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

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 ...

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

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 same …

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 ...

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 linear …

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?

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} \right) e^ { …