linear algebra - if $T: V\to V$ and $ dim (KerT)+dim (ImT)=dimV
Mar 29, 2023 · $KerT+ImT=dimV$ ? Is this possible? $Ker T, Im T$ are subspaces of $V$ and $dimV$ is a just a...
real analysis - Why doesn't IMT hold for all compact sets ...
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 …
V = ImT \\oplus \\ KerT - Mathematics Stack Exchange
Linear Tranformation that preserves Direct sum V = ImT ⊕ KerT Ask Question Asked 12 years, 11 months ago Modified 12 years, 11 months ago
Prove that $T^*$ is injective iff $ImT$ Is dense
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?
Give an example of a linear map $T$ such that $\dim …
Jan 1, 2020 · This is completely correct. This will give a linear map with the properties you're asked for. I think that it is a bit too general to actually be "an example". I think it would be …
Finding $Im(T)$ and $Ker(T)$ of the following linear transformation
Let $$T:\\mathbb{R}^4\\to\\mathbb{R}^3$$ $$T(x,y,z,w)=(x-y+z-w,x+y,z+w)$$ I need to find $\\operatorname{Ker}(T),\\operatorname{Im}(T)$ and the basis of them and to ...
Finding the basis of ker (T) and im (T) - Mathematics Stack Exchange
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 ...
Find a basis for KerT and ImT (T is a linear transformation)
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?
SageMath: Orthogonal projection of $\mathbb {C}^3$ onto a …
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 …
Is it true that if $T$ is a linear operator on a finite-dimensional ...
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.