# Linear Algebra: Isomorphism Proof

Here are the initials:

$\nu$ is a vector space

$\beta=\begin{Bmatrix}b_{1},\cdots,b_{n}\end{Bmatrix}$ is a basis for $\nu$

Argument:

coordinate mapping $\Phi_{\beta }:\nu \rightarrow \mathbb{R}^{n}$ defined by sending a vector $v$ $\epsilon$ $\nu$ to its coordinate vector $\begin{bmatrix}v\end{bmatrix}_{\beta}\epsilon\mathbb{R}^{n}$ is an isomorphism between $\nu$ and $\mathbb{R}^{n}$.

Questions:

1) Explain why each vector $v$ $\epsilon$ $\nu$ gets sent to exactly one coordinate vector $\begin{bmatrix}v\end{bmatrix}_{\beta}\epsilon\mathbb{R}^{n}$

2) Show how $\Phi_{\beta }:\nu \rightarrow \mathbb{R}^{n}$ respects vector addition and scalar multiplication.

3) Explain why the linear transformation $\Phi_{\beta }:\nu \rightarrow \mathbb{R}^{n}$ is one-to-one.

4) Explain why the linear transformation $\Phi_{\beta }:\nu \rightarrow \mathbb{R}^{n}$ is onto.

I'm not to sure how to prove all this. I can show it if I had numbers but with arbitrary terms I'm not sure how to do it. Thanks.

1: If $[v]_\beta = x$ and $[v]_\beta = y$, show that $[0]_\beta = x-y$. By the definition of a basis, show that this means that $x - y = 0$, so that $x = y$. Thus, $v$ can only get sent to one coordinate vector.

2: Prove that the mapping $\Phi_\beta$ is linear as you would normally prove that a map is linear.

3: Similar to 1. If $v_1$ and $v_2$ get sent to the same coordinate vector, then $v_1 - v_2$ gets sent to the $0$-vector. Use the definition of a basis to show that $v_1 - v_2$ has to be zero, which in turn implies that $v_1 = v_2$.

4: Starting with $x \in \Bbb R^n$, find a $v \in \nu$ such that $\Phi_\beta(v) = x$.

• So for part 1 would I show that since the only solution to 0 is the trivial solution? Each scalar has to be 0? – KFC Apr 1 '15 at 0:19
• That's right ${}$ – Omnomnomnom Apr 1 '15 at 0:20
• But how does that explain that it can only be sent to one coordinate vector? – KFC Apr 1 '15 at 0:21
• Ah, I made a mistake. I'll have it fixed in 2 minutes – Omnomnomnom Apr 1 '15 at 0:23
• Fixed. It's a similar argument, you see... – Omnomnomnom Apr 1 '15 at 0:25