For a given basis $B \subset V$ of a vector space $V$, any vector $\vec{x} \in V$ can be written in a unique form of

$$ \vec{x} = \lambda_1 \vec{b_1} + \cdots + \lambda_n \vec{b_n} $$

That means that for a given $B$, there is only one set of $\{\lambda_1, \cdots, \lambda_n\} \subset \R$ that can be used to construct $\vec{x}$. From this unique set of coefficients, we construct the coordinate vector of $\vec{x}$ relative to $B$

$$ (\vec{x})_{B} = \begin{bmatrix} \lambda_1 \\ \vdots \\ \lambda_n \end{bmatrix} \in \R^n $$

<aside> ⭐ Notice by using a basis $B \subset V$, we can translate any vector $\vec{x}$ from an abstract vector space $V$ into a concrete vector space $\R^n$ where $n = \dim(V)$. This is super cool!!! This will enable us apply the concrete ideas of matrix algebra to abstract linear transformations

</aside>

Another key fact

<aside> ⭐ Structural properties/vector calculations are preserved when switching from $V$ to $\R^n$. Also taking the coordinate vector of $\vec{x}$ relative to $B$ is a linear tranformation from $V$ to $\R^n$

</aside>

• Example: Is $\{x^2 + 2x + 1, x^2, x + 1\} \in \mathbb{P_2}(\R)$ linearly independent?

<aside> 🚨 Warning: Different bases will give different coordinate systems

</aside>

There are an infinite number of bases you could choose for a given vector space (e.g. in $\R^2$ you have $\{ \vec{e_1}, \vec{e_2}\}$ as well as $\{<1, 1>, <-1, 1>\}$ — any two vectors going in different directions). Each basis will produce different coordinate vectors

How can we switch between coordinate systems?

We have $\vec{x} \in V$ and two bases — $B = \{\vec{b_1}, \cdots, \vec{b_n}\}, C = \{\vec{c_1}, \cdots, \vec{c_n}\} \subset V$.

We can express the coordinate vector of $\vec x$ relative to $B$, $(\vec x)_B$, or $\vec x$ relative to $C$, $(\vec x)_C$. If $V = \R^n$, we can easily solve for these coordinate vectors using the matrix operations we developed at the beginning of the course. We just reduce the augmented matrix

$$ \begin{equation} \left[ \begin{array}{ccc|r} | & & | & |\\ \vec b_1 & \cdots & \vec b_n & \vec x\\ | & & | & | \end{array} \right] \end{equation} $$

So, we can do $\vec x \rightarrow (\vec x)_B$ and $\vec x \rightarrow (\vec x)_C$ , but how do we get from $(\vec x)_B \rightarrow (\vec x)_C$?

• Proof

To calculate the full matrix $P_{C \leftarrow B}$ which maps $(\vec x)_B \rightarrow (\vec x)_C$, we can just repeat the operations in (2), augmenting the matrix of vectors of $C$ with the full matrix of vectors of $B$ and reducing. Meaning