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

Jul 19, 2021 · For the most part, you're correct. Part (a) is fairly straightforward; B:= {1, x,x2} B:= {1, x, x 2} is the standard basis we use for this space after all. You should probably verify that it is, indeed, a …

Mar 1, 2015 · Let $T:V→W$ be linear transformation and V have a finite dimension. Show that $ImT^t=(kerT)°$ I have to prove it by mutual inclusion. I have proven the first ...

May 26, 2023 · This means we have $v \in (ImT^*)^\bot$ and therfore we have $KerT \subseteq (ImT^*)^\bot$. For the other side, consider $0 \neq v \in (ImT^*)^\bot$, (which exists from the same …

Jun 15, 2019 · Find a basis for KerT and ImT (T is a linear transformation) Ask Question Asked 6 years, 8 months ago Modified 6 years, 8 months ago

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 ...

Linear Alegbra - Find Base for ImT and KerT Ask Question Asked 11 years, 3 months ago Modified 11 years, 3 months ago

The prefactors p r e f a c t o r s can only be determined from the 'exact evaluation' ( see the @Brightsun fine answer ) but the qualitative evaluation yields the correct asymptotic behaviours m−−−√ …