A question about minimal polynomials

[Artusartos]

Let $A \in M_n(F)$ and $v \in F^n$.

Also...$[g \in F[x] : g(A)(v)=0] = Ann_A (v)$ is an ideal in F[x], called the annihilator of v with respect to A. We know that $g \in Ann_A(v)$ if and only if f|g in F[x]. Let $V = Span(v, Av, A^2v, ... , A^{k-1}v).$. V is the smallest A-invariant subspace containing v. We denote the fact by writing V=F[x]v. This corresponds to the F[x]-module structure on $F^n$ induced by multiplication by A. We also know that $v, Av, A^2v, ... , A^{k-1}v$ is a basis, B, of v.

Now...

A = $\begin{bmatrix} 2 & 0 & 0 \\0 & 3 & 0 \\0 & 0 & 4 \end{bmatrix}$

v = $\begin{bmatrix} 1 \\1 \\0 \end{bmatrix}$

This is what the professor told us:

You need to find the matrix $[v | Av | ... | A^{k-1}v | A^kv]$ and then row reduce it to echelon form...He said that we would get a matrix of the form:

$\begin{bmatrix} I_k & X \\0 & 0\end{bmatrix}$...I couldn't really write this matrix the way I wanted to...but this is basically what he said. The first few columns and rows will give the identity matrix $I_k$. Then you will have zeros below $I_k$. On the right hand side of $I_k$, you will have a few columns (call it "X"), and zeros below it.

He said that the minimal polynomial of T depends on the first column of X.

So this is what I did:


Let N = $[v | Av | ... | A^{k-1}v | A^kv]$ = $\begin{bmatrix} 1 & 2 & 4 & 8 \\1 & 3 & 9 & 27 \\0 & 0 & 0 & 0 \end{bmatrix}$

After row reducing, I got:

$\begin{bmatrix} 1 & 0 & -6 & -30 \\0 & 1 & 5 & 19 \\0 & 0 & 0 & 0 \end{bmatrix}$

So the first 2 rows and 2 columns of this matrix is $I_2$, and there are zeros below it. Then we have X=$\begin{bmatrix} -6 & -30 \\5 & 19 \end{bmatrix}$. According to what the professor said, the first column of X determins the minimal polynomial of T. So the minimal polynomial is 6 + 5x + x^2, right?

So, what is T?

The definition that he gave for T is "$T: V \rightarrow V$ be induced by multiplication by A: T(w) = Aw for $w \in V$.

In this case the basis is the first two columns of N. So, basically, they are v and Av. Since T is induced by multiplication by A...T has the columns A(v) and A(Av)=A^2v

so...

...we have T = $[Av | A^2v]$ = $\begin{bmatrix} 2 & 4 \\3 & 19 \end{bmatrix}$

But this matrix does not satisfy the minimal polynomial. Can anybody please help me with this? Did I do something wrong?

Thanks in advance

 

[Ray Vickson]

Let $A \in M_n(F)$ and $v \in F^n$.

Also...$[g \in F[x] : g(A)(v)=0] = Ann_A (v)$ is an ideal in F[x], called the annihilator of v with respect to A. We know that $g \in Ann_A(v)$ if and only if f|g in F[x]. Let $V = Span(v, Av, A^2v, ... , A^{k-1}v).$. V is the smallest A-invariant subspace containing v. We denote the fact by writing V=F[x]v. This corresponds to the F[x]-module structure on $F^n$ induced by multiplication by A. We also know that $v, Av, A^2v, ... , A^{k-1}v$ is a basis, B, of v.

Now...

A = $\begin{bmatrix} 2 & 0 & 0 \\0 & 3 & 0 \\0 & 0 & 4 \end{bmatrix}$

v = $\begin{bmatrix} 1 \\1 \\0 \end{bmatrix}$

This is what the professor told us:

You need to find the matrix $[v | Av | ... | A^{k-1}v | A^kv]$ and then row reduce it to echelon form...He said that we would get a matrix of the form:

$\begin{bmatrix} I_k & X \\0 & 0\end{bmatrix}$...I couldn't really write this matrix the way I wanted to...but this is basically what he said. The first few columns and rows will give the identity matrix $I_k$. Then you will have zeros below $I_k$. On the right hand side of $I_k$, you will have a few columns (call it "X"), and zeros below it.

He said that the minimal polynomial of T depends on the first column of X.

So this is what I did:


Let N = $[v | Av | ... | A^{k-1}v | A^kv]$ = $\begin{bmatrix} 1 & 2 & 4 & 8 \\1 & 3 & 9 & 27 \\0 & 0 & 0 & 0 \end{bmatrix}$

After row reducing, I got:

$\begin{bmatrix} 1 & 0 & -6 & -30 \\0 & 1 & 5 & 19 \\0 & 0 & 0 & 0 \end{bmatrix}$

So the first 2 rows and 2 columns of this matrix is $I_2$, and there are zeros below it. Then we have X=$\begin{bmatrix} -6 & -30 \\5 & 19 \end{bmatrix}$. According to what the professor said, the first column of X determins the minimal polynomial of T. So the minimal polynomial is 6 + 5x + x^2, right?

So, what is T?

The definition that he gave for T is "$T: V \rightarrow V$ be induced by multiplication by A: T(w) = Aw for $w \in V$.

In this case the basis is the first two columns of N. So, basically, they are v and Av. Since T is induced by multiplication by A...T has the columns A(v) and A(Av)=A^2v

so...

...we have T = $[Av | A^2v]$ = $\begin{bmatrix} 2 & 4 \\3 & 19 \end{bmatrix}$

But this matrix does not satisfy the minimal polynomial. Can anybody please help me with this? Did I do something wrong?

Thanks in advance

I don't know what you are doing, but I understand you want to find the monic polynomial p of least degree such that p(A) annihilates v, so you want to find the c_i that give you
$$c_0v + c_1 Av + c_2 A^2v + c_3 A^3 v = 0$$
or
$$c_0 + 2c_1 + 4c_2 + 8c_3 = 0\\ c_0 + 3c_1 + 9c_2 + 27c_3 = 0.$$
Row reduction gives us a solution of some c in terms of the others:
$$c_0 = 6c_2 + 30c_3, \; c_1 = -5c_2 -19c_3.$$
So, if we put c_3 = 0 and c_2 = 1 we get c_0 = 6 and c_1 = -5, hence the minimial polynomial of v is p(x) = x^2 - 5x + 6.

RGV

 

[Artusartos]

I don't know what you are doing, but I understand you want to find the monic polynomial p of least degree such that p(A) annihilates v, so you want to find the c_i that give you
$$c_0v + c_1 Av + c_2 A^2v + c_3 A^3 v = 0$$
or
$$c_0 + 2c_1 + 4c_2 + 8c_3 = 0\\ c_0 + 3c_1 + 9c_2 + 27c_3 = 0.$$
Row reduction gives us a solution of some c in terms of the others:
$$c_0 = 6c_2 + 30c_3, \; c_1 = -5c_2 -19c_3.$$
So, if we put c_3 = 0 and c_2 = 1 we get c_0 = 6 and c_1 = -5, hence the minimial polynomial of v is p(x) = x^2 - 5x + 6.

RGV

Thanks a lot for answering
Actually, that's the same thing that I got...I accidentally wrote x^2 + 5x +6 instead of x^2 - 5x + 6. But the problem that I'm having is that...when I substitute T for x, I do not get zero...why?

 

[Ray Vickson]

Thanks a lot for answering
Actually, that's the same thing that I got...I accidentally wrote x^2 + 5x +6 instead of x^2 - 5x + 6. But the problem that I'm having is that...when I substitute T for x, I do not get zero...why?

I suspect that T and A are two ways of talking about the same thing: T is a transformation and A is its matrix representation with respect to a particular basis.

So, you are saying that you do not get A^2 - 5A + 6I = 0. Well, that just means that p(A) does not annihilate the whole vector space. That's OK: it DOES annihilate the two-dimensional subspace span{v, Av, A^2v} = span{v,Av} for the particular v you started with. To annihilate the whole space, you would need to find the minimal polynomial q for another starting vector w (so that q(A)w = 0) and then (maybe) take the least common multiple of the polynomials p(x) and q(x), or something similar.

RGV