# A calculus of the absurd

##### 22.3.3 Bases
###### 22.3.3.1 Change of basis

Let us suppose that we have two sets of basis vectors for the same vector space $$V$$. It doesn’t really matter what we call them, but $$\mathcal {B}$$ and $$\mathcal {C}$$ are names as good as any. These vectors can be written in the form

\begin{align} & \mathcal {B} = \{\beta _1, \beta _2, ..., \beta _{\dim (V)}\} \\ & \mathcal {C} = \{\gamma _1, \gamma _2, ..., \gamma _{\dim (V)}\} \end{align}

For any vector $$\mathbf {v} \in V$$ we can always write it in the co-ordinate system $$\mathcal {B}$$ by writing the vector as a linear combination133133 This is always possible because $$\mathbf {B}$$ is a basis for $$V$$. of the vectors in $$\mathcal {B}$$. We can write this as

$$[\mathbf {v}]_{\mathcal {B}} = \begin{pmatrix} v_1 \\ v_2 \\ ... \\ v_{\dim (V)} \end {pmatrix}$$

Where $$v_1, v_2, ..., v_{\dim (V)}$$ are such that

$$\mathbf {v} = v_1 \beta _1 + v_2 \beta _2 + ... + v_{\dim (V)} \beta _{\dim (V)}$$

That is, they are the coefficients needed to write $$\mathbf {v}$$ as a linear combination of $$\mathcal {B}$$. This also helps to understand why for example the vector space of $$2 \times 2$$ symmetric matrices134134 i.e. those in the form
$$\begin{pmatrix} a & b \\ b & c \end {pmatrix}$$
is three dimensional; we can write every matrix as a vector of dimension $$3 \times 1$$ where each coefficient denotes what to multiply each basis vector by to obtain our specific vector.

But what if we want to find a way to translate $$[\mathbf {v}]_{\mathcal {C}}$$ into $$[\mathbf {v}]_{\mathcal {B}}$$? This is actually doable using a single matrix. Here’s how. We start by applying the definition of $$[\mathbf {v}]_{\mathcal {B}}$$, that is, we have that $$[\mathbf {v}]_{\mathcal {B}} = [v_1, v_2, ..., v_{\dim (V)}]$$ if and only if

$$\mathbf {v} = v_1 \beta _1 + v_2 \beta _2 + ... + v_{\dim (V)} \beta _{\dim (V)}$$

To find $$[\mathbf {v}]_{\mathcal {C}}$$, it is sufficient to find the $$\mathbf {v}$$ in terms of the basis vectors in $$\mathbf {c}$$. How do we do this? A straightforward approach is to write every vector in $$\mathcal {B}$$ in terms of those in $$\mathcal {C}$$ and then to substitute for them, which removes all the $$\mathcal {B}$$-vectors and means that we instead have $$\mathcal {C}$$-vectors.

Because $$\mathcal {B}$$ and $$\mathcal {C}$$ are both basis for $$V$$, we can write every vector in $$\mathcal {V}$$ in terms of those in $$C$$.

\begin{align} & \beta _1 = \alpha _{1, 1} \gamma _1 + \alpha _{2, 1} \gamma _2 + ... + \alpha _{\dim (V), 1} \gamma _{\dim (V)} \\ & \beta _2 = \alpha _{1, 2} \gamma _1 + \alpha _{2, 2} \gamma _2 + ... + \alpha _{\dim (V), 2} \gamma _{\dim (V)} \\ & ... \\ & \beta _{\dim (V)} = \alpha _{1, \dim (V)} \gamma _1 + \alpha _{2, \dim (V)} \gamma _2 + ... + \alpha _{\dim (V), \dim (V)} \gamma _{\dim (V)} \\ \end{align}

We can then substitute this into the linear combination of $$\mathbf {v}$$ in terms of the basis vectors in $$\mathcal {B}$$, giving

\begin{align} \mathbf {v} &= v_1 \left (\alpha _{1, 1} \gamma _1 + \alpha _{2, 1} \gamma _2 + ... + \alpha _{\dim (V), 1} \gamma _{\dim (V)}\right ) \\ &\quad + v_2 \left (\alpha _{1, 2} \gamma _1 + \alpha _{2, 2} \gamma _2 + ... + \alpha _{\dim (V), 2} \gamma _{\dim (V)}\right ) \\ &\quad + ... \\ &\quad + v_{\dim (V)} \left (\alpha _{1, \dim (V)} \gamma _1 + \alpha _{2, \dim (V)} \gamma _2 + ... + \alpha _{\dim (V), \dim (V)} \gamma _{\dim (V)}\right ) \end{align}

This looks scary, but we just need to stick to the definitions and keep our goal in mind; writing $$\mathbf {v}$$ in terms of all the $$\gamma$$. We can move things around to obtain

\begin{align} \mathbf {v} &= \left (v_1 \alpha _{1, 1} + v_2 \alpha _{2, 1} + ... + v_{\dim (V)} \alpha _{\dim (V), 1}\right ) \gamma _1 \\ &\quad + \left (v_2 \alpha _{1, 2} + v_2 \alpha _{2, 2} + ... + v_{\dim (V)} \alpha _{\dim (V), 2}\right ) \gamma _2 \\ &\quad + ... \\ &\quad + \left (v_2 \alpha _{1, \dim (V)} + v_2 \alpha _{2, \dim (V)} + ... + v_{\dim (V)} \alpha _{\dim (V), \dim (V)}\right ) \gamma _{\dim (V)} \end{align}

Therefore, we have that

\begin{align} [\mathbf {v}]_{\mathcal {B}} &= \begin{pmatrix} v_1 \alpha _{1, 1} + v_2 \alpha _{2, 1} + ... + v_{\dim (V)} \alpha _{\dim (V), 1} \\ v_2 \alpha _{1, 2} + v_2 \alpha _{2, 2} + ... + v_{\dim (V)} \alpha _{\dim (V), 2} \\ ... \\ v_{\dim (V)} \alpha _{1, \dim (V)} + v_{\dim (V)} \alpha _{2, \dim (V)} + ... + v_{\dim (V)} \alpha _{\dim (V), \dim (V)} \end {pmatrix} \\ &= \begin{pmatrix} \alpha _{1, 1} & \alpha _{2, 1} & ... & \alpha _{\dim (V), 1} \\ \alpha _{1, 2} + \alpha _{2, 2} & ... & \alpha _{\dim (V), 2} \\ ... \\ \alpha _{1, \dim (V)} & \alpha _{2, \dim (V)} & ... & \alpha _{\dim (V), \dim (V)} \end {pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ ... \\ v_{\dim (V)} \end {pmatrix} \\ &= \begin{pmatrix} \alpha _{1, 1} & \alpha _{2, 1} & ... & \alpha _{\dim (V), 1} \\ \alpha _{1, 2} & \alpha _{2, 2} & ... & \alpha _{\dim (V), 2} \\ ... \\ \alpha _{1, \dim (V)} & \alpha _{2, \dim (V)} & ... & \alpha _{\dim (V), \dim (V)} \end {pmatrix} [\mathbf {v}]_{\mathcal {C}} \end{align}

Alternatively - the change of basis matrix has as its $$k$$th column the scalars needed to write the $$k$$th element of the one basis as a linear combination of the others.