- #1
Incand
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So the theorem says:
Suppose that ##U## and ##V## are finite dimensional vector spaces, and that ##T:U\to V##, ##S: V \to W##. Then
##\text{dim Ker }ST \le \text{dim Ker }S + \text{dim Ker }T##.
Proof:
Set ##U_0 = \text{Ker }ST## and ##V_0 = \text{Ker }S##. ##U_0## and ##V_0## are subspaces of ##U## and ##V##. Since ##ST=0## on ##U_0## we have ##T(U_0) \subset V_0##. We can then consider ##T## and ##S## as mappings defined on ##U_0## respectively ##V_0##. Since ##\text{Ran }T## (Range, column space, image etc.) is a subspace to ##V_0## we have ##\text{dim Ran }T \le \text{dim }V_0 = \text{dim Ker }S##. But according to the Rank-nullity theorem, ##\text{dim Ker }T + \text{dim Ran }T = \text{dim }U_0##.
Hence ##\text{dim Ker }ST = \text{dim }U_0 = \text{dim Ker T} + \text{dim Ran }T \le \text{dim Ker }T + \text{dim Ker }S##.
(I translated this so it's possible worse worded than originally.)
I can't really follow the steps in this proof, no matter how many times I look through it I don't follow all the steps. Starting it with
##U_0## and ##V_0## are subspaces of ##U## and ##V##
The word "and"" confuse me here. It's supposed to mean that ##U_0## to ##U## and ##V_0## to ##V## and nothing else right?
We can then consider ##T## and ##S## as mappings defined on ##U_0## respectively ##V_0##.
What does this mean? That for our purposes we can consider ##T:U_0 \to V## and ##S:V_0 \to W##? Why?
Since ##\text{Ran }T## is a subspace to ##V_0##
I understand this to be true if we indeed consider ##T:U_0 \to V## since by definition ##T(U_0)## maps into ##V_0## but not for ##U## in general.
And then ##\text{dim Ker }T + \text{dim Ran }T = \text{dim }U_0##. I don't get how we're allowed to only consider ##U_0## instead of ##U##.
Suppose that ##U## and ##V## are finite dimensional vector spaces, and that ##T:U\to V##, ##S: V \to W##. Then
##\text{dim Ker }ST \le \text{dim Ker }S + \text{dim Ker }T##.
Proof:
Set ##U_0 = \text{Ker }ST## and ##V_0 = \text{Ker }S##. ##U_0## and ##V_0## are subspaces of ##U## and ##V##. Since ##ST=0## on ##U_0## we have ##T(U_0) \subset V_0##. We can then consider ##T## and ##S## as mappings defined on ##U_0## respectively ##V_0##. Since ##\text{Ran }T## (Range, column space, image etc.) is a subspace to ##V_0## we have ##\text{dim Ran }T \le \text{dim }V_0 = \text{dim Ker }S##. But according to the Rank-nullity theorem, ##\text{dim Ker }T + \text{dim Ran }T = \text{dim }U_0##.
Hence ##\text{dim Ker }ST = \text{dim }U_0 = \text{dim Ker T} + \text{dim Ran }T \le \text{dim Ker }T + \text{dim Ker }S##.
(I translated this so it's possible worse worded than originally.)
I can't really follow the steps in this proof, no matter how many times I look through it I don't follow all the steps. Starting it with
##U_0## and ##V_0## are subspaces of ##U## and ##V##
The word "and"" confuse me here. It's supposed to mean that ##U_0## to ##U## and ##V_0## to ##V## and nothing else right?
We can then consider ##T## and ##S## as mappings defined on ##U_0## respectively ##V_0##.
What does this mean? That for our purposes we can consider ##T:U_0 \to V## and ##S:V_0 \to W##? Why?
Since ##\text{Ran }T## is a subspace to ##V_0##
I understand this to be true if we indeed consider ##T:U_0 \to V## since by definition ##T(U_0)## maps into ##V_0## but not for ##U## in general.
And then ##\text{dim Ker }T + \text{dim Ran }T = \text{dim }U_0##. I don't get how we're allowed to only consider ##U_0## instead of ##U##.