Rank-nullity theorem
Let
and thus the sum of the rank and the nullity equals the dimension of
In full generality, this is downstream of AC.
Proof
By ^Existence we have
whence π = k e r β‘ π β ( k e r β‘ π ) π . Let d i m β‘ π = n u l l i t y β‘ π + d i m β‘ ( ( k e r β‘ π ) π ) . Note π π = π βΎ ( k e r β‘ π ) π is monic since π π . Let k e r β‘ π π = ( k e r β‘ π ) β© ( k e r β‘ π ) π = { 0 } . Since π π£ β i m β‘ π for π£ = π’ + π’ π and π’ β k e r β‘ π we have π’ β ( k e r β‘ π ) π π π£ = π π’ + π π’ π = π π’ π = π π π’ π β i m β‘ ( π π ) hence
so i m β‘ ( π ) β i m β‘ ( π π ) is an isomorphism It follows immediately that π π : ( k e r β‘ π ) π β i m β‘ π . r a n k β‘ π + n u l l i t y β‘ π = d i m β‘ π
Corollaries
Footnotes
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2008. Advanced Linear Algebra, p. 63 β©