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# Two finite dimensional vector spaces over the same field are isomorphic iff they are of the same dimensions

## 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. â€‹$$\exists$$â€‹ a function â€‹$$f:U\to V$$â€‹Â which is one-one, onto and linear transformation.
Claim : â€‹$$\dim U=\dim V$$â€‹
Consider, â€‹$$\dim U=n$$â€‹
Let â€‹$$S=\{\alpha_1, \alpha_2, … , \alpha_n\}$$â€‹Â be a basis of U.
We prove that â€‹$$S’=\{f(\alpha_1), f(\alpha_2), … , f(\alpha_n)\}$$â€‹Â is a basis of V.
For this first we prove that, S’ is linearly independent.
Consider,
â€‹$$a_1f(\alpha_1)+a_2f(\alpha_2)+ … +a_nf(\alpha_n)=0$$â€‹, where â€‹$$a_i\in \mathbb{R}$$â€‹Â â€‹

$\implies f(a_1\alpha_1)+f(a_2\alpha_2)+ … +f(a_n\alpha_n)=0,$

â€‹(â€‹$$\because$$â€‹Â f is linear)

â€‹$$\implies f(a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n)=0$$â€‹, (â€‹$$\because$$â€‹Â f is linear)
â€‹$$\implies a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n=0$$â€‹, (â€‹$$\because$$â€‹Â f is linear and one-one)
Since, â€‹$$S=\{\alpha_1, \alpha_2, … , \alpha_n\}$$â€‹Â is a basis of U,
â€‹$$\implies a_1=a_2= … =a_n=0$$â€‹
â€‹$$\therefore S’=\{f(\alpha_1), f(\alpha_2), … , f(\alpha_n)\}$$â€‹Â is linearly independent. ……. (1)
Now, to prove that, â€‹$$S’=\{f(\alpha_1), f(\alpha_2), … , f(\alpha_n)\}$$â€‹Â spans V.
i.e. to prove that, every vector in V can be expressed as a linear combination of â€‹$$\{f(\alpha_1), f(\alpha_2), … , f(\alpha_n)\}$$â€‹.
Let v be any arbitrary element of V.
As â€‹$$f:U\to V$$â€‹ is onto, for â€‹$$v\in V, \exists \ \alpha\in U$$â€‹ such that â€‹$$f(\alpha)=v$$â€‹.Â
As â€‹$$\alpha\in U$$â€‹ and â€‹$$S=\{\alpha_1, \alpha_2, … , \alpha_n\}$$â€‹ is a basis of U, â€‹$$\therefore \exists \ a_1, a_2, , … , a_n\in \mathbb{R}$$â€‹ such thatÂ
â€‹$$\alpha=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n$$â€‹Â
Now, â€‹$$v=f(\alpha)$$â€‹
â€‹$$=f(a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n)$$â€‹
â€‹$$=a_1f(\alpha_1)+a_2f(\alpha_2)+ … +a_nf(\alpha_n)=0$$â€‹, (â€‹$$\because$$â€‹ f is linear)
â€‹$$\therefore$$â€‹Â every vector in V can be expressed as a linear combination of elements in S’.
â€‹$$\therefore$$â€‹Â L(S’)=V …….. (2)
From (1) and (2),
The set S’ forms basis of V.
â€‹$$\therefore \dim V=n$$â€‹
Hence, â€‹$$\dim U=n=\dim V$$â€‹
Conversely,
Suppose that, â€‹$$\dim U=n=\dim V$$â€‹
Let â€‹$$B=\{\alpha_1, \alpha_2, … , \alpha_n\}$$â€‹ be a basis of U and â€‹$$B’=\{\beta_1, \beta_2, … , \beta_n\}$$â€‹ be a basis of V.
Claim : â€‹$$U\cong V$$â€‹
As â€‹$$\alpha \in U$$â€‹ and â€‹$$\{\alpha_1, \alpha_2, … , \alpha_n\}$$â€‹ is a basis of U,
â€‹$$\alpha=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n , \forall \ a_i\in \mathbb{R}$$â€‹
For this â€‹$$\alpha$$â€‹, define â€‹$$f:U\to V$$â€‹ as â€‹$$f(\alpha)=a_1\beta_1+a_2\beta_2+ … +a_n\beta_n, \forall \ \alpha\in U$$â€‹
â€‹$$f(\alpha)=f(a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n)$$â€‹
Â  Â  Â  Â  Â  Â  â€‹$$=a_1\beta_1+a_2\beta_2+ … +a_n\beta_n$$â€‹
Let â€‹$$\alpha, \beta$$â€‹ be any two vectors in U and â€‹$$\{\alpha_1, \alpha_2, … , \alpha_n\}$$â€‹ is a basis of U.Â
â€‹$$\therefore \alpha$$â€‹ and â€‹$$\beta$$â€‹ can be expressed as linear combination of â€‹$$\{\alpha_1, \alpha_2, … , \alpha_n\}$$â€‹.
Let â€‹$$\alpha=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n$$â€‹ and â€‹$$\beta=b_1\alpha_1+b_2\alpha_2+ … +b_n\alpha_n$$â€‹, where â€‹$$a_i, b_i \in \mathbb{R}$$â€‹
To prove that, f is well defined.
Let â€‹$$\alpha=\beta$$â€‹â€‹

$\implies a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n=b_1\alpha_1+b_2\alpha_2+ … +b_n\alpha_n$

$\implies (a_1-b_1)\alpha_1+(a_2-b_2)\alpha_2+ … +(a_n-b_n)\alpha_n=0 \}$

$\implies a_1-b_1=0, a_2-b_2=0, … , a_n-b_n=0,$

â€‹Â (â€‹$$\because \{\alpha_1, \alpha_2, … , \alpha_n\}$$â€‹ is linearly independent)Â Â
â€‹$$\implies a_1=b_1, a_2=b_2, … , a_n=b_n$$â€‹Â
â€‹$$\therefore f(\alpha)=f(a_1\alpha_1+ … +a_n\alpha_n)$$â€‹Â
Â  Â  Â  Â  Â  Â  Â  Â  â€‹$$=f(b_1\alpha_1+ … +b_n\alpha_n)$$â€‹
Â  Â  Â  Â  Â  Â  Â  Â  â€‹$$=f(\beta)$$â€‹
$$\therefore$$Â f is well defined.
Now, to prove that f is one-one.
Consider, â€‹$$f(\alpha)=f(\beta)$$â€‹â€‹

$\implies f(a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n)=f(b_1\alpha_1+b_2\alpha_2+ … b_n\alpha_n)$

$\implies a_1\beta_1+a_2\beta_2+ … +a_n\beta_n=b_1\beta_1+b_2\beta_2+ … +b_n\beta_n$

$\implies (a_1-b_1)\beta_1+(a_2-b_2)\beta_2+ … +(a_n-b_n)\beta_n=0$

$\implies a_1-b_1=0, a_2-b_2=0, … , a_n-b_n=0$

â€‹â€‹$$\implies a_1=b_1, a_2=b_2, … , a_n=b_n$$â€‹

â€‹$$\implies \alpha=\beta$$â€‹
â€‹$$\therefore$$â€‹Â f is one-one.
Now, to prove that f is onto.
Consider, â€‹$$v\in V, \implies v=a_1\beta_1+a_2\beta_2+ … +a_n\beta_n$$â€‹, for some â€‹$$a_i\in \mathbb{R}$$â€‹Â
Let â€‹$$u=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n\in U$$â€‹Â
â€‹$$\therefore f(u)=f(a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n)$$â€‹
Â  Â  Â  Â  Â  Â  Â  Â  â€‹$$=a_1\beta_1+a_2\beta_2+ … +a_n\beta_n$$â€‹
Â  Â  Â  Â  Â  Â  Â  Â  â€‹$$=v$$â€‹
â€‹$$\therefore$$â€‹Â 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 â€‹$$\{\alpha_1, \alpha_2, … , \alpha_n\}$$â€‹ is a basis of U,Â
â€‹$$x=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n$$â€‹ and â€‹$$y=b_1\alpha_1+b_2\alpha_2+ … +b_n\alpha_n$$â€‹, where â€‹$$a_i, b_i \in \mathbb{R}$$â€‹â€‹

$\therefore x+y=(a_1+b_1)\alpha_1+(a_2+b_2)\alpha_2+ … +(a_n+b_n)\alpha_n$

â€‹â€‹$$\implies f(x+y)$$â€‹Â  Â  Â  Â  Â  Â Â  â€‹

$=f[(a_1+b_1)\alpha_1+(a_2+b_2)\alpha_2+ … +(a_n+b_n)\alpha_n]$

$=(a_1+b_1)\beta_1+(a_2+b_2)\beta_2+ … +(a_n+b_n)\beta_n$

$=a_1\beta_1+b_1\beta_1+a_2\beta_2+b_2\beta_2+ … +a_n\beta_n+b_n\beta_nâ€‹$

$=a_1\beta_1+a_2\beta_2+ … +a_n\beta_n+b_1\beta_1+b_2\beta_2+ … +b_n\beta_n$

â€‹Â Â  Â  Â  Â  Â  Â  â€‹$$=f(x)+f(y)$$â€‹
Now, to prove that, â€‹$$f(cx)=cf(x)$$â€‹
Consider, â€‹$$c\in \mathbb{R}$$â€‹ and â€‹$$x=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n\in U$$â€‹
â€‹$$\therefore f(cx)=f[c(a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n)]$$â€‹
â€‹$$=f(ca_1\alpha_1+ca_2\alpha_2+ … +ca_n\alpha_n)$$â€‹
$$=ca_1\beta_1+ca_2\beta_2+ … +ca_n\beta_n$$â€‹
â€‹$$=c(a_1\beta_1+a_2\beta_2+ … +a_n\beta_n)$$â€‹
$$=cf(a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n)$$â€‹
â€‹$$=cf(x)$$â€‹
Hence, â€‹$$U\cong V$$â€‹

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