Do these two statements imply an underlying induction proof?

[zenterix]

Homework Statement

Suppose ##T\in\mathcal{L}(V)##.
Suppose ##U## is a subspace of ##V## invariant under ##T##.
Prove that ##U## is invariant under ##p(T)## for every polynomial ##p\in\mathcal{P}(\mathbb{F})##.

Relevant Equations

My question is about writing proofs.

Here is one proof

$$\forall u\in U\implies Tu\in U\subset V\implies T^2u\in U\implies \forall m\in\mathbb{N}, T^m\in U\tag{1}$$

Is the statement above actually a proof that ##\forall m\in\mathbb{N}, T^m\in U## or is it just shorthand for "this can be proved by induction"?
In other words, for the proof to be rigorous, would (1) have to be expanded out into an induction proof?

Then

$$\forall u\in U\implies p(T)(u)=\sum\limits_{i=0}^na_i T^iu\tag{2}$$

The righthand sum is a linear combination of vectors in ##U##. Thus, ##p(T)u\in U## and so ##U## is invariant under ##p(T)##.

Here again I have the same doubt. To make the argument as rigorous as possible, does (2) and the sentence above need to be made into an induction argument?

 

[PeroK]

At this level and given the simplicity of the argument I would say you only need to mention induction.

Note, however, that ##\forall u \in U \Rightarrow## is not a standard construction. It should be either ##\forall u \in U:## or ##u \in U \Rightarrow##.

 

[Mark44]

Here is one proof

$$\forall u\in U\implies Tu\in U\subset V\implies T^2u\in U\implies \forall m\in\mathbb{N}, T^m\in U\tag{1}$$

The above doesn't make sense to me.
As already mentioned, ##\forall u\in U## is not a statement. I.e., it's neither true nor false. Did you mean this: ##\forall u\in U, Tu \in U \subset V##?

 

[zenterix]

@Mark44

In English, I want to say

If a vector ##u## is in ##U## then ##Tu## is in ##U##.

Another way to say this

Let ##u\in V##.

Suppose ##u\in U##.
(...)
Then ##Tu\in U##

Thus, I can infer the conditional

##u\in U\implies Tu\in U##.

And since I used a generic ##u\in V## then I can write

##\forall u [(u\in V\ \text{and}\ u\in U) \implies Tu\in U]##.

Now, if it is implicit that all the vectors I am considering are in ##V## then perhaps I could write just

##\forall u (u\in U\implies Tu\in U)##

When I used ##\forall u\in U## I meant ##\forall u, u\in U##.

I guess at this point I do this in my own proofs as a shorthand to not write everything out, but you are right, ##\forall## is a quantifier and expresses the quantity of things that satisfy some condition. The latter (the condition) must be present to have a statement. In my case it is ##u\in U\implies Tu\in U##.

Note that the chain of conditionals has all sorts of my shorthand notation.

For example, the second conditional should be

##\forall u (Tu\in U \implies T^2u\in U)##

The third conditional is actually

##\forall u (u \in U\implies [\forall m (m\in\mathbb{N}\implies T^m\in U)])##

So the question now is: do people really write out such logic statements without using any kind of shorthand notation?

I mean, as I am writing out the proof, I could write out all the statements in a way that obeys the rules of first order logic, but it would take longer.

 

[WWGD]

If ##T(U) \subset U##, the same would be the case for ##T^n(U)##, as ##T^2(U)=T( T(U))## would be operating on U itself*. All thats left is to verify it for the scaling . Maybe you can first define a basis for U to accomplish that.

*A copy of U, to be more accurate, but the point stands.

 

[PeroK]

I mean, as I am writing out the proof, I could write out all the statements in a way that obeys the rules of first order logic, but it would take longer.

If you have a good linear algebra textbook, you should see the appropriate mathematical notation.

Something like this is definitely not needed.

##\forall u (u \in U\implies [\forall m (m\in\mathbb{N}\implies T^m\in U)])##

And, you make it so complicated, that your notation is starting to break down.

 

[zenterix]

Here is an example of how my textbook (Axler Linear Algebra Done Right) writes two statements (I am paraphrasing)

1) For all ##m\times n## matrices ##A## and ##C## we have ##(A+C)^t=A^t+C^t##.

2) For all ##m\times n## matrices ##A## and ##C## and all ##\lambda\in\mathbb{F}##, we have ##(\lambda A)^t=\lambda A^t##.

And how these statements would look with only what I think we can call first-order logic (I am not totally sure if the notation below uses elements of logic beyond first-order logic, which is what I learned, but not in the context of math):

1) ##\forall\ m\ \forall\ n (A\in\mathbb{F}^{m,n}, C\in\mathbb{F}^{m,n}\implies (A+C)^t=A^t+C^t)##

2) ##\forall\ m\ \forall\ n\ \forall\ \lambda (\lambda\in\mathbb{F}, A\in\mathbb{F}^{m,n} \implies (\lambda A)^t=\lambda A^t)##

And here is how I have been writing this in my own notes

1) ##\forall A,C\in \mathbb{F}^{m,n}\implies (A+C)^t=A^t+C^t##

2) ##\forall \lambda\in\mathbb{F}, \forall A\in\mathbb{F}^{m,n} \implies (\lambda A)^t=\lambda A^t##

 

[zenterix]

Here is an attempt to rewrite

##\forall u\in U\implies Tu\in U\subset V\implies T^2u\in U\implies\forall m\in\mathbb{N}, T^m\in U##.

First with mostly English.

Since ##U## is invariant under the linear operator ##T##, then every ##u\in U## is mapped to ##Tu\in U\subset V##.

Similarly, if ##Tu\in U## then ##T^2u\in U##.

(..induction argument)

Therefore, by induction, we conclude that for all natural numbers ##m## we have ##T^m\in U##.

Now with mostly symbols.

Let ##u\in U##.

Then ##Tu\in U##.

Then ##T^2u\in U##.

(..induction argument)

By induction, we can show that ##\forall\ m (m\in\mathbb{N}\implies T^mu\in U)##.

Finally, with slightly altered notation I have been using.

##\forall u (u\in U\implies Tu\in U \implies T^2u\in U \xrightarrow{\text{induction}} \forall m(m\in\mathbb{N} \implies T^m\in U))##

Now, I would usually write just

##\forall u (u\in U\implies Tu\in U \implies T^2u\in U \xrightarrow{\text{induction}} T^m\in U, \forall\ m\in\mathbb{N})##

I can see how my notation is confusing to anyone else reading the argument.

For myself, when sketching a proof, what matters most to me is building the argument and so this last notation is what I usually use. But I recognize that it doesn't make sense in a rigorous sense.

But it makes sense in terms of expressing my own reasoning. I can always rewrite it. Is this not a thing?