Mar 29, 2023 · But how can i proof that KerT ∩ ImT = {0} K e r T ∩ I m T = {0} or disprove it linear-algebra linear-transformations Share Cite

Apr 8, 2020 · 0 In my college's notes, it says for all compact sets, extreme value theorem holds but intermediate value theorem doesn't. I wonder why since I think the original proof of IMT for …

Linear Tranformation that preserves Direct sum V = ImT ⊕ KerT Ask Question Asked 12 years, 11 months ago Modified 12 years, 11 months ago

Jan 1, 2020 · Let's do an easy variation on your example, where V =R5 V = R 5 and W =R2 W = R 2. Define T T by T(x1,x2,x3,x4,x5) = (x1,x3) T (x 1, x 2, x 3, x 4, x 5) = (x 1, x 3).. Then …

Dec 21, 2014 · The title of your question does not really match the actual question (maybe the statement of the current question is used to prove the result in the title?). Is this intended?

Jun 15, 2019 · I managed to find the basis and the dimension for ImT I m T pretty easily, however how do I formally prove the dimension and the basis for KerT K e r T?

Dec 13, 2024 · Now, my problem arises when I evaluate P_imT with specific values of a,b,c (in this case, the standard basis of $\mathbb {C}^3$) in order to obtain the columns of the …

If you knew that $\ker T \cap \operatorname {im} T= \emptyset$, then you'd have a proof. But this isn't true, and you can easily find an example in small dimensional spaces.

Jul 19, 2021 · for part d, would elaborate by showing that the image of $T$ is equal to the span of $\ {1,x\}$. Since you already know that $1$ and $x$ are linearly independent ...

May 22, 2018 · I haven't had much experience with infinite dimensional vector spaces, and I was working on a problem that asks to prove that for a finite dimensional vector space V V, and …