/sci/ - Science & Math

[Anonymous No.6314968]

Hi guys, i have this proof for =>.
<span class="math">nullT\bigoplusimageT=V means that nullT\capImageT={0} and nullT+imageT=V.[/spoiler]
since the eigenvalue λ=0, then (T-0I)(v)=0, so Tv=0 and this will compose the null space for V. I have no idea where to go from now. i actually think what ive shown above might not be true in fact.

[Anonymous No.6314970]

<span class="math">nullT\bigoplus imageT=V means that nullT\capImageT={0} and nullT+imageT=V.[/spoiler]

[Anonymous No.6314973]

<span class="math">nullT\bigoplusimageT=V means that nullT\cap ImageT={0} and nullT+imageT=V.[/spoiler]

[Anonymous No.6314980]

<span class="math">nullT\bigoplus imageT=V means that nullT\cap ImageT={0} and nullT+imageT=V.[/spoiler]

[Anonymous No.6314983]

<span class="math">nullT\bigoplus imageT=V[/spoiler] means that <span class="math">nullT\capI mageT={0}[/spoiler] and <span class="math">nullT+imageT=V.[/spoiler]

[Anonymous No.6314991]

[Anonymous No.6314988]

<span class="math">nullT\bigoplus imageT=V[/spoiler] means that <span class="math">nullT\cap ImageT={0}[/spoiler] and <span class="math">nullT+imageT=V.[/spoiler]

[Anonymous No.6315013]

Anyway bro, if V decomposes as direct sum im(T)+ker(T), then let T^nv = 0 for smallest possible positive n. Then T^(n-1)v is in both the kernel and image of T, and so is 0. It follows that n-1 is not positive, ie n = 1.

Conversely, if K_0 = E_0 and v is in the intersection of Im(T) and null(T), then Tu = v and Tv = 0, so T^2u = 0 => 0 = Tu = v. So im(T) and ker(T) are disjoint, and dimension theorems tell you that V decomposes as a direct sum of these two.

Sorry for no latex.

[Anonymous No.6315022]

>>6315013
Thanks for the proof bro.
I have one more question. i can post the whole theorem if you want. If The minimal polynomial consists of one eigenvalue, does it mean that the vector space is decomposable?

[Anonymous No.6315029]

>>6315022
decomposable as Im(T)+Null(T)? You can take a matrix T with zeros everywhere except a 1 in upper right corner. Then T^2 = 0 and so x^2 is the minimal polynomial with single root, but you will see that kernel of T is a subset of its image, so such decomposition is impossible. Of course this follows from your previous exercise, for in this case K_0 is not equal to E_0.

[Anonymous No.6315037]

>>6315029
i think that what theyre asking is that if the eigenvalue is not equal to zero.
Correct me if im wrong, but if the minimal polynomial has only one lambda as an eigenvalue, then there is nothing shooting into the null space, and there will be no way to that we can decompose V as we wont be able to write <span class="math">W_1\cap W_2={0}[/spoiler]

[Anonymous No.6315060]

>>6315037
Well, consider diagonal matrices which are scalar multiples of the identity. These correspond to linear operators which have only one eigenvalue but they decompose 'totally', as a sum of n 1-dimensional invariant subspaces. So what you claim isnt true. The decomposition doesnt necessarily have to involve Im(T) and Ker(T).

[Anonymous No.6315061]

>>6315013
The image and kernel of an operator are never disjoint, they both contain 0.

[Anonymous No.6315067]

>>6315061
Yes thanks bro, but I think everyone and their mothers understood anyway.

[Anonymous No.6316540]

>>6315060
>>6315067
>>6315037
Ok guys ive typed up this proof, can you check whether it makes sense pls
Im going to bump the thread in the morning again(its 3:30 now)

Assume that the minimal polynomial is <span class="math">p(x)=(p(x))^n[/spoiler]
If<span class="math">V=V_1 \bigoplus V_2[/spoiler] then <span class="math">P_{V_1}(x)[/spoiler] and <span class="math">P_{V_2}(x)[/spoiler] must divide p(x) as p(x) annihilates both <span class="math">V_1[/spoiler] and <span class="math">V_2[/spoiler]
But p is irreducible, so it must be that <span class="math">V_i=0[/spoiler] for either i=1, or i=2

now Assume V is indecomposable. If min polynomial p(x) has 2 or more distinct roots, then <span class="math">\lambda I-T[/spoiler] has at least 2 divisors. But this means that V is decomposable.

[Anonymous No.6316835]

bump

[Anonymous No.6317076]

Bump