Basis change via matrices – "Math for Non-Geeks"

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In this article, you will learn about basis change via matrices. Basis change matrices can be used to convert coordinates with respect to a given basis into coordinates with respect to another basis. This is particularly useful for matrices of linear maps, which are always taken with respect to two specific bases.

We have seen in the article on bases that every finite-dimensional vector space has a basis. This means if ${\displaystyle V}$ is an ${\displaystyle n}$-dimensional ${\displaystyle K}$-vector space, then there is a basis ${\displaystyle B=\{b_{1},\ldots ,b_{n}\}}$ of ${\displaystyle V}$. Every vector ${\displaystyle v\in V}$ can therefore be written uniquely as a linear combination of the basis vectors ${\displaystyle b_{1},\ldots ,b_{n}}$, i.e. ${\displaystyle v=\sum _{i=1}^{n}\lambda _{i}b_{i}}$ with unique ${\displaystyle \lambda _{1},\ldots ,\lambda _{n}\in K}$.

We also know that vector spaces usually have more than one basis. Let ${\displaystyle C=\{c_{1},\ldots ,c_{n}\}}$ be a second basis of ${\displaystyle V}$. Then we can also write ${\displaystyle v}$ uniquely as a linear combination of ${\displaystyle c_{i}}$, i.e. ${\displaystyle v=\sum _{i=1}^{n}\mu _{i}c_{i}}$ with unique coefficients ${\displaystyle \mu _{1},\ldots ,\mu _{n}\in K}$.

We therefore have two representations of the vector ${\displaystyle v}$. Using the basis ${\displaystyle B}$ we get the representation ${\displaystyle v=\sum _{i=1}^{n}\lambda _{i}b_{i}}$ and using the basis ${\displaystyle C}$ we get ${\displaystyle v=\sum _{i=1}^{n}\mu _{i}c_{i}}$.

How can we convert the basis representation with respect to ${\displaystyle B}$ of the vector ${\displaystyle v}$ into the representation with respect to ${\displaystyle C}$?

This question is particularly interesting in the context of matrices of linear maps, as we will see below in the section Application of basis change via matrices. Mapping matrices allow us to calculate with coordinates instead of vectors of ${\displaystyle V}$. However, the coordinates of a vector always depend on the chosen basis in ${\displaystyle V}$. We want a simple way to convert the coordinates of any vector in ${\displaystyle V}$ with respect to a basis ${\displaystyle B}$ into coordinates with respect to another basis ${\displaystyle C}$.

To answer this question, we start with a simpler special case. We consider ${\displaystyle K^{n}}$ as a vector space and set ${\displaystyle B=(e_{1},\ldots ,e_{n})}$ as the (ordered) standard basis. Let further ${\displaystyle C=(c_{1},\ldots ,c_{n})}$ be any ordered basis of ${\displaystyle K^{n}}$. Since matrices of linear maps depend on the order of the basis vectors, we have to use ordered bases ${\displaystyle B}$ and ${\displaystyle C}$.

Let ${\displaystyle v=(x_{1},\ldots ,x_{n})^{T}=\sum _{i=1}^{n}x_{i}e_{i}}$ be a vector for whom we know the coordinates with respect to the standard basis ${\displaystyle B}$. The vector ${\displaystyle v\in K^{n}}$ can be written in the basis ${\displaystyle C}$ as ${\displaystyle v=\lambda _{1}c_{1}+\cdots +\lambda _{n}c_{n}}$ for unique ${\displaystyle \lambda _{1},\ldots ,\lambda _{n}\in K}$. How can we calculate the coordinates ${\displaystyle \lambda _{1},\dots ,\lambda _{n}\in K}$ of ${\displaystyle v}$ with respect to ${\displaystyle C}$ simply from the coordinates ${\displaystyle x_{1},\dots ,x_{n}}$ of ${\displaystyle v}$ with respect to the standard basis ${\displaystyle B}$?

To do this, we need to describe the mapping ${\displaystyle K^{n}\to K^{n}}$, which maps each vector ${\displaystyle v=(x_{1},...,x_{n})^{T}\in K^{n}}$ to its coordinate vector ${\displaystyle (\lambda _{1},\dots ,\lambda _{n})^{T}\in K^{n}}$ with respect to ${\displaystyle C}$. This is done by the coordinate mapping ${\displaystyle k_{C}:K^{n}\to K^{n}}$, which is a linear map that we know from the article on isomorphims.

In order to describe ${\displaystyle k_{C}}$, we calculate its matrix ${\displaystyle M_{\mathrm {Std} }^{\mathrm {Std} }(k_{C})}$ with respect to the standard basis ${\displaystyle B=(e_{1},\ldots ,e_{n})}$. By using matrix-vector multiplication in ${\displaystyle K^{n}}$, we then obtain the coordinate vector ${\displaystyle (\lambda _{1},\ldots ,\lambda _{n})^{T}}$ by multiplying ${\displaystyle v=(x_{1},\ldots ,x_{n})^{T}}$ from the left by ${\displaystyle M_{\mathrm {Std} }^{\mathrm {Std} }(k_{C})}$.

To calculate the matrix ${\displaystyle M_{\mathrm {Std} }^{\mathrm {Std} }(k_{C})}$, we need to determine ${\displaystyle k_{C}(e_{1}),\ldots ,k_{C}(e_{n})}$. These will then be the columns of ${\displaystyle M_{\mathrm {Std} }^{\mathrm {Std} }(k_{C})}$. We are therefore looking for the coordinates of ${\displaystyle e_{1},\ldots ,e_{n}}$ with respect to ${\displaystyle C}$, so we have to write these as a linear combination of vectors in ${\displaystyle C}$. This gives us ${\displaystyle n}$ equations ${\displaystyle {\begin{aligned}e_{1}=&\sum _{i=1}^{n}a_{i1}c_{i}\\&\vdots \\e_{n}=&\sum _{i=1}^{n}a_{in}c_{i}\end{aligned}}}$ where ${\displaystyle a_{ij}}$ are the coordinates we are looking for. The coefficients ${\displaystyle a_{ij}}$ can be determined by solving a linear system of equations.

Then ${\displaystyle k_{C}(e_{j})=(a_{1j},a_{2j},\ldots ,a_{nj})^{T}}$ for ${\displaystyle j=1,\ldots ,n}$. This gives us the matrix ${\displaystyle M_{\mathrm {Std} }^{\mathrm {Std} }(k_{C})={\begin{pmatrix}a_{11}&a_{12}&\ldots &a_{1n}\\a_{21}&a_{22}&\ldots &a_{2n}\\\vdots &\vdots &\ddots &\vdots \\a_{n1}&a_{n2}&\ldots &a_{nn}\end{pmatrix}}.}$ We obtain ${\displaystyle M_{\mathrm {Std} }^{\mathrm {Std} }(k_{C})y=k_{C}(y)}$ for all ${\displaystyle y\in K^{n}}$. The required coefficients ${\displaystyle \lambda _{1},\dots ,\lambda _{n}}$ are therefore obtained by ${\displaystyle M_{\mathrm {Std} }^{\mathrm {Std} }(k_{C}){\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}={\begin{pmatrix}\lambda _{1}\\\vdots \\\lambda _{n}\end{pmatrix}}.}$

In a general finite-dimensional vector space ${\displaystyle V}$, unlike in ${\displaystyle K^{n}}$, there is no standard basis. In this situation, we have two ordered bases ${\displaystyle B=(b_{1},\dots ,b_{n})}$ and ${\displaystyle C=(c_{1},\dots ,c_{n})}$. Usually, we are then given an arbitrary vector ${\displaystyle v\in V}$ as a linear combination ${\displaystyle v=x_{1}b_{1}+\dots +x_{n}b_{n}}$ with respect to the basis ${\displaystyle B}$ with ${\displaystyle x_{1},\dots ,x_{n}\in K}$. The coefficients ${\displaystyle x_{1},\ldots ,x_{n}}$ are also called the coordinates of ${\displaystyle v}$ with respect to ${\displaystyle B}$. Correspondingly, the coordinates with respect to ${\displaystyle C}$ are the scalars ${\displaystyle \lambda _{1},\dots ,\lambda _{n}\in K}$ with ${\displaystyle v=\lambda _{1}c_{1}+\dots +\lambda _{n}c_{n}}$.

We are looking for a method to convert the coordinates ${\displaystyle x_{1},\ldots ,x_{n}}$ with respect to ${\displaystyle B}$ of any vector ${\displaystyle v\in V}$ into the coordinates ${\displaystyle \lambda _{1},\dots ,\lambda _{n}}$ with respect to ${\displaystyle C}$. For this, we need a mapping ${\displaystyle K^{n}\to K^{n}}$, which sends ${\displaystyle (x_{1},\dots ,x_{n})^{T}}$ to ${\displaystyle (\lambda _{1},\dots ,\lambda _{n})^{T}}$.

We already know the coordinate mappings ${\displaystyle k_{B}:V\to K^{n}}$ with ${\displaystyle k_{B}(v)=(x_{1},\dots ,x_{n})^{T}\in K^{n}}$ and ${\displaystyle k_{C}:V\to K^{n}}$ with ${\displaystyle k_{C}(v)=(\lambda _{1},\dots ,\lambda _{n})^{T}}$. From ${\displaystyle (x_{1},\dots ,x_{n})^{T}\in K^{n}}$ we want to obtain the vector ${\displaystyle (\lambda _{1},\dots ,\lambda _{n})^{T}\in K^{n}}$. The coordinate mappings are isomorphisms. So ${\displaystyle k_{B}^{-1}:K^{n}\to V}$ maps the vector ${\displaystyle (x_{1},\dots ,x_{n})^{T}}$ to ${\displaystyle v}$ and ${\displaystyle k_{C}:V\to K^{n}}$ maps ${\displaystyle v}$ to ${\displaystyle (\lambda _{1},\dots ,\lambda _{n})^{T}}$. If we first execute ${\displaystyle k_{B}^{-1}}$ and then ${\displaystyle k_{C}}$, we obtain a mapping that sends ${\displaystyle (x_{1},\dots ,x_{n})^{T}}$ to ${\displaystyle (\lambda _{1},\dots ,\lambda _{n})^{T}}$.

Our desired transformation is therefore realized by the linear map ${\displaystyle k_{C}\circ k_{B}^{-1}:K^{n}\to K^{n}}$. As above for the situation in ${\displaystyle K^{n}}$, we can then determine the matrix of this linear map in ${\displaystyle K^{n}}$ with respect to the standard basis. This matrix is given by ${\displaystyle M_{\mathrm {Std} }^{\mathrm {Std} }(k_{C}\circ k_{B}^{-1})}$. If we remember the article on matrices of linear maps, however, this matrix is just ${\displaystyle M_{C}^{B}(\operatorname {id} _{V})}$, because ${\displaystyle k_{C}\circ k_{B}^{-1}=k_{C}\circ \operatorname {id} _{V}\circ k_{B}^{-1}}$.

It also makes intuitive sense that the matrix executing the basis change from ${\displaystyle B}$ to ${\displaystyle C}$ is given exactly by ${\displaystyle M_{C}^{B}(\operatorname {id} _{V})}$ representing the identity from basis ${\displaystyle B}$ to ${\displaystyle C}$. This is because, if we multiply the coordinate vector ${\displaystyle k_{B}(v)}$ of ${\displaystyle v\in V}$ with respect to ${\displaystyle B}$ from the left with ${\displaystyle M_{C}^{B}(\operatorname {id} _{V})}$, then we obtain exactly the coordinate vector of ${\displaystyle \operatorname {id} _{V}(v)=v}$ with respect to ${\displaystyle C}$, just by definition of the representing matrix. That is, ${\displaystyle k_{C}(v)=M_{C}^{B}(\operatorname {id} _{V})\cdot k_{B}(v)}$ for all ${\displaystyle v\in V}$. The matrix ${\displaystyle M_{C}^{B}(\operatorname {id} _{V})}$ therefore converts coordinates with respect to ${\displaystyle B}$ into coordinates with respect to ${\displaystyle C}$. This is exactly what a basis change matrix does.

The basis change matrix has many other names. It is also referred to in the literature as a transition matrix, basis transition matrix, transformation matrix or coordinate change matrix.

We can find a matrix ${\displaystyle M_{C}^{B}(f)}$ for every linear map ${\displaystyle f\colon V\to W}$ between two finite-dimensional vector spaces, with respect to bases ${\displaystyle B}$ and ${\displaystyle C}$. However, this matrix depends on ${\displaystyle B}$ and ${\displaystyle C}$, and their order. If we choose other bases ${\displaystyle B'}$ or ${\displaystyle C'}$, we will very likely get a different matrix. We can see this in the following example:

${\displaystyle f:\mathbb {R} ^{2}\to \mathbb {R} ^{2},\quad {\begin{pmatrix}x\\y\end{pmatrix}}\mapsto {\begin{pmatrix}x+y\\2y\end{pmatrix}}}$

Consider a linear map ${\displaystyle f:V\to W}$ and two ordered bases ${\displaystyle B}$ and ${\displaystyle B'}$ of ${\displaystyle V}$ as well as ${\displaystyle C}$ and ${\displaystyle C'}$ of ${\displaystyle W}$. We are asking now: How can we convert the matrix ${\displaystyle M_{C}^{B}(f)}$ into ${\displaystyle M_{C'}^{B'}(f)}$?

In the following, we will consider why the formula in this theorem is correct and how we arrived at it.

From the definition of the matrix of a linear map we know that for all vectors ${\displaystyle x\in K^{n}}$, we have ${\displaystyle M_{C}^{B}(f)x=k_{C}\circ f\circ k_{B}^{-1}(x)}$ and ${\displaystyle M_{C'}^{B'}(f)x=k_{C'}\circ f\circ k_{B'}^{-1}(x)}$. We can visualize this equation in a diagram:

In these two diagrams, it doesn't matter which way you go. For example, it does not matter whether we use ${\displaystyle f}$ to go directly from ${\displaystyle V}$ to ${\displaystyle W}$ or take the detour via ${\displaystyle K^{n}}$ and ${\displaystyle K^{m}}$. If the same map is constructed along each path, this is referred to as a commutative diagram.

We can join the two diagrams together:

This diagram is also commutative. That means, if you have a fixed start and end point, it still doesn't matter which path you take in the diagram. If we start at the top left at ${\displaystyle K^{n}}$, it doesn't matter which path we use to get to ${\displaystyle K^{m}}$ at the bottom left. We can get from ${\displaystyle K^{n}}$ to ${\displaystyle K^{m}}$ via ${\displaystyle x\mapsto M_{C'}^{B'}(f)x}$, or using first ${\displaystyle k_{B}\circ k_{B'}^{-1}:K^{n}\to K^{n}}$, then ${\displaystyle x\mapsto M_{C}^{B}(f)x}$ and finally ${\displaystyle k_{C'}\circ k_{C}^{-1}:K^{m}\to K^{m}}$.

Consequently, the map ${\displaystyle K^{n}\to K^{m},\ x\mapsto M_{C'}^{B'}(f)x}$ is equal to the combination of the maps ${\displaystyle k_{B}\circ k_{B'}^{-1}}$, ${\displaystyle x\mapsto M_{C}^{B}(f)x}$, and ${\displaystyle k_{C'}\circ k_{C}^{-1}}$. We have now seen that the ${\displaystyle x\mapsto M_{C}^{B}(f)x}$ can be transformed into the map ${\displaystyle x\mapsto M_{C'}^{B'}(f)x}$. Originally, however, we wanted to transform the matrix ${\displaystyle M_{C}^{B}(f)}$ into the matrix ${\displaystyle M_{C'}^{B'}(f)}$. How do we get from the map ${\displaystyle K^{n}\to K^{m},\ x\mapsto M_{C'}^{B'}(f)x}$ back to the matrix ${\displaystyle M_{C'}^{B'}(f)\in K^{m\times n}}$?

The matrix ${\displaystyle M_{C'}^{B'}(f)}$ looks complicated. We therefore consider how we can answer this question for a general matrix ${\displaystyle A\in K^{m\times n}}$. We consider the linear map ${\displaystyle L_{A}:K^{n}\to K^{m},\ x\mapsto Ax}$ associated with ${\displaystyle A}$. The matrix of ${\displaystyle L_{A}}$ with respect to the standard bases of ${\displaystyle K^{n}}$ and ${\displaystyle K^{m}}$ is again ${\displaystyle A}$. Let us now plug in the matrix ${\displaystyle M_{C'}^{B'}(f)}$ for ${\displaystyle A}$. The matrix of the linear map ${\displaystyle x\mapsto M_{C'}^{B'}(f)x}$ with respect to the standard bases is exactly ${\displaystyle M_{C'}^{B'}(f)}$.

As we have already seen, the map ${\displaystyle x\mapsto M_{C'}^{B'}(f)x}$ is equal to the combination of the three maps ${\displaystyle k_{B}\circ k_{B'}^{-1}}$, ${\displaystyle x\mapsto M_{C}^{B}(f)x}$, and ${\displaystyle k_{C'}\circ k_{C}^{-1}}$. Therefore, the matrix of the combination of ${\displaystyle k_{B}\circ k_{B'}^{-1}}$, ${\displaystyle x\mapsto M_{C}^{B}(f)x}$, and ${\displaystyle k_{C'}\circ k_{C}^{-1}}$ corresponds to ${\displaystyle M_{C'}^{B'}(f)}$ with regard to the standard bases.

However, we can also determine the matrix of the concatenation in another way. In the article on matrix multiplication, we saw that concatenation between linear maps correspond exactly to the multiplication of the respective matrices. Therefore, we write down the matrices of the concatenated linear maps individually and then multiply them.

• As we have already seen for ${\displaystyle M_{C'}^{B'}(f)}$, the matrix of ${\displaystyle x\mapsto M_{C}^{B}(f)x}$ with respect to the standard bases of ${\displaystyle K^{n}}$ and ${\displaystyle K^{m}}$ is again ${\displaystyle M_{C}^{B}(f)}$.
• We have already derived the matrix of ${\displaystyle k_{C'}\circ k_{C}^{-1}}$ above; it is ${\displaystyle M_{C'}^{C}(\operatorname {id} )}$. This is exactly the basis change matrix ${\displaystyle T_{C'}^{C}}$.
• Similarly, the matrix of ${\displaystyle k_{B}\circ k_{B'}^{-1}}$ is given by the basis change matrix ${\displaystyle T_{B}^{B'}=M_{B}^{B'}(\operatorname {id} )}$.

If we multiply these three matrices, we obtain ${\displaystyle M_{C'}^{B'}(f)}$: ${\displaystyle M_{C'}^{B'}(f)=T_{C'}^{C}M_{C}^{B}(f)T_{B}^{B'}}$ So ${\displaystyle M_{C'}^{B'}(f)}$ can be calculated from ${\displaystyle M_{C}^{B}(f)}$ by left multiplication with ${\displaystyle T_{C'}^{C}}$ and right multiplication with ${\displaystyle T_{B}^{B'}}$.

We now know, how we can convert matrices of a linear map with respect to different bases into each other. Let's look at the example above again. We consider the linear map ${\displaystyle f:\mathbb {R} ^{2}\to \mathbb {R} ^{2},\quad {\begin{pmatrix}x\\y\end{pmatrix}}\mapsto {\begin{pmatrix}x+y\\2y\end{pmatrix}}}$ as well as the ordered bases ${\displaystyle B=(e_{1},e_{2})}$, ${\displaystyle C=((1,1)^{T},(1,0)^{T})}$, and ${\displaystyle C'=((1,2)^{T},(1,0)^{T})}$. We have already calculated the matrix ${\displaystyle M_{C}^{B}(f)}$: ${\displaystyle M_{C}^{B}(f)={\begin{pmatrix}0&2\\1&-1\end{pmatrix}}}$ We want to determine ${\displaystyle M_{C'}^{B}(f)}$ by matrix multiplication, i.e., by ${\displaystyle M_{C'}^{B}(f)=T_{C'}^{C}M_{C}^{B}(f)T_{B}^{B}}$. We have to determine ${\displaystyle T_{B}^{B}}$ and ${\displaystyle T_{C'}^{C}}$. Now, ${\displaystyle T_{B}^{B}=I_{2}}$, since the basis ${\displaystyle B}$ does not change. Now let us turn to computing the basis change matrix ${\displaystyle T_{C'}^{C}}$: We know that ${\displaystyle T_{C'}^{C}=M_{C'}^{C}(\operatorname {id} )}$. In order to determine this matrix, we need to express the basis vectors of ${\displaystyle C}$ in the basis ${\displaystyle C'}$:

${\displaystyle {\begin{aligned}{\begin{pmatrix}1\\1\end{pmatrix}}&={\frac {1}{2}}\cdot {\begin{pmatrix}1\\2\end{pmatrix}}+{\frac {1}{2}}\cdot {\begin{pmatrix}1\\0\end{pmatrix}}\\[0.5em]{\begin{pmatrix}1\\0\end{pmatrix}}&=0\cdot {\begin{pmatrix}1\\2\end{pmatrix}}+1\cdot {\begin{pmatrix}1\\0\end{pmatrix}}.\end{aligned}}}$ Hence, ${\displaystyle T_{C'}^{C}={\begin{pmatrix}{\frac {1}{2}}&0\\{\frac {1}{2}}&1\end{pmatrix}}.}$ Therefore ${\displaystyle M_{C'}^{B}(f)=T_{C'}^{C}M_{C}^{B}(f)={\begin{pmatrix}{\frac {1}{2}}&0\\{\frac {1}{2}}&1\end{pmatrix}}\cdot {\begin{pmatrix}0&2\\1&-1\end{pmatrix}}={\begin{pmatrix}0&1\\1&0\end{pmatrix}}.}$ You may convince yourself that this result agrees with the result above.

Consider the bases ${\displaystyle B'=\left({\begin{pmatrix}2\\1\end{pmatrix}},{\begin{pmatrix}1\\1\end{pmatrix}}\right)\quad {\text{and}}\quad B=\left({\begin{pmatrix}0\\-1\end{pmatrix}},{\begin{pmatrix}2\\3\end{pmatrix}}\right)}$ of ${\displaystyle \mathbb {R} ^{2}}$, as well as the bases ${\displaystyle C=\left({\begin{pmatrix}1\\-2\\3\end{pmatrix}},{\begin{pmatrix}0\\0\\-1\end{pmatrix}},{\begin{pmatrix}2\\1\\1\end{pmatrix}}\right)\quad {\text{and}}\quad C'=\left({\begin{pmatrix}1\\1\\1\end{pmatrix}},{\begin{pmatrix}1\\1\\0\end{pmatrix}},{\begin{pmatrix}1\\0\\0\end{pmatrix}}\right)}$ of ${\displaystyle \mathbb {R} ^{3}}$. Let ${\displaystyle f:\mathbb {R} ^{2}\to \mathbb {R} ^{3}}$ be a map with the following matrix with respect to ${\displaystyle B}$ and ${\displaystyle C}$: ${\displaystyle M_{C}^{B}(f)={\begin{pmatrix}5&-3\\2&4\\0&-2\end{pmatrix}}}$

We want to determine the matrix of ${\displaystyle f}$ with respect to the bases ${\displaystyle B'}$ and ${\displaystyle C'}$. This can be done by matrix multiplication ${\displaystyle M_{C'}^{B'}(f)=T_{C'}^{C}M_{C}^{B}(f)T_{B}^{B'}}$. To do so, we must first calculate the basis change matrices ${\displaystyle T_{B}^{B'}}$ and ${\displaystyle T_{C'}^{C}}$.