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More precisely, the fundamental theorem of algebra asserts that every non-constant polynomial equation with real or complex coefficients has a solution which is a complex number. For example, the equation (+) = has no real solution, because the square of a real number cannot be negative, but has the two nonreal complex solutions + and .
In the above example of the complex numbers viewed as a two-dimensional algebra over the real numbers, the one-dimensional real line is a subalgebra. A left ideal of a K -algebra is a linear subspace that has the property that any element of the subspace multiplied on the left by any element of the algebra produces an element of the subspace.
The fundamental example of a linear complex structure is the structure on R 2n coming from the complex structure on C n.That is, the complex n-dimensional space C n is also a real 2n-dimensional space – using the same vector addition and real scalar multiplication – while multiplication by the complex number i is not only a complex linear transform of the space, thought of as a complex ...
In multilinear algebra, one considers multivariable linear transformations, that is, mappings that are linear in each of a number of different variables. This line of inquiry naturally leads to the idea of the dual space , the vector space V* consisting of linear maps f : V → F where F is the field of scalars.
In linear algebra and operator theory, the resolvent set of a linear operator is a set of complex numbers for which the operator is in some sense "well-behaved". The resolvent set plays an important role in the resolvent formalism .
The Fredholm alternative is the statement that, for every non-zero fixed complex number, either the first equation has a non-trivial solution, or the second equation has a solution for all (). A sufficient condition for this statement to be true is for K ( x , y ) {\displaystyle K(x,y)} to be square integrable on the rectangle [ a , b ] × [ a ...
The set of algebraic numbers is countable, [4] [5] and therefore its Lebesgue measure as a subset of the complex numbers is 0 (essentially, the algebraic numbers take up no space in the complex numbers). That is to say, "almost all" real and complex numbers are transcendental. All algebraic numbers are computable and therefore definable and ...
n(2n+1) u(n) square complex matrices A satisfying A = −A ∗, with Lie bracket the commutator Note: this is not a complex Lie algebra n 2: su(n) n≥2 square complex matrices A with trace 0 satisfying A = −A ∗, with Lie bracket the commutator Yes Yes Note: this is not a complex Lie algebra n 2 −1
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