I am self-studying Linear Algebra using Axler Linear Algebra Done Right 3rd Edition. Usually I use the problems from the 2nd edition because there is a solution manual available whereas the …

Considering the integrand as the Fourier transform of a tempered distribution, it makes sense then to write $$ \int_ {m}^ {+\infty}\sqrt {E^2-m^2}e^ {-iEt}dE = -\frac {\partial^2} {\partial t^2}\int_ …

You will write down a matrix with the desired $\ker$, and any matrix represents a linear map :) No, you want to think geometrically. The key thing is that the kernel is the orthogonal complement …

Here is another version of the Cantor-Lebesgue theorem that is a little more general than the one stated in the OP: Theorem (Cantor-Lebesgue): Let $ (c_n:n\in\mathbb ...

Then, can I assume $ (0,0,1)$ not inside $ImT$ because it's not linearly independent of $ (1,0,0), (0,1,0)$ and also assume that $ (0,0,1) \in KerT$ because it's not in $ImT$ and thus there is a …

Let $X$ and $Y$ be Banach spaces and $T\\in B(X,Y)$ be Fredholm. Then there is $S\\in B(Y,X)$ such that $ST=I+K_{1}$ and $TS=I+K_{2}$ where $K_{1},K_{2}$ are compact ...

Feb 1, 2012 · @Listing: First of all, there is a continuous surjective map from $\mathbb {R}$ to $\mathbb {R}^2$: you seem to have misread the link you quoted. (There is no continuous …

Nov 22, 2021 · $=\frac {1} {2\pi} \int_ {-\pi}^ {\pi}P_r (\theta -t)\displaystyle \sum_ {m=-N}^ {N} c_me^ {imt}dt$ $=\frac {1} {2\pi} \int_ {-\pi}^ {\pi}\displaystyle \sum_ {m=-N}^ {N} c_me^ …

So I think the the image is two dimensional, right? Since dim (kerT) + dim (imT)=dim ($\mathbb {R}^6$) and since the image is 2, does it mean that the kernel of the linear map must be 4?

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