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
\(\seteqnumber{0}{22.}{82}\)\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
\(\seteqnumber{0}{22.}{84}\)\begin{equation} [\mathbf {v}]_{\mathcal {B}} = \begin{pmatrix} v_1 \\ v_2 \\ ... \\ v_{\dim (V)} \end {pmatrix} \end{equation}
Where \(v_1, v_2, ..., v_{\dim (V)}\) are such that
\(\seteqnumber{0}{22.}{85}\)\begin{equation} \mathbf {v} = v_1 \beta _1 + v_2 \beta _2 + ... + v_{\dim (V)} \beta _{\dim (V)} \end{equation}
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
\(\seteqnumber{0}{22.}{86}\)
\begin{equation}
\begin{pmatrix} a & b \\ b & c \end {pmatrix}
\end{equation}
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
\(\seteqnumber{0}{22.}{87}\)\begin{equation} \mathbf {v} = v_1 \beta _1 + v_2 \beta _2 + ... + v_{\dim (V)} \beta _{\dim (V)} \end{equation}
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 \(C\).
\(\seteqnumber{0}{22.}{88}\)\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
\(\seteqnumber{0}{22.}{93}\)\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
\(\seteqnumber{0}{22.}{97}\)\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
\(\seteqnumber{0}{22.}{101}\)\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.