matrices,real numbers ...

hi,everyone.I have a doubt.It regards the relations among real numbers ,vectors of R^n( n natural >1), matrices with real or complex inputs.i would like to know (even without proof) if :

a) taken a in R^n(n natual >0) , b in R^m(m natual >0) ,with n different from m,does it results a different from b ?

b) taken a in C^n(n natual >0) , b in R^(2*n) ,does it results a different from b ?

c)taken a in MR^(m,n)(m,n natua numbersl >0) , b in R^k(k natual >0) ,does it results a different from b ?

Comments

b)i know the proposition is true for n=1 because it results C=R^2 .For n >1 i only know there exists an oblivious bijection from C^n to R^(2*n),i.e. C^n and R^(2*n) have the same cardinality.

c) i paid attenction particulary on the case

c')m=n=1 , k=1

c'') n=1, k=m

c''')m=1, k=n

Thanks to everyone could help me.Bye !

Re: matrices,real numbers ...

I don't understand exactly what you mean by 'different' by each of these cases based on your comments, but I'll try to guess that you mean something like 'comparability'.

a) It can be shown that there is a natural bijection between the real numbers R, and the cross product R x R. It follows that there is a bijection between R^n and R^m by induction since R^n for any n has the same cardinality.

b) I would say that a bijection that is as natural as the one in your comment works. Given a set of real values $\displaystyle a_p$, $\displaystyle 0 \leq p \leq 2n$ not necessarily distinct, let A be an arbitrary element of C^n, for n fixed

If $\displaystyle a_1 +a_2 i$ is the first entry, $\displaystyle a_3+ a_4i$ is your second entry, and so on, let B be a bijection which sends A to B with B a vector of elements, $\displaystyle a_1,a_2, ... a_{2n}$ (you're just splitting the cells of A into 2 new cells of B)

An inductive argument can work to generalize.

c) I'll have to think more on c, because I'm not sure if it's the case that there exists a bijection between matrices and vectors.

Re: matrices,real numbers ...

there is a natural bijection (it's actually a linear isomorphism) between Mat_{F}(m,n) and F^{mn}, for any field F.

Re: matrices,real numbers ...

Quote:

Originally Posted by

**TheDifferentialProability** hi,everyone.I have a doubt.It regards the relations among real numbers ,vectors of R^n( n natural >1), matrices with real or complex inputs.i would like to know (even without proof) if :

a) taken a in R^n(n natual >0) , b in R^m(m natual >0) ,with n different from m,does it results a different from b ?

Since a and b are members of sets of different **kinds** of things, yes they are different. Of course, if n< m, we can "identify" a vector $\displaystyle <a_1, a_2, ..., a_n>$ in $\displaystyle R^n$ with the vector $\displaystyle <a_1, a_2, ..., a_n, 0, 0,..., 0>$ in $\displaystyle R^m$. But that involves a choice of how to make the identification. We could, as easily, identify it with $\displaystyle <0, 0, ..., 0, a_1, a_2, ..., a_n>$. We would say that one is "isomorphic" to a subset of the other.

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b) taken a in C^n(n natual >0) , b in R^(2*n) ,does it results a different from b ?

Again, we could **choose** to identity $\displaystyle <a_1+ b_1i, a_2+ b_2i, ..., a_n+ b_ni>$ with $\displaystyle <a_1, b_1, a_2, b_2, ..., a_n, b_n>$ or with $\displaystyle <a_1, a_2, ..., a_n, b_1, b_2, ..., b_n>$. The two spaces are isomorphic with many different possible isomophisms.

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c)taken a in MR^(m,n)(m,n natua numbersl >0) , b in R^k(k natual >0) ,does it results a different fro0m b ?

Yes. Although, again, one can be thought of as "isomorphic" to a subspace of the other.

Quote:

Comments

b)i know the proposition is true for n=1 because it results C=R^2 .For n >1 i only know there exists an oblivious bijection from C^n to R^(2*n),i.e. C^n and R^(2*n) have the same cardinality.

c) i paid attenction particulary on the case

c')m=n=1 , k=1

c'') n=1, k=m

c''')m=1, k=n

Thanks to everyone could help me.Bye !

You need to understand the difference between "are the same" and "are isomorphic".

Re: matrices,real numbers ...

Thanks to everyone.Hallsofivy , given two sets A,B it results one and only one between "A= B" and " A=B is false" .The most common method to see if A=B or "A=B is false " is the logic equivalence between the proposition 1) A=B and the proposition 2)A is a subset of B and B is a subset of A. Particulary ,being matrices ,elements of R^n ,C^n sets( the last two sets are built by the Kuratowski definition) , taken two of those sets ,saying A and B, it results " A=B" or this is false.

Said this , i know ,given two sets A and B , when we say" A and B have the same cardinality or ,given two vectorial spaces (A,field K,+(1),*(1)) and (B,field K,+(2),*(2)) when we say " (A,field K,+(1),*(1)) and (B,field K,+(2),*(2)) are isomorphic" but those are another problems respect to the original problem .

Hallosofivy ,in your post you say " Since a and b are members of sets of different **kinds** of things, yes they are different " . I don't know what you intend with the expression "kinds of things" :i suppose you intend " different sets"(i.e. not equal) .In this case the propostion you have written is false :an example is given by the the line of R^2 having "direction vector" (1,2) containg (0,0) and by the line of R^2 having "direction vector" (1,3) containing (0,0): the two sets are not equal but the two lines admit the (0,0) element.