description/proof of rank-nullity law for linear map between finite-dimensional vectors spaces
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of rank of linear map between vectors spaces.
- The reader knows a definition of nullity of linear map between vectors spaces.
- The reader admits the proposition that for any vectors space, the bases have the same cardinality.
Target Context
- The reader will have a description and a proof of the rank-nullity law for linear map between finite-dimensional vectors spaces.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\text{ the fields }\}\)
\(V_1\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(V_2\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(f\): \(: V_1 \to V_2\), \(\in \{\text{ the linear maps }\}\)
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Statements:
\(Dim (V_1) = Rank (f) + Nullity (f)\)
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2: Proof
Whole Strategy: Step 1: take any basis for \(f (V_1)\), \((f (v_{1, 1}) = b_{2, 1}, ..., f (v_{1, r}) = b_{2, r})\), and any basis for \(Ker (f)\), \((b_{1, 1}, ..., b_{1, k})\); Step 2: see that \((b_{1, 1}, ..., b_{1, k}, v_{1, 1}, ..., v_{1, r})\) is a basis for \(V_1\).
Step 1:
Let us take any basis for \(f (V_1)\), \((f (v_{1, 1}) = b_{2, 1}, ..., f (v_{1, r}) = b_{2, r})\): there may be some multiple options for \(v_{1, j}\) to \(b_{2, j}\), but choose any of the options.
Let us take any basis for \(Ker (f)\), \((b_{1, 1}, ..., b_{1, k})\).
Step 2:
Let us see that \((b_{1, 1}, ..., b_{1, k}, v_{1, 1}, ..., v_{1, r})\) is a basis for \(V_1\).
Let \(r^1 b_{1, 1} + ... + r^k b_{1, k} + s^1 v_{1, 1} + ... + s^r v_{1, r} = 0\), where \(r^j, s^j \in F\).
\(f (r^1 b_{1, 1} + ... + r^k b_{1, k} + s^1 v_{1, 1} + ... + s^r v_{1, r}) = f (0) = 0\).
\(= r^1 f (b_{1, 1}) + ... + r^k f (b_{1, k}) + s^1 f (v_{1, 1}) + ... + s^r f (v_{1, r}) = r^1 0 + ... + r^k 0 + s^1 b_{2, 1} + ... + s^r b_{2, r}\).
That implies that \(s^1 = ... = s^r = 0\).
So, \(r^1 b_{1, 1} + ... + r^k b_{1, k} = 0\).
That implies that \(r^1 = ... = r^k = 0\).
So, \((b_{1, 1}, ..., b_{1, k}, v_{1, 1}, ..., v_{1, r})\) is linearly independent.
Let \(v \in V_1\) be any.
\(f (v) = c^1 b_{2, 1} + ... c^r b_{2, r}\).
\(v - (c^1 v_{1, 1} + ... + c^r v_{1, r}) \in Ker (f)\), because \(f (v - (c^1 v_{1, 1} + ... + c^r v_{1, r})) = f (v) - (c^1 f (v_{1, 1}) + ... + c^r f (v_{1, r})) = f (v) - (c^1 b_{2, 1} + ... + c^r b_{2, r}) = 0\).
So, \(v - (c^1 v_{1, 1} + ... + c^r v_{1, r}) = d^1 b_{1, 1} + ... d^k b_{1, k}\).
So, \(v = d^1 b_{1, 1} + ... d^k b_{1, k} + c^1 v_{1, 1} + ... + c^r v_{1, r}\).
So, \((b_{1, 1}, ..., b_{1, k}, v_{1, 1}, ..., v_{1, r})\) spans \(V_1\).
So, \((b_{1, 1}, ..., b_{1, k}, v_{1, 1}, ..., v_{1, r})\) is a basis for \(V_1\).
By the proposition that for any vectors space, the bases have the same cardinality, \(Dim (V_1) = Rank (f) + Nullity (f)\).
