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Find a basis for KerT and ImT (T is a linear transformation)
$\begingroup$ You can, for instance, use the "dimensions thm", and you have that $5=\dim(Ker) + \dim(Im)$ , so you know which is the dimension of the kernel
Find Base for ImT and KerT - Mathematics Stack Exchange
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V = ImT \\oplus \\ KerT - Mathematics Stack Exchange
I'm trying to prove the following: $\ \ \ Im L \oplus KerL = V \ \ $ when $\ \ \ L = T^k \ \ \ $ When: $\mathbb T: V \rightarrow V \ $ and $ ImT \oplus KerT =V$ My take on this is below I would like to know if my proof is a proper one? If not can you point me to where it fails? Any comment's and suggestions would be much appreciated
linear algebra - Proof that columns of an invertible matrix are . . .
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reference request - What are usual notations for surjective, injective . . .
Update: In the category of sets, an epimorphism is a surjective map and a monomorphism is an injective map As is mentioned in the morphisms question, the usual notation is $\\rightarrowtail$ or $\\
Methods for choosing $u$ and $dv$ when integrating by parts?
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What does a having pivot in every row tell us? What about a pivot in . . .
$\begingroup$ why does having a pivot in every row necessarily mean Ax=b has at least one solution? Even if there weren't pivots in every row, couldn't we still have solutions—for eg, if A = [4 5 6 ; 0 0 0] and b= [5 ; 0] then we have 4*x_1 + 5*x_2 + 6*x_3 = 5, which does give at least one solution (x_2 and x_3 in particular are free variables), but A doesn't have a pivot in every row