anonymous
  • anonymous
Linear Algebra : Cⁿ(complex number)is a vector space over field R(real number).
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
can i get some help please .......
anonymous
  • anonymous
what is the question asking to be found about the vector space?
anonymous
  • anonymous
i need proof....

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
i've only ever bothered proving this as part of a relation to a matrix. I think you have to show it's closed under addition and multiplication and to itself, using z = a +ib with a and b in the reals, show for complex y and z, y+z and y*z are complex. I've only ever bothered showing by relation to the matrix: Z = [a -b] <=> z = a+ib [b a] you then use matrix algebra to show Z <=> z in closure of addition and multiplication
anonymous
  • anonymous
oh and scalar
anonymous
  • anonymous
thank you

Looking for something else?

Not the answer you are looking for? Search for more explanations.