# How do I prove this? [Linear Algebra]

1. Nov 5, 2004

### AKG

Let $T : V \to V$ be a linear operator on an n-dimensional vector space V. Assume that the characteristic polynomial of T splits over F (the field underlying V). Prove that the following are equivalent:
1. There exists a vector $x \in V$ such that $\{x,T(x),T^2(x),\dots ,T^{n - 1}(x)\}$ is linearly independent.
2. The characteristic polynomial of T is equal to $(-1)^n$ times the minimal polynomial of T.
3. There exists a basis $\beta$ such that

$$[T]_{\beta} = \left (\begin{array}{ccccc}0 & 0 & \dots & 0 & (-1)^{n - 1}a_0\\1 & 0 & \dots & 0 & (-1)^{n - 1}a_1\\0 & 1 & \dots & 0 & (-1)^{n - 1}a_2\\ \vdots & \vdots & \ddots & \vdots & \vdots \\0 & 0 & \dots & 1 & (-1)^{n - 1}a_{n - 1}\end{array}\right )$$

where $(-1)^nt^n + a_{n - 1}t^{n - 1} + \dots + a_1t + a_0$ is the characteristic polynomial of T.
(Hint: For proving that (2) implies (3), it is helpful to show that the Jordan canonical form of T is the same as the Jordan canonical form of the matrix given in part (3)).
I can prove (1) implies (2). In my book, the section on Jordan canonical form comes before the section on minimal polynomials, and then the section on minimal polynomials says very little about Jordan canonical form, so the hint given to prove (2) implies (3) isn't helping me too much. Any hints and suggestions would be helpful, thanks.

Last edited: Nov 5, 2004
2. Nov 5, 2004

### AKG

Can this even be proved? In my book, a proof for (1) implies (2) is basically given (which is why I can do it), but it says nothing along the lines of (2) implies (1). That is, if it is in fact true that (1) iff (2), why would my book only prove (1) implies (2) and not even suggest that the converse is true?

Furthermore, I can't seem to figure out what the minimal polynomial tells you about the Jordan canonical form at all, so I can't see why (2) would imply (3). The only possible useful facts I know are that the Jordan canonical form of an operator is the same as the Jordan canonical form of the matrix of that operator with respect to some basis, and that if two operators', S's and T's, characteristic polynomials split, then they have the same Jordan canonical form if and only if they are similar to each other, i.e. there exists an invertible operator U such that $S = U^{-1}TU$. Let's call the matrix labeled $[T]_{\beta}$ A, i.e. $A = [T]_{\beta}$. Let $\beta \prime$ be a Jordan canonical basis for T. Then $[T]_{\beta \prime}$ is the Jordan canonical form for T. Now, if I compute the characteristic polynomial of A, I suspect I will get it to look something like the characteristic polynomial of T, and therefore I know it will split. Now, if I can show that the Jordan canonical form of A is the same as $[T]_{\beta \prime}$, then that's implies that A is similar to $[T]_{\beta \prime}$. If this is true, then $A = U^{-1}[T]_{\beta \prime}U$. U can be seen as a change of basis matrix, specifically, a change of basis matrix from some basis $\beta$ to $\beta \prime$. This implies that there is some basis $\beta$ such that $[T]_{\beta} = A$, as required.

The problem with the above is getting the Jordan canonical form of the operator T satisfying condition (2). Like I said, all I have, as far as I can tell, are some facts about the characteristic polynomial and minimal polynomial of T, but I don't think I have enough to say anything about it's Jordan canonical form. Is there something I'm missing.

And I haven't even looked at how I would go about proving (3) implies (1). Glancing at it, I'm even more lost for that part than I am for the part proving (2) implies (3).

Please! Any help will be appreciated!

3. Nov 13, 2004

### lurflurf

I remember freshman year fun stuff!
You are really good with all the fancy symbol markups. I am not, but I hope I can still be clear.

It seems you define the characteristic polynomial as det(A-I*t) so we have all the (-1)^n stuff and that you define the minimal polynomial to be monic. I hope when you talk about what section are before others that you read all of them, I assume minimal polynomial is defined as
a monic poly mp so that mp(T)=0 and if p(T)=0 for some poly p, then mp divides p. I we use characteristic of T splits over F only to assure T has n eigen values in F (counting multiplicity).

I think
3=>1=>2=>3 should do it

3=>1
let x=e(1) the vector [1,0,...,0]
then $$[T]_{\beta}$$^k*e(1)=e(k+1) k=0,...,n-1
(this is clear from the simple basis)
clearly {e(1),...,e(n)} is linearly independent
so 1 is true in the basis B hence true in any basis
qed

1=>2
1 implies that no nontrivial polynomials in T of order less than n are equal to the 0 operator. So deg(minimal polynomial)=n
deg(characteristic polynomial)=n so they are equal up to multiplication by a constant. By the definitions the constant is (-1)^n
qed

2=>3
we will show $$[T]_{\beta}$$ and T have the same jordan conical form, thus they are similar, hence a basis between them exist. We know $$[T]_{\beta}$$ will have all of its Jordan blocks full. That is if lamda is an eigenvalue of multiplicity m Nullity(($$[T]_{\beta}$$-lamda*I)^k)=k k=0,1,...,m, and the jordan block is J_m(lamda). We know this because even though we do not yet know that the basis B will exist given (2) we do know from above that (1) and (2) hold (for $$[T]_{\beta}$$) if $$[T]_{\beta}$$ is defined as it is in (3) Thus $$[T]_{\beta}$$ and T will have the same Jordan form iff they have the same characteristic function.
we see
$$[T]_{\beta}$$^n*e(1)=(-1)^k Sum[a(k)e(k+1),{k,0,n-1}]
putting
e(k+1)=$$[T]_{\beta}$$^k+*e(1) k=0,...,n-1 as before
we see
((-1)^n $$[T]_{\beta}$$^n+Sum[a(k)$$[T]_{\beta}$$^k,{k,0,n-1}])*e(1)=0
appying $$[T]_{\beta}$$^k k=0,...,n-1 to both sides and knowing polynomials in $$[T]_{\beta}$$ commute we have
((-1)^n $$[T]_{\beta}$$^n+Sum[a(k)$$[T]_{\beta}$$^k,{k,0,n-1}])*$$[T]_{\beta}$$^k*e(1) k=0,...,n-1
or
$$[T]_{\beta}$$^n+Sum[a(k)$$[T]_{\beta}$$^k,{k,0,n-1}])*e(k)=0 k=1,...,n
{e(1),...,e(n)} is a basis so
$$[T]_{\beta}$$^n+Sum[a(k)$$[T]_{\beta}$$^k,{k,0,n-1}])=0
thus the two characteristic polys agree $$[T]_{\beta}$$ and T are similar (since same Jordan form) so the basis exists

Last edited: Nov 13, 2004
4. Nov 13, 2004

### AKG

Thanks anyways, I did it already. I wasn't able to do it before because I needed to know something that hadn't been taught at that point, nor was it in the book, namely "We know $[T]_{\beta}$ will have all of its Jordan blocks full."