Mar 29, 2023 · $KerT+ImT=dimV$ ? Is this possible? $Ker T, Im T$ are subspaces of $V$ and $dimV$ is a just a...

Note that the Row Space gives you the rank of A A and in your last matrix you have three linearly independent vectors. That implies the Rank(A) = 3 R a n k (A) = 3 which is the dimension of the …

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 ...

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 f: [a, b] → …

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?

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

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 linear …

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 projection matrix …

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?

Dec 5, 2013 · Let $T:\mathbb {R}^n \to \mathbb {R}^m$ and $S:\mathbb {R}^m \to \mathbb {R}^l$ be linear maps. I have to show that: $S∘T=0$ if and only if Im$T \subset$ ker$S$ Can ...