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

Linear Tranformation that preserves Direct sum $ V = ImT \oplus \ KerT $ Ask Question Asked 12 years, 11 months ago Modified 12 years, 10 months ago

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?

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 ...

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

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 …

Apr 25, 2025 · $$ \frac {c_m} { (kn)^m} = \frac {1} {2\pi} \int_0^ {2\pi} \left (1 + \frac {k^2} {kn}e^ {it} + \dots \right)^n e^ {-imt}dz \\ \to_ {n\to \infty} \frac {1} {2\pi} \int_0^ {2\pi} \exp\left ( ke^ {it} …

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 …

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 …

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 $V$, and …