Prove Invariance of Intersection Under T

[evilpostingmong]

Homework Statement

Suppose T is contained in the set of linear transformations from
V to V. Prove that the intersection of any collection of subspaces of
V invariant under T is invariant under T.

Homework Equations

The Attempt at a Solution

Choose a basis for V. This basis is <v1...vn>.
Two possible subspaces for V are A and B with basis
<v1...vi> for A and <vi-1...vn> for B (2$$\leq$$i$$\leq$$n)
and we will assume that both are invariant under T.
A basis for the intersection of the bases is <vi-1, vi>.
Because the subspaces A and B are invariant under T, and
vi-1+vi is a linear combination within the span of both invariant subspaces
under T, vi-1+vi is mapped from A to A and from B to B thus it goes
from A and B to A and B, the intersection of both subspaces.

 

[CompuChip]

I think you can give the proof without choosing a basis for V, which I personally find more appealing. To do this, recall that a subset A c V is invariant under T whenever T(A) c A.

Now you want to show that $T(A \cap B) \subset A \cap B$. You can do this by showing that it is simultaneously a subset of both A and B.

 

[evilpostingmong]

Sounds good. Let x$$\in$$A$$\cap$$B
Distributing T, we get T(A$$\cap$$B)=T(A)$$\cap$$T(B).
Since x$$\in$$A, T(x)$$\in$$T(A) and since x$$\in$$(B),
T(x)$$\in$$T(B) thus T(x)$$\in$$T(A)$$\cap$$T(B)
and since both A and B are invariant under T (thus T(x) x$$\in$$A gets mapped to
A and T(x) x$$\in$$B gets mapped to B, for T(A) is in A and T(B) is in B for A and B to be invariant), T(x)=cx$$\in$$A$$\cap$$B.

 

[CompuChip]

Sounds good. Let x$$\in$$A$$\cap$$B

Good start.

Distributing T, we get T(A$$\cap$$B)=T(A)$$\cap$$T(B).

Probably that's true by linearity, although I don't even think you need that in the proof.
Let's call this statement (*).

Since x$$\in$$A, T(x)$$\in$$T(A) and
since x$$\in$$B, T(x)$$\in$$T(B) thus
T(x)$$\in$$T(A)$$\cap$$T(B)
and since both A and B are invariant under T (thus T(x) x$$\in$$A gets mapped to
A and T(x) x$$\in$$B gets mapped to B, for T(A) is in A and T(B) is in B for A and B to be invariant),

That part is entirely correct (try finding where you actually needed statement (*), because I couldn't ).

T(x)=cx$$\in$$A$$\cap$$B.

This may be written down a bit sloppily, with respect to your rigour before. From T(x) being in T(A) you can conclude that it is also in A, and from T(x) being in T(B) it is also in B. So T(x) is in A and B at the same time. QED

Apart from those two minor details, well done!

 

[evilpostingmong]

Thanks compuchip, I need to work on writing more elegant proofs.
Sometimes I need to read my own a few times to get them, lol.

 

[CompuChip]

That's ok, luckily this is a classic example of "practice makes perfect".
So just write a lot of them.

And if you read them and it's hard for you to decipher what you meant, you really need to re-write them :)