Change of base matrix and invert-ability

Question

Let's say you want to express the vectors $u_1, u_2, u_3$ in the basis $\mathbb{B_v}$

$u_1 = a_1v_1 + a_2v_2 + a_3v_3$

$u_2 = b_1v_1 + b_2v_2 + b_3v_3$

$u_3 = c_1v_1 + c_2v_2 + c_3v_3$

If $w_{\mathbb{B_u}} = (\alpha. \beta, \gamma)$ i.e $w_{\mathbb{B_u}} = \alpha u_1 + \beta u_2 + \gamma u_3$

we can write $w_{\mathbb{B_u}}$ as

$$\left( \begin{array}{3} \alpha a_1+ \beta b_1 + \gamma c_1 = 0 \\ \alpha a_2+ \beta b_2 + \gamma c_3 = 0 \\ \alpha a_3 + \beta b_3 + \gamma c_3 = 0 \end{array}\right ) = \left( \begin{array}{3} a_1, b_1, c_1 \\ a_2, b_2, c_3\\ a_3, b_3, c_3\end{array}\right ) \left( \begin{array}{3} \alpha \\ \beta \\ \gamma \end{array}\right )$$

This means that the matrix: $$ A =\left( \begin{array}{3} a_1, b_1, c_1 \\ a_2, b_2, c_3\\ a_3, b_3, c_3\end{array}\right )$$

This is the change of basis matrix from the base $\mathbb{B_u}$ to $\mathbb{B_v}$.

Now my textbook claims that this matrix is: "Obviously invertible since the columns are linearly independent, this is due to the fact that the coordinates come from coordinate vectors from linearly independent vectors.

Now this last part is what I have trouble understanding.

For example:

That the coordinates in the columns in themselves are linearly independent is not hard to understand since they were expressed in a base $\mathbb{B_u}$, like: $u_1 = a_1v_1 + a_2v_2 + a_3v_3$.

But that the columns in $A$ are linearly independent from each other must be harder to assert?

As far as I understand the coordinates for $\left( a_1, a_2, a_3 \right )$ could be the exact same as (or a multiple of) $\left( b_1, b_2, b_3 \right )$ and hence the vectors $u_1$ and $u_2$ would be linearly dependent?

Wouldn't this make the base change matrix $A$ not invertible?

Obviously there is something here that I have missed...

Answer 1

Generally, if $u_1, u_2, \ldots, u_n$ are linearly independent, then their coordinate vectors with respect to a different basis $\mathcal{B}$ are still linearly independent. I'll denote these coordinate vectors as $[u_i]_{\mathcal{B}}$.

Proof: Recall that $[au + bv]_{\mathcal{B}} = a[u]_{\mathcal{B}} + b[v]_{\mathcal{B}}$ for $a, b \in \mathbb{R}$ and vectors $u, v$. Thus, if $c_1 [u_1]_{\mathcal{B}} + c_2 [u_2]_{\mathcal{B}} + \ldots + c_n [u_n]_{\mathcal{B}} = 0$, then we have $[c_1u_1 + c_2u_2 + \ldots + c_nu_n]_{\mathcal{B}} = 0$. Generally, if $[u]_{\mathcal{B}} = 0$, this means that $u = 0 v_1 + 0 v_2 + \ldots 0 v_n = 0$. Thus, we have shown that $c_1 u_1 + \ldots + c_n u_n = 0$. Of course, since the $u_i$ are linearly independent, this happens exactly when $c_1 = c_2 = \ldots = c_n = 0$. This proves that the $[u_i]_{\mathcal{B}}$ are linearly independent, as claimed.

Answer 2

This is an instance of a more general theorem.

If $V$ is a $n$-dimensional vector space over field $K$, then $V$ is isomorphic to $K^n$. In other words, there's a bijective linear map $$ \mathfrak{C} \,: V \to K^n \text{.} $$ In fact, there are many such maps $\mathfrak{C}$ - for each pair of bases $(v_i)$ of $V$ and $(b_i)$ of $K^n$, requiring that $\mathfrak{C}(v_i) = b_i$ defines a different such map. And, as for all invertible linear maps, the image of a linearly independent set of vectors is again linearly independent!

In particular, if $(b_i)$ is the canonical basis of $K^n$, i.e. if $$ b_i = (\underbrace{0,\ldots,0}_{\text{$i-1$ times}},1,0,\ldots,0) \in K^n $$ and if $B = (v_i)$ is a basis of $V$ then $$ \mathfrak{C}_B \,:\, V \to K^n \,:\, v_i \mapsto b_i $$ is the coordinatization map corresponding to the basis $B$ of $B$.

So if we're talking about the coordinates $(c_i)$ of some vector $x \in V$ with respect to a basis $B$ of $V$, then we're really talking about the vector $c \in K^n$, where $c = \mathfrak{C}_B(x)$. So it follows, from the bold sentence above, that the image of a linearly independent set of vectors in $V$ under $\mathfrak{C}_B$ is a linearly independent set of vectors in $K^n$.