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Finding $Im(T)$ and $Ker(T)$ of the following linear transformation Note that the Row Space gives you the rank of A A and in your last matrix you have three linearly independent vectors That implies the Rank(A) = 3 R a n k (A) = 3 which is the dimension of the Im(A) I m (A) Thus the image of A A is a three dimensional subspace of R3 R 3 which is R3 R 3 In order to find the Ker(A) K e r (A) , you solve, AX = 0 A X = 0 which implies X = t(1, −1, −1, 1)T X
real analysis - Why doesnt IMT hold for all compact sets . . . 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 f: [a, b] → R f: [a, b] → R will suit for all f: S → R f: S → R with S S compact
Example of linear transformation on infinite dimensional vector space 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 V, and linear transformation T: V → V T: V → V, V = imT + kerT V = imT ⨁ kerT V = i m T + k e r T V = i m T ⨁ k e r T
Prove that $T^*$ is injective iff $ImT$ Is dense 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?
SageMath: Orthogonal projection of $\mathbb {C}^3$ onto a subspace. 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 projection matrix P_B Th issue is that this supposedly projection matrix I obtain is not even idempotent