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Theorem:Â
Two finite dimensional vector spaces over the same field are isomorphic iff they are of the same dimensions.
Proof:
Let U and V be two finite dimensional real vector spaces which are isomorphic.
i.e. ​\( \exists \)​ a function ​\( f:U\to V \)​ which is one-one, onto and linear transformation.
Claim : ​\( \dim U=\dim V \)​
Consider, ​\( \dim U=n \)​
Let ​\( S=\{\alpha_1, \alpha_2, … , \alpha_n\} \)​ be a basis of U.
We prove that ​\( S’=\{f(\alpha_1), f(\alpha_2), … , f(\alpha_n)\} \)​ is a basis of V.
For this first we prove that, S’ is linearly independent.
Consider,
​\( a_1f(\alpha_1)+a_2f(\alpha_2)+ … +a_nf(\alpha_n)=0 \)​, where ​\( a_i\in \mathbb{R} \)​ ​
\[ \implies f(a_1\alpha_1)+f(a_2\alpha_2)+ … +f(a_n\alpha_n)=0, \]
​(​\( \because \)​ f is linear)
​\( \implies f(a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n)=0 \)​, (​\( \because \)​ f is linear)
​\( \implies a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n=0 \)​, (​\( \because \)​ f is linear and one-one)
Since, ​\( S=\{\alpha_1, \alpha_2, … , \alpha_n\} \)​ is a basis of U,
​\( \implies a_1=a_2= … =a_n=0 \)​
​\( \therefore S’=\{f(\alpha_1), f(\alpha_2), … , f(\alpha_n)\} \)​ is linearly independent. ……. (1)
Now, to prove that, ​\( S’=\{f(\alpha_1), f(\alpha_2), … , f(\alpha_n)\} \)​ spans V.
i.e. to prove that, every vector in V can be expressed as a linear combination of ​\( \{f(\alpha_1), f(\alpha_2), … , f(\alpha_n)\} \)​.
Let v be any arbitrary element of V.
As ​\( f:U\to V \)​ is onto, for ​\( v\in V, \exists \ \alpha\in U \)​ such that ​\( f(\alpha)=v \)​.Â
As ​\( \alpha\in U \)​ and ​\( S=\{\alpha_1, \alpha_2, … , \alpha_n\} \)​ is a basis of U, ​\( \therefore \exists \ a_1, a_2, , … , a_n\in \mathbb{R} \)​ such thatÂ
​\( \alpha=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n \)​Â
Now, ​\( v=f(\alpha) \)​
​\( =f(a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n) \)​
​\( =a_1f(\alpha_1)+a_2f(\alpha_2)+ … +a_nf(\alpha_n)=0 \)​, (​\( \because \)​ f is linear)
​\( \therefore \)​ every vector in V can be expressed as a linear combination of elements in S’.
​\( \therefore \)​ L(S’)=V …….. (2)
From (1) and (2),
The set S’ forms basis of V.
​\( \therefore \dim V=n \)​
Hence, ​\( \dim U=n=\dim V \)​
Conversely,
Suppose that, ​\( \dim U=n=\dim V \)​
Let ​\( B=\{\alpha_1, \alpha_2, … , \alpha_n\} \)​ be a basis of U and ​\( B’=\{\beta_1, \beta_2, … , \beta_n\} \)​ be a basis of V.
Claim : ​\( U\cong V \)​
As ​\( \alpha \in U \)​ and ​\( \{\alpha_1, \alpha_2, … , \alpha_n\} \)​ is a basis of U,
​\( \alpha=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n , \forall \ a_i\in \mathbb{R} \)​
For this ​\( \alpha \)​, define ​\( f:U\to V \)​ as ​\( f(\alpha)=a_1\beta_1+a_2\beta_2+ … +a_n\beta_n, \forall \ \alpha\in U \)​
​\( f(\alpha)=f(a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n) \)​
      ​\( =a_1\beta_1+a_2\beta_2+ … +a_n\beta_n \)​
Let ​\( \alpha, \beta \)​ be any two vectors in U and ​\( \{\alpha_1, \alpha_2, … , \alpha_n\} \)​ is a basis of U.Â
​\( \therefore \alpha \)​ and ​\( \beta \)​ can be expressed as linear combination of ​\( \{\alpha_1, \alpha_2, … , \alpha_n\} \)​.
Let ​\( \alpha=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n \)​ and ​\( \beta=b_1\alpha_1+b_2\alpha_2+ … +b_n\alpha_n \)​, where ​\( a_i, b_i \in \mathbb{R} \)​
To prove that, f is well defined.
Let ​\( \alpha=\beta \)​​
\[ \implies a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n=b_1\alpha_1+b_2\alpha_2+ … +b_n\alpha_n \]
\[ \implies (a_1-b_1)\alpha_1+(a_2-b_2)\alpha_2+ … +(a_n-b_n)\alpha_n=0 \} \]
\[ \implies a_1-b_1=0, a_2-b_2=0, … , a_n-b_n=0, \]
​ (​\( \because \{\alpha_1, \alpha_2, … , \alpha_n\} \)​ is linearly independent) Â
​\( \implies a_1=b_1, a_2=b_2, … , a_n=b_n \)​Â
​\( \therefore f(\alpha)=f(a_1\alpha_1+ … +a_n\alpha_n) \)​Â
        ​\( =f(b_1\alpha_1+ … +b_n\alpha_n) \)​
        ​\( =f(\beta) \)​
\( \therefore \)Â f is well defined.
Now, to prove that f is one-one.
Consider, ​\( f(\alpha)=f(\beta) \)​​
\[ \implies f(a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n)=f(b_1\alpha_1+b_2\alpha_2+ … b_n\alpha_n) \]
\[ \implies a_1\beta_1+a_2\beta_2+ … +a_n\beta_n=b_1\beta_1+b_2\beta_2+ … +b_n\beta_n \]
\[ \implies (a_1-b_1)\beta_1+(a_2-b_2)\beta_2+ … +(a_n-b_n)\beta_n=0 \]
\[ \implies a_1-b_1=0, a_2-b_2=0, … , a_n-b_n=0 \]
​​\( \implies a_1=b_1, a_2=b_2, … , a_n=b_n \)​
​\( \implies \alpha=\beta \)​
​\( \therefore \)​ f is one-one.
Now, to prove that f is onto.
Consider, ​\( v\in V, \implies v=a_1\beta_1+a_2\beta_2+ … +a_n\beta_n \)​, for some ​\( a_i\in \mathbb{R} \)​Â
Let ​\( u=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n\in U \)​Â
​\( \therefore f(u)=f(a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n) \)​
        ​\( =a_1\beta_1+a_2\beta_2+ … +a_n\beta_n \)​
        ​\( =v \)​
​\( \therefore \)​ f is onto.
Now, to prove that, f is linear transformation.
Let x, y be any arbitrary elements of U.Â
To prove that, f(x+y)=f(x)+f(y)Â
As ​\( \{\alpha_1, \alpha_2, … , \alpha_n\} \)​ is a basis of U,Â
​\( x=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n \)​ and ​\( y=b_1\alpha_1+b_2\alpha_2+ … +b_n\alpha_n \)​, where ​\( a_i, b_i \in \mathbb{R} \)​​
\[ \therefore x+y=(a_1+b_1)\alpha_1+(a_2+b_2)\alpha_2+ … +(a_n+b_n)\alpha_n \]
​​\( \implies f(x+y) \)​       ​
\[ =f[(a_1+b_1)\alpha_1+(a_2+b_2)\alpha_2+ … +(a_n+b_n)\alpha_n] \]
\[ =(a_1+b_1)\beta_1+(a_2+b_2)\beta_2+ … +(a_n+b_n)\beta_n \]
\[ =a_1\beta_1+b_1\beta_1+a_2\beta_2+b_2\beta_2+ … +a_n\beta_n+b_n\beta_n​ \]
\[ =a_1\beta_1+a_2\beta_2+ … +a_n\beta_n+b_1\beta_1+b_2\beta_2+ … +b_n\beta_n \]
​       ​\( =f(x)+f(y) \)​
Now, to prove that, ​\( f(cx)=cf(x) \)​
Consider, ​\( c\in \mathbb{R} \)​ and ​\( x=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n\in U \)​
​\( \therefore f(cx)=f[c(a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n)] \)​
​\(=f(ca_1\alpha_1+ca_2\alpha_2+ … +ca_n\alpha_n) \)​
\( =ca_1\beta_1+ca_2\beta_2+ … +ca_n\beta_n \)​
​\( =c(a_1\beta_1+a_2\beta_2+ … +a_n\beta_n) \)​
\( =cf(a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n) \)​
​\( =cf(x) \)​
Hence, ​\( U\cong V \)​
Also Read:Â First isomorphism theorem for vector space, homomorphism, kernel and isomorphism of vector space
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