Consider Example 12 in section 6.2 (page 311) Study each space closely and attem

Question

Consider Example 12 in section 6.2 (page 311) Study each space closely and attempt to understand how each is "the same" as the other: Isomorphic Vector Spaces The following vector spaces are isomorphic to each other. R4 = 4-space M4, 1 = space of all 4 Times 1 matrices M2,2 = space of all 2 Times 2 matrices P3 = space of all polynomials of degree 3 or less V = {(x1, x2, x3, x4, 0): xi is a real number} (subspace of R5) Example 12 tells you that the elements in these spaces behave in the same way as vectors. What mapping can you define to go from one to another? Is this mapping unique? Things that may seem more obvious in one space would have an analog in its isomorphic spaces, yet these concepts may not seem so obvious. Try to make the complete mental translation from one space to another. Can you now explain with great detail why the real line R1 and the Cartesian plane R2 are "not the same"? How would you explain this to a mathematician?

Explanation / Answer

each of the given spaces will have 4 basis define a function which is one to one and onto from basis of one to another for eg in case of a) and d basis of a) is (0,0,0,1),(0,0,1,0),(0,1,0,0),(1,0,0,0) basis of d) 1,x,x^2,x^3 now define a function f(0,0,0,1)=1 f(0,0,1,0)=x f(0,1,0,0)=x^2 f(1,0,0,0)=x^3 let f(x+y)=f(x)+f(y) f(xy)=f(x)f(y) then now any element in R4 can be written as a(0,0,0,1)+b(0,0,1,0)+c(0,1,0,0)+d(1,0,0,0) then f(a,b,c,d)=a+bx+cx^2+dx^3 now clearly this function is one one and onto not only for basis but also for the whole space to prove it let f(a,b,c,d)=f(a1,b1,c1,d1) then a+bx+cx^2+dx^3=a1+b1x+c1x^2+d1x^3 which means a=a1 b=b1 c=c1 d=d1 and its onto because for any a+bx+cx^2+dx^3 there exists (a,b,c,d) in R4 b) no this mapping is not unique we can have many mappings in the example of and b we can define any one one onto mapping now no of such mappings are equal to no of one one onto mappings possible c) R1 is not equal to R2 as we have only one basis of R1 i.e 1 and of R2 we have 2 basis (1,0)(0,1) so we cant have one one onto function from basis of R1 to R2 hence they are not isomorphic and hence not same

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