# Linear Algebra help!

• Feb 29th 2008, 07:47 PM
x3non25
Linear Algebra help!
Hi I'm new here and I have a bit of trouble understanding a problem that was given to us:

Let V and W be finite-dimensional vector spaces and let T:V->W be an isomorphism. Let Vo be a subspace of V. Prove that 1) T(Vo) is a subspace of W, and 2) dim (Vo)=dim(T(Vo))

I would appreciate any help. thank you
• Mar 1st 2008, 04:14 AM
kalagota
Quote:

Originally Posted by x3non25
Hi I'm new here and I have a bit of trouble understanding a problem that was given to us:

Let V and W be finite-dimensional vector spaces and let T:V->W be an isomorphism. Let Vo be a subspace of V. Prove that 1) T(Vo) is a subspace of W, and 2) dim (Vo)=dim(T(Vo))

I would appreciate any help. thank you

the first one is easy..
i)
clearly, $\displaystyle T(V_0) \subseteq W$ by taking an arbitrary element of $\displaystyle V_0 \subseteq V$ and apply the mapping.

ii)
suppose $\displaystyle u,w \in T(V_0)$, then there exist $\displaystyle v_1, v_2 \in V_0$ such that $\displaystyle u=T(v_1)$ and $\displaystyle w=T(v_2)$.
now, $\displaystyle u+w = T(v_1)+T(v_2) = T(v_1 + v_2)$ since $\displaystyle T$ is an isomorphism. and since $\displaystyle v_1, v_2 \in V_0$, so does $\displaystyle v_1 + v_2$. thus $\displaystyle u+v = T(v_1 + v_2) \in T(V_0)$

iii)
Let $\displaystyle a \in F$ and $\displaystyle v \in T(V_0)$. Then $\displaystyle \exists v_0 \in V_0$ such that $\displaystyle v = T(v_0)$
now, $\displaystyle av = aT(v_0) = T(av_0)$ and just like the reasoning above, $\displaystyle av = T(av_0) \in T(V_0)$.

for the second one..
i)
let dim$\displaystyle (V_0) = n$. take $\displaystyle \left\{ v_1, v_2, ..., v_n \right\}$ be a basis for $\displaystyle V_0$. Let $\displaystyle w \in T(V_0)$. then, there exists a $\displaystyle v \in V_0$ such that $\displaystyle w = T(v)$, say $\displaystyle v = a_1v_1 + a_2v_2 + ... + a_nv_n$..

$\displaystyle w = T(v) = T(a_1v_1 + a_2v_2 + ... + a_nv_n) = T(a_1v_1) + T(a_2v_2) + ... T(a_nv_n)$
$\displaystyle =a_1T(v_1) + a_2T(v_2) + ... + a_nT(v_n)$

thus $\displaystyle \left\{ T(v_1), T(v_2), ..., T(v_n)\right\}$ is a spanning set for $\displaystyle T(V_0)$

ii)
suppose $\displaystyle a_1T(v_1) + a_2T(v_2) + ... + a_nT(v_n) = 0$ (the zero vector for $\displaystyle T(V_0)$).
then,
$\displaystyle a_1T(v_1) + a_2T(v_2) + ... + a_nT(v_n) = T(a_1v_1 + a_2v_2 + ... + a_nv_n) = 0$
$\displaystyle \Longleftrightarrow a_1v_1 + a_2v_2 + ... + a_nv_n = 0$ (the zero vector for $\displaystyle V_0$).. since $\displaystyle \left\{ v_1, v_2, ..., v_n \right\}$ be a basis for $\displaystyle V_0$,
$\displaystyle a_1v_1 + a_2v_2 + ... + a_nv_n = 0 \Longleftrightarrow a_i = 0 \, \, \forall \, i = 1,...,n$..

this implies that $\displaystyle \left\{ T(v_1), T(v_2), ..., T(v_n)\right\}$ is a basis for $\displaystyle T(V_0)$. Therefore dim$\displaystyle T(V_0) = n$.. QED