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?