Let [itex]A \in M_n(F)[/itex] and [itex]v \in F^n[/itex].(adsbygoogle = window.adsbygoogle || []).push({});

Also...[itex][g \in F[x] : g(A)(v)=0] = Ann_A (v)[/itex] is an ideal in F[x], called the annihilator of v with respect to A. We know that [itex]g \in Ann_A(v)[/itex] if and only if f|g in F[x]. Let [itex]V = Span(v, Av, A^2v, ... , A^{k-1}v).[/itex]. 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 [itex]F^n[/itex] induced by multiplication by A. We also know that [itex]v, Av, A^2v, ... , A^{k-1}v[/itex] is a basis, B, of v.

Now...

A = [itex]\begin{bmatrix} 2 & 0 & 0 \\0 & 3 & 0 \\0 & 0 & 4 \end{bmatrix}[/itex]

v = [itex]\begin{bmatrix} 1 \\1 \\0 \end{bmatrix}[/itex]

This is what the professor told us:

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

[itex]\begin{bmatrix} I_k & X \\0 & 0\end{bmatrix}[/itex]...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 [itex]I_k[/itex]. Then you will have zeros below [itex]I_k[/itex]. On the right hand side of [itex]I_k[/itex], 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 = [itex][v | Av | ... | A^{k-1}v | A^kv][/itex] = [itex]\begin{bmatrix} 1 & 2 & 4 & 8 \\1 & 3 & 9 & 27 \\0 & 0 & 0 & 0 \end{bmatrix}[/itex]

After row reducing, I got:

[itex]\begin{bmatrix} 1 & 0 & -6 & -30 \\0 & 1 & 5 & 19 \\0 & 0 & 0 & 0 \end{bmatrix}[/itex]

So the first 2 rows and 2 columns of this matrix is [itex]I_2[/itex], and there are zeros below it. Then we have X=[itex]\begin{bmatrix} -6 & -30 \\5 & 19 \end{bmatrix}[/itex]. 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 "[itex]T: V \rightarrow V[/itex] be induced by multiplication by A: T(w) = Aw for [itex]w \in V[/itex].

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 = [itex][Av | A^2v][/itex] = [itex]\begin{bmatrix} 2 & 4 \\3 & 19 \end{bmatrix}[/itex]

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

Thanks in advance

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# Homework Help: A question about minimal polynomials

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