Theorem:
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 \( f(\alpha)=v \).
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 that
\( \alpha=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n \)
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.
\( \therefore \alpha \) 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)
\( \implies a_1=b_1, a_2=b_2, … , a_n=b_n \)
\( \therefore f(\alpha)=f(a_1\alpha_1+ … +a_n\alpha_n) \)
\( =f(b_1\alpha_1+ … +b_n\alpha_n) \)
\( =f(\beta) \)
\( \therefore \) 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 \( a_i\in \mathbb{R} \)
Let \( u=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n\in U \)
\( \therefore f(u)=f(a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n) \)
\( =a_1\beta_1+a_2\beta_2+ … +a_n\beta_n \)
\( =v \)
\( \therefore \) f is onto.
Now, to prove that, f is linear transformation.
Let x, y be any arbitrary elements of U.
To prove that, f(x+y)=f(x)+f(y)
As \( \{\alpha_1, \alpha_2, … , \alpha_n\} \) is a basis of U,
\( x=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n \) 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: First isomorphism theorem for vector space, homomorphism, kernel and isomorphism of vector space
