# Diemsion of Hom

• Feb 7th 2009, 03:17 AM
Showcase_22
Diemsion of Hom
I started by making $\displaystyle e_1, e_2,........,e_n$ a basis of U and $\displaystyle f_1,f_2,.....,f_m$ a basis of V.

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

$\displaystyle T(e_1),T(e_2),......,T(e_r)$ where $\displaystyle r \leq n$. Let $\displaystyle T(e_i)=f_i$ 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.
• Feb 7th 2009, 07:59 AM
Isomorphism
Quote:

Originally Posted by Showcase_22
I started by making $\displaystyle e_1, e_2,........,e_n$ a basis of U and $\displaystyle f_1,f_2,.....,f_m$ a basis of V.

Prove that the lin. transfs $\displaystyle T_{ij}$ such that $\displaystyle T_{ij} (e_i) = f_j$(actually they are just maps according to definition, but you can extend it linearly), form a basis of $\displaystyle \text{Hom (U,V)}$
• Feb 7th 2009, 09:43 AM
ThePerfectHacker
Quote:

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

Find the dimension and describe a basis of $\displaystyle Hom_K(U,V)$.
Let $\displaystyle T \in \hom_K (U,V)$ and $\displaystyle (u_1,...,u_n)$ be a basis for $\displaystyle U$ and $\displaystyle (v_1,...,v_m)$ be a basis for $\displaystyle V$. This means that $\displaystyle T(u_j) = \Sigma_i a_{ij}v_i$. Let $\displaystyle M(T)$ be the matrix $\displaystyle (a_{ij})$ i.e. $\displaystyle M(T)$ is the matrix associated with $\displaystyle T$. We see that $\displaystyle M(T_1+T_2) = M(T_1)+M(T_2)$ and $\displaystyle M(kT) = kM(T)$. Therefore, $\displaystyle F: \hom_K(U,V)\to M_{m\times n}(K)$ defined by $\displaystyle F(T) = M(T)$ is a linear transformation between the vector space $\displaystyle \hom_K(U,V)$ and $\displaystyle M_{m\times n}(K)$ - the $\displaystyle m\times n$ matrices over $\displaystyle K$. This linear transformation is one-to-one because if $\displaystyle F(T) = 0$ it means $\displaystyle M(T) = 0$ and so $\displaystyle T(u_j) = 0$ for each $\displaystyle 1\leq j \leq n$. All elements in $\displaystyle U$ are linear combinations of $\displaystyle u_j$ and so if $\displaystyle T(u_j)=0$ then it means $\displaystyle T$ is the zero linear transformation. Thus, $\displaystyle \ker (T) = \{ \bold{0}\}$ and that implies $\displaystyle T$ is one-to-one. It remains to show that $\displaystyle F$ is onto. Thus, $\displaystyle F$ provides a vector space isomorphism between $\displaystyle \hom_K(U,V)$ and $\displaystyle M_{m\times n}(K)$. Now, $\displaystyle M_{m\times n}(K)$ has dimension $\displaystyle mn$ with the basis $\displaystyle \{ \bold{e}_{ij}\}$ where $\displaystyle \bold{e}_{ij}$ is a matrix having $\displaystyle 1$ in $\displaystyle ij$ location and $\displaystyle 0$'s everywhere else. Thus, $\displaystyle \text{dim}( \hom_k(U,V) ) = mn$.
• Feb 8th 2009, 03:10 AM
Showcase_22
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 $\displaystyle Hom_K(U,V)$ and $\displaystyle M_{m \times n}$ have the same basis?
• Feb 8th 2009, 11:19 AM
ThePerfectHacker
Quote:

Originally Posted by Showcase_22
Is it true that bijectivity implies that the dimensions are the same?

If $\displaystyle U$ and $\displaystyle V$ are isomorphic (bijective linear transformation) then they have the same dimension.
• Feb 8th 2009, 12:21 PM
Showcase_22
Can you argue that F is surjective because every linear map can be represented as a matrix, since $\displaystyle Hom_K(U,V)$ is the set of all these linear maps then a row in the matrix $\displaystyle M _{m \times n}(K)$ can be mapped to the linear map it represents.

Hence it must be surjective.