what is the condition to show a Euclidian space

**yht0251** · January 2nd 2010, 02:37 PM

see the attachment, it is easy to show that is a inner product, but what condition needed to fulfill to show that forms a a Euclidian space?

**HallsofIvy** · January 3rd 2010, 02:07 AM

Originally Posted by yht0251

see the attachment, it is easy to show that is a inner product, but what condition needed to fulfill to show that forms a a Euclidian space?

What is the definition of "Euclidean space" given in your text book? Those are the conditions you need to prove. I think that your problem should be to show this space is isomorphic to a Euclidean space rather than "is" one- unless your text book has a very general definition of Euclidean space.

**yht0251** · January 3rd 2010, 02:44 AM

Originally Posted by HallsofIvy

What is the definition of "Euclidean space" given in your text book? Those are the conditions you need to prove. I think that your problem should be to show this space is isomorphic to a Euclidean space rather than "is" one- unless your text book has a very general definition of Euclidean space.

well, there is no definition of 'euclidean space' in our handouts(we don't have a text book this course). this question comes from my coursework which usually ask questions outside the handouts by assuming people know that.as you mentioned this question may need to show this space is isomorphic to euclidean space, but how? thanks for replying the post

**HallsofIvy** · January 3rd 2010, 04:17 AM

The standard definition of "Euclidean space" is just $R^m$ for some n- that is, the set of ordered m-tuples of real numbers with both addition and scalar multiplication defined "coordinate-wise"- that is $(x_1, x_2, \cdot\cdot\cdot, x_m)+ (y_1, y_2, \cdot\cdot\cdot, y_m)= (x_1+y_1, x_2+y_2, \cdot\cdot\cdot, x_m+y_m)$ and $a(x_1,x_2, \cdot\cdot\cdot, x_m)= (ax_1, ax_2, \cdot\cdot\cdot, ax_m)$ .

There is a "standard basis" for $R^m$ . It consists of $\{(1, 0, \cot\cdot\cdot, 0), (0, 1, \cdot\cdot\cdot, 0),\cdot\cdot\cdot, (0, 0, \cdot\cdot\cdot, 1)\}$ - that is, each basis vector has "1" in one place and "0" everywhere else.

There is also a "standard basis" for Pn, the space of polynomials of degree at most n: $\{1, x, x^2, \cdot\cdot\cdot, x^n\}$ . The standard way to show that one vector space is isomorphic to another is to show that you can map one basis into the other. Be a little be careful here: while $R^m$ has dimension m, Pn has dimension n+1. For example, a basis for P2 is $\{1, x, x^2\}$ which contains three vectors and so P2 has dimension 3. To show that $R^3$ is isomorphic to $P2$ , map 1 to (1, 0, 0), x to (0, 1, 0), and $x^2$ to (0, 0, 1). That would then map any $a+ bx+ cx^2$ to (a, b, c).

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