MHB Tricky Linear Algebra Question. To show that an operator is 'cyclic'.

[caffeinemachine]

Hello MHB,
I am stuck at this problem for quite a long time now.

Problem. Let $F_p$ denote the field of $p$ elements, where $p$ is prime. Let $n$ be a positive integer. Let $V$ be the vector space $(F_p)^n$ over the field $F_p$. Let $GL_n(F_p)$ denote the set of all the invertible linear transformations on $V$. Then note that each element $T\in GL_n(F_p)$ is a permutation of the elements in $V$. Show that there is an element $T\in GL_n(F_p)$ such that the cycle decomposition of $T$ has a cycle of length $p^n-1$.Attempt.
Write $s=p^n=|V|$. Let $T\in\mathcal L(V)$ and $v\in V$. We say that $T$ is good with $v$ if $\{Tv,\ldots,T^{s-1}(v)\}=V\setminus\{0\}$.

Claim 1: If $T\in \mathcal L(V)$ is good with a vector $v\in V$ then $T\in GL_n(F_p)$.
Proof: Let $K=\{v,Tv,\ldots,T^{s-2}(v)\}$. Then $T(K)=V\setminus\{0\}$ since $T$ is good with $v$. Thus $T$ is surjective and hence invertible.

Claim 2: If $T$ is good with a vector $v\in V$ then $T^kv=v\Rightarrow k\geq s$.
Proof: Let $k$ be the smallest positive integer with the property that $T^kv=v$. Now $\{Tv,\ldots,T^{s-1}v\}=\{Tv,\ldots,T^{k-1}v\}$. The LHS has cardinality $s-1$ while the RHS has the cardinality $k-1$. Hence $s=k$ and we are done.

Claim 3: If $T$ is good with a vector $v$ then $T^sv=v$.
Proof: Clearly $T^sv\in V\setminus\{0\}$. So let $T^sv=T^iv$ for some $i$ in $\{1,\ldots,s-1\}$. This gives $T^{s-i}v=v$ and form the above claim $s-i\geq s$ and hence $i\leq 0$, which is a contradiction.

Claim 4: If $T$ is good with $v\in V$ then $T$ is good with each $u\in V\setminus\{0\}$.
Proof: Let $u\in V\setminus\{0\}$. Since $T$ is good with $v$, we have $u=T^iv$ for some $i \in \{1,\ldots,s-1\}$. Let $T^ku=u$ and $k$ be minimum, positive integer with this property. Then $T^{i+k}v=T^{i}v$. This leads to $T^k v=v$ and hence $k\geq s$. This settles that $T$ is good with $u$.

Claim 5: Let $T\in \mathcal L(V)$ is good with a vector $v\in V$ if and only if $T^k-I$ is invertible for each $k\in\{1,\ldots,s-1\}$.
Proof: Let $T$ be good with a vector $v\in V$ but $T^k-I$ is not invertible for a $k\in\{1,\ldots,s-1\}$. Then there is a vector $u\in V\setminus\{0\}$ such that $(T^k-I)u=0$. This means $T^ku=u$ and hence $T$ is not good with $u$. But this implies that $T$ isn't good with $v$ either and hence this direction of the claim in settled. The other direction is trivial.

Now coming back to the problem.
The question asks us to establish that there is some $T\in GL_n(F_p)$ such that $T$ is good with some vector $v\in V$. By Claim 5 this is equivalent to showing that there is a $T$ in $\mathcal L(V)$ such that $T^k-I$ is invertible for each $k\in\{1,\ldots,s-1\}$. To do this may be it will help to use the fact that an operator is invertible if and only if its matrix with respect to any basis has determinant non zero. I am unable to make progress here. Can anybody help?

 

[Deveno]

This is an interesting problem, and my solution is a bit circuitous (I apologize for that).

I will show that it suffices to exhibit an element [math]T \in GL_n(F_p)[/math], of order [math]p^n - 1[/math].

First of all, it should be clear from your opening remarks that we can consider [math]GL_n(F_p)[/math] as a subgroup of [math]S_{p^n - 1}[/math] (the symmetric group on [math]p^n - 1[/math] letters), so that ANY such element [math]\ T[/math] must be a [math](p^n - 1)[/math]-cycle.

Now, consider a primitive element [math]\xi \in (F_{p^n})^{\ast}[/math]. Let its minimal polynomial (necessarily of degree n) be:

[math]m(x) = c_0 + c_1x + \cdots + c_{n-1}x^{n-1} + x^n \in F_p[x][/math].

We have the matrix:

[math]T = \begin{bmatrix}0&0&\dots&0&-c_0\\1&0&\dots&0&-c_1\\0&1&\dots&0&-c_2\\ \vdots&\vdots&\ddots&\vdots&\vdots\\0&0&\dots&1&-c_{n-1} \end{bmatrix}[/math]

with [math]m(T) = 0[/math] (see: Companion matrix - Wikipedia, the free encyclopedia).

Since [math]\xi[/math] is an eigenvalue of [math]T[/math], we have for any eigenvector [math]v \in (F_{p^n})^n[/math]:

[math]T^kv = \xi^kv \neq v[/math] for [math]k = 1,2,\dots,p^n-1[/math].

This shows that the order of [math]T[/math] is at least [math]p^n - 1[/math].

Since the order of [math]T[/math] by Lagrange divides [math](p^n - 1)![/math], we conclude that:

[math]|T| = p^n - 1[/math]

(In all fairness, this proof just turns the problem into a different one: that the field with [math]p^n[/math] elements has a primitive element, for example see: http://www.math.wm.edu/~vinroot/430S13MultFiniteField.pdf ).

As a concrete example: consider [math]p = 3, n = 2[/math]. We have:

[math]F_9 \cong \Bbb Z_3(u)[/math], where [math]u[/math] is a root of [math]x^2 + 1[/math] (in other words, [math]u[/math] is a square root of 2). A primitive element for [math]\Bbb Z_3(u)[/math] is [math]u+1[/math]:

[math](u+1)^2 = u^2 + 2u + 1 = 2 + 2u + 1 = 2u[/math],
[math](u+1)^4 = (2u)^2 = 2^2u^2 = u^2 = 2[/math],
[math](u+1)^8 = 2^2 = 1[/math].

The minimal polynomial for [math]u+1[/math] over [math]\Bbb Z_3[/math] is:

[math]m(x) = x^2 + x + 2[/math].

The companion matrix in this case is:

[math]T = \begin{bmatrix}0&1\\1&2 \end{bmatrix}[/math], and indeed:

[math]T^2 + T + 2I = \begin{bmatrix}0&1\\1&2 \end{bmatrix}^2 + \begin{bmatrix}0&1\\1&2 \end{bmatrix} + \begin{bmatrix}2&0\\0&2 \end{bmatrix}[/math]

[math]= \begin{bmatrix}1&2\\2&2 \end{bmatrix} + \begin{bmatrix}0&1\\1&2 \end{bmatrix} + \begin{bmatrix}2&0\\0&2 \end{bmatrix} = \begin{bmatrix}0&0\\0&0 \end{bmatrix}[/math].

Identifying [math]u+1[/math] with the vector [math](1,1)[/math] (in the basis: [math]\{1,u\}[/math]), we see that:

[math]T(1,1) = (1,0)[/math]
[math]T(1,0) = (0,1)[/math]
[math]T(0,1) = (1,2)[/math]
[math]T(1,2) = (2,2)[/math]
[math]T(2,2) = (2,0)[/math]
[math]T(2,0) = (0,2)[/math]
[math]T(0,2) = (2,1)[/math]
[math]T(2,1) = (1,1)[/math]

So listing the elements of [math](F_9)^{\ast}[/math] as [math]\{(1,0),(2,0),(0,1),(1,1),(2,1),(0,2),(1,2),(2,2) \}[/math] we may identify [math]T[/math] with the 8-cycle:

[math](1\ 3\ 7\ 8\ 2\ 6\ 5\ 4) \in S_8[/math]

and [math]\langle T \rangle[/math] is clearly a cyclic subgroup of order 8 in [math]GL(2,3)[/math].

 

[Deveno]

I had to rush off while composing this, and reflecting upon it during my shopping, I realized there is an error.

For some reason, I had thought [math]p^n - 1[/math] was prime, which is clearly not so, so it is a false conclusion to invoke Lagrange to show that:

[math]|T|[/math] divides [math](p^n - 1)![/math] implies [math]|T| \leq p^n - 1[/math].

Fortunately, this is easily remedied:

Consider the algebra [math]F_p[T][/math]. Since [math]T[/math] satisfies a polynomial of degree n, [math]F_p[T][/math] has at most [math]p^n - 1[/math] non-zero elements. Since the powers of [math]T[/math] are among these, and since [math]T[/math] is an element of a finite group ([math]GL_n(F_p)[/math]), we conclude that the order of [math]T[/math] can be at most [math]p^n - 1[/math], and therefore that [math]|T| = p^n - 1[/math], as before.

 

[caffeinemachine]

I think what you are doing is not work in $(F_{p})^n$ as a vector space over $F_p$ but work in $F_{p^n}$ as a vector space over $F_p$. Showing the existence of a linear operator $T:F_{p^n}\to F_{p^n}$ which is good with a vector in $F_{p^n}$ (see my original post for the definition of goodness ) suffices.
For this solution to make sense it must be true that $F_p$ is embedded in $F_{p^n}$. I had not known about this fact since I have not studied finite field explicitly yet but I confirmed that this is indeed true. If you have an easy proof of this then please post it. Meanwhile I'll try to prove it myself.
If I am wrong here then please stop me right away because then I have clearly misunderstood you solution.

This is an interesting problem, and my solution is a bit circuitous (I apologize for that).

I will show that it suffices to exhibit an element [math]T \in GL_n(F_p)[/math], of order [math]p^n - 1[/math].

First of all, it should be clear from your opening remarks that we can consider [math]GL_n(F_p)[/math] as a subgroup of [math]S_{p^n - 1}[/math] (the symmetric group on [math]p^n - 1[/math] letters), so that ANY such element [math]\ T[/math] must be a [math](p^n - 1)[/math]-cycle.

Now, consider a primitive element [math]\xi \in (F_{p^n})^{\ast}[/math]. Let its minimal polynomial (necessarily of degree n) be:

[math]m(x) = c_0 + c_1x + \cdots + c_{n-1}x^{n-1} + x^n \in F_p[x][/math].

So here you are using the fact that $F_p$ is embedded in $F_{p^n}$ right?

We have the matrix:

[math]T = \begin{bmatrix}0&0&\dots&0&-c_0\\1&0&\dots&0&-c_1\\0&1&\dots&0&-c_2\\ \vdots&\vdots&\ddots&\vdots&\vdots\\0&0&\dots&1&-c_{n-1} \end{bmatrix}[/math]

with [math]m(T) = 0[/math] (see: Companion matrix - Wikipedia, the free encyclopedia).

Okay...

Since [math]\xi[/math] is an eigenvalue of [math]T[/math]...

An eigenvalue of any linear operator must come for the ground field of the vector space, isn't it? Here our ground field is $F_p$. I don't understand how can an element of $F_{p^n}$ be an eigenvalue?

... we have for any eigenvector [math]v \in (F_{p^n})^n[/math]:

I think you mean $v\in F_{p^n}$ and not $(F_{p^n})^n$. Am I right?

[math]T^kv = \xi^kv \neq v[/math] for [math]k = 1,2,\dots,p^n-1[/math].

This shows that the order of [math]T[/math] is at least [math]p^n - 1[/math].

Since the order of [math]T[/math] by Lagrange divides [math](p^n - 1)![/math], we conclude that:

[math]|T| = p^n - 1[/math]

I don't understand this because of the doubt I had above about $\xi$ being an eigenvalue and also a little bit unsure about $v$.

(In all fairness, this proof just turns the problem into a different one: that the field with [math]p^n[/math] elements has a primitive element, for example see: http://www.math.wm.edu/~vinroot/430S13MultFiniteField.pdf ).

Yes. I am comfortable with the cyclic nature of the group of units of any finite field.

As a concrete example: consider [math]p = 3, n = 2[/math]. We have:

[math]F_9 \cong \Bbb Z_3(u)[/math], where [math]u[/math] is a root of [math]x^2 + 1[/math] (in other words, [math]u[/math] is a square root of 2). A primitive element for [math]\Bbb Z_3(u)[/math] is [math]u+1[/math]:

[math](u+1)^2 = u^2 + 2u + 1 = 2 + 2u + 1 = 2u[/math],
[math](u+1)^4 = (2u)^2 = 2^2u^2 = u^2 = 2[/math],
[math](u+1)^8 = 2^2 = 1[/math].

The minimal polynomial for [math]u+1[/math] over [math]\Bbb Z_3[/math] is:

[math]m(x) = x^2 + x + 2[/math].

The companion matrix in this case is:

[math]T = \begin{bmatrix}0&1\\1&2 \end{bmatrix}[/math], and indeed:

[math]T^2 + T + 2I = \begin{bmatrix}0&1\\1&2 \end{bmatrix}^2 + \begin{bmatrix}0&1\\1&2 \end{bmatrix} + \begin{bmatrix}2&0\\0&2 \end{bmatrix}[/math]

[math]= \begin{bmatrix}1&2\\2&2 \end{bmatrix} + \begin{bmatrix}0&1\\1&2 \end{bmatrix} + \begin{bmatrix}2&0\\0&2 \end{bmatrix} = \begin{bmatrix}0&0\\0&0 \end{bmatrix}[/math].

Identifying [math]u+1[/math] with the vector [math](1,1)[/math] (in the basis: [math]\{1,u\}[/math]), we see that:

[math]T(1,1) = (1,0)[/math]
[math]T(1,0) = (0,1)[/math]
[math]T(0,1) = (1,2)[/math]
[math]T(1,2) = (2,2)[/math]
[math]T(2,2) = (2,0)[/math]
[math]T(2,0) = (0,2)[/math]
[math]T(0,2) = (2,1)[/math]
[math]T(2,1) = (1,1)[/math]

So listing the elements of [math](F_9)^{\ast}[/math] as [math]\{(1,0),(2,0),(0,1),(1,1),(2,1),(0,2),(1,2),(2,2) \}[/math] we may identify [math]T[/math] with the 8-cycle:

[math](1\ 3\ 7\ 8\ 2\ 6\ 5\ 4) \in S_8[/math]

and [math]\langle T \rangle[/math] is clearly a cyclic subgroup of order 8 in [math]GL(2,3)[/math].

I am trying to understand this. I'll get back in some time.

Thanks.

 

[johng1]

Hi,
I think once you know a little about finite fields, this is an easy problem.
Let GF(pⁿ) be the Galois field with pⁿ elements. Then GF(pⁿ) is a vector space V of dimension n over the field GF(p). So GL(n,K) with K=GF(p) is your linear group. Furthermore the multiplicative group of GF(pⁿ) is cyclic with generator say a. The order of a in this multiplicative group is then pⁿ-1 . Define the linear transformation T of V by T(v)=av for all v in V. Then clearly when T is viewed as a permutation on V, T is the product of a 1 cycle and a pⁿ-1 cycle.
I took this solution directly from B. Huppert, Endliche Gruppen I, p. 187

 

[caffeinemachine]

Hi,
I think once you know a little about finite fields, this is an easy problem.
Let GF(pⁿ) be the Galois field with pⁿ elements. Then GF(pⁿ) is a vector space V of dimension n over the field GF(p). So GL(n,K) with K=GF(p) is your linear group. Furthermore the multiplicative group of GF(pⁿ) is cyclic with generator say a. The order of a in this multiplicative group is then pⁿ-1 . Define the linear transformation T of V by T(v)=av for all v in V. Then clearly when T is viewed as a permutation on V, T is the product of a 1 cycle and a pⁿ-1 cycle.
I took this solution directly from B. Huppert, Endliche Gruppen I, p. 187

I understand this solution. Thanks.

Can you look into my approach see if you can make it work?

It's hard for me to abandon my approach because I had made good progress with it.

 

[Deveno]

I think what you are doing is not work in $(F_{p})^n$ as a vector space over $F_p$ but work in $F_{p^n}$ as a vector space over $F_p$. Showing the existence of a linear operator $T:F_{p^n}\to F_{p^n}$ which is good with a vector in $F_{p^n}$ (see my original post for the definition of goodness ) suffices.

You have a good eye...In fact, when working with eigenvalues and eigenvectors, it is often necessary to enlarge the base field...for example the matrix:

[math]\begin{bmatrix}\cos \theta&-\sin\theta \\ \sin\theta&\cos\theta \end{bmatrix}[/math]

has no real eigenvectors (except when [math]\theta = 0[/math] or [math]\pi[/math]), but it DOES have complex eigenvectors.

For this solution to make sense it must be true that $F_p$ is embedded in $F_{p^n}$. I had not known about this fact since I have not studied finite field explicitly yet but I confirmed that this is indeed true. If you have an easy proof of this then please post it. Meanwhile I'll try to prove it myself.
If I am wrong here then please stop me right away because then I have clearly misunderstood you solution.

[math]F_{p^n}[/math] can be realized as the unique (up to isomorphism) splitting field of the polynomial [math]x^{p^n} - x \in F_p[x][/math].

So here you are using the fact that $F_p$ is embedded in $F_{p^n}$ right?Okay...An eigenvalue of any linear operator must come for the ground field of the vector space, isn't it? Here our ground field is $F_p$. I don't understand how can an element of $F_{p^n}$ be an eigenvalue?

No, this is not true (see my example above). I used the larger field to show that if [math]T^k \neq I[/math] in this larger field, then it certainly doesn't have order k considered back in the base field.

I think you mean $v\in F_{p^n}$ and not $(F_{p^n})^n$. Am I right?

No, I meant in the larger field...the eigenvalues in this case don't exist in the smaller field (because of the irreducibility of m(x)).

I don't understand this because of the doubt I had above about $\xi$ being an eigenvalue and also a little bit unsure about $v$.

One can prove that the roots of m(x) are the eignevalues of the companion matrix using induction on n, for example see here:

homework - characteristic polynomial of companion matrix - Mathematics Stack Exchange

Yes. I am comfortable with the cyclic nature of the group of units of any finite field.I am trying to understand this. I'll get back in some time.

Thanks.

I would like to mention here that johng's comment is also dead on the money...what I did is show how one can specifically pull this back "to matrix form". The key point is that multiplication in an extension field [math]E[/math] over a base field [math]F[/math] is [math]F[/math]-linear, due to the field axioms. One generally has this result for any extension ring over a central sub-ring (a fact that is put to good use in studying algebras).

The interplay of linear algebra, polynomials, and fields in the finite case is almost magical to me. The finite general linear groups are fascinating, and somewhat mysterious.