Diemsion of Hom

**Showcase_22** · February 7th 2009, 03:17 AM

I started by making $e_1, e_2,........,e_n$ a basis of U and $f_1,f_2,.....,f_m$ a basis of V.

Transforming the basis of U by T gives $T(e_1),T(e_2),......,T(e_n)$ . These vectors may not be linearly independent so sifting them gives:

$T(e_1),T(e_2),......,T(e_r)$ where $r \leq n$ . Let $T(e_i)=f_i$ which gives me the basis of V.

Is this the correct way to start? I really can't see another way of doing this question.

EDIT: Sorry for mistyping "Dimension" as the title.

**Isomorphism** · February 7th 2009, 07:59 AM

Originally Posted by Showcase_22

I started by making $e_1, e_2,........,e_n$ a basis of U and $f_1,f_2,.....,f_m$ a basis of V.

Prove that the lin. transfs $T_{ij}$ such that $T_{ij} (e_i) = f_j$ (actually they are just maps according to definition, but you can extend it linearly), form a basis of $\text{Hom (U,V)}$

**ThePerfectHacker** · February 7th 2009, 09:43 AM

Let U and V be vector spaces of dimensions n and m over K, and let $Hom_K(U,V)$ be the vector space over K of all linear maps from U to V.

Find the dimension and describe a basis of $Hom_K(U,V)$ .

Let $T \in \hom_K (U,V)$ and $(u_1,...,u_n)$ be a basis for $U$ and $(v_1,...,v_m)$ be a basis for $V$ . This means that $T(u_j) = \Sigma_i a_{ij}v_i$ . Let $M(T)$ be the matrix $(a_{ij})$ i.e. $M(T)$ is the matrix associated with $T$ . We see that $M(T_1+T_2) = M(T_1)+M(T_2)$ and $M(kT) = kM(T)$ . Therefore, $F: \hom_K(U,V)\to M_{m\times n}(K)$ defined by $F(T) = M(T)$ is a linear transformation between the vector space $\hom_K(U,V)$ and $M_{m\times n}(K)$ - the $m\times n$ matrices over $K$ . This linear transformation is one-to-one because if $F(T) = 0$ it means $M(T) = 0$ and so $T(u_j) = 0$ for each $1\leq j \leq n$ . All elements in $U$ are linear combinations of $u_j$ and so if $T(u_j)=0$ then it means $T$ is the zero linear transformation. Thus, $\ker (T) = \{ \bold{0}\}$ and that implies $T$ is one-to-one. It remains to show that $F$ is onto. Thus, $F$ provides a vector space isomorphism between $\hom_K(U,V)$ and $M_{m\times n}(K)$ . Now, $M_{m\times n}(K)$ has dimension $mn$ with the basis $\{ \bold{e}_{ij}\}$ where $\bold{e}_{ij}$ is a matrix having $1$ in $ij$ location and $0$ 's everywhere else. Thus, $\text{dim}( \hom_k(U,V) ) = mn$ .

**Showcase_22** · February 8th 2009, 03:10 AM

Is it true that bijectivity implies that the dimensions are the same?

What's the best way to prove that it's surjective? We're not really sure how to start.

Do both $Hom_K(U,V)$ and $M_{m \times n}$ have the same basis?

**ThePerfectHacker** · February 8th 2009, 11:19 AM

Originally Posted by Showcase_22

Is it true that bijectivity implies that the dimensions are the same?

If $U$ and $V$ are isomorphic (bijective linear transformation) then they have the same dimension.

**Showcase_22** · February 8th 2009, 12:21 PM

Can you argue that F is surjective because every linear map can be represented as a matrix, since $Hom_K(U,V)$ is the set of all these linear maps then a row in the matrix $M _{m \times n}(K)$ can be mapped to the linear map it represents.

Hence it must be surjective.

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