$KerT+ImT=dimV$ ? Is this possible? $Ker T, Im T$ are subspaces of $V$ and $dimV$ is a just a...
linear algebra - if $T: V\to V$ and $ dim (KerT)+dim (ImT)=dimV $ can i ...
Since $ImT$ and $KerT$ are a direct sum we have $ KerS = \ { \ 0 \ } $ and since we have $$ dimImS \ + \ dimKerS \ = dimV $$ we get $$ V = Im S \Rightarrow \ V = KerS \oplus \ ImS$$
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 ...
1) "and therefore since $V = ImT^* \oplus (ImT^*)^\bot$, we have $ (ImT^*)^\bot=\ {0\}$" can be replaced with "and therefore $ (ImT^*)^\bot=V^\perp=\ {0\}$". 2) $Find a basis for KerT and ImT (T is a linear transformation) Ask Question Asked 6 years, 10 months ago Modified 6 years, 10 months ago
Find a basis for KerT and ImT (T is a linear transformation)
You may check your basis for $ImT$. $\dim Im T $ must be $2$.
$$ \frac {c_m} { (kn)^m} = \frac {1} {2\pi} \int_0^ {2\pi} \left (1 + \frac {k^2} {kn}e^ {it} + \dots \right)^n e^ {-imt}dz \ \to_ {n\to \infty} \frac {1} {2\pi} \int_0^ {2\pi} \exp\left ( ke^ {it} \right) e^ {-imt}dz \ = \frac {1} {2\pi i} \int_ { (0)^+} \frac {\exp (ks)} {s^ {m+1}}ds = \frac {k^m} {m!}. $$