B^2 = \begin{bmatrix}3&5&6\\1&2&3\\1&1&1\end{bmatrix} \\<br /> B^3 = \begin{bmatrix} 8&12&15\\3&5&6\\2&3&4\end{bmatrix}<br />
Form the vectors
I \leftrightarrow v_0 = \begin{bmatrix}1&0&0&0&1&0&0&0&1\end{bmatrix}\\<br /> A \leftrightarrow v_1 = \begin{bmatrix}1&2&3&1&0&0&0&1&1\end{bmatrix}\\<br /> A^2 \leftrightarrow v_2 = \begin{bmatrix}3&5&6&1&2&3&1&1&1\end{bmatrix}\\<br /> A^3 \leftrightarrow v_3 = \begin{bmatrix}8&12&15&3&5&6&2&3&4\end{bmatrix}<br />
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}\\<br /> v_2' = v_2 -3v_0 = \begin{bmatrix}0&5&6&1&-1&3&1&1&-2\end{bmatrix}\\<br /> v_3' = v_3 - 8v_0 = \begin{bmatrix}0&12&15&3&-3&6&2&3&-4\end{bmatrix}\\<br />
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}\\<br /> v_3'' = v_3' - 6v_1' = \begin{bmatrix}0&0&-3&-3&3&6&2&-3&-4\end{bmatrix}<br />
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:
<br /> v_3''' = v_3'' - [(-3)/(-3/2)]v_2'' = v_3'' -2 v_2'' = <br /> \begin{bmatrix}0&0&0&0&0&0&0&0&0\end{bmatrix}<br />
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