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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. 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€‹

Also Read:Â First isomorphism theorem for vector space, homomorphism, kernel and isomorphism of vector space

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