\(\bf Ax=b\) IS A CHANGE OF BASIS:


We can consider \(A\) as transformation \(A: \mathbb R^n \rightarrow \mathbb R^m\) for an \(m \times n\) matrix.

From the Kahn Academy series:

If we have a basis, \(B=\{\vec v_1, \vec v_2, \dots, v_k\}\) the coordinates of a vector \(\vec a\) with respect to this basis can be expressed as:

\(\large [\vec a]_B=\begin{bmatrix}c_1\\c_2\\\vdots\\c_k\end{bmatrix}\)

Equivalently,

\(\large c_1\vec v_1 + c_2\vec v_2+\dots+c_k \vec v_k=\underset{\color{blue}{\text{change of basis matrix 'C'}}}{\underbrace{\color{blue}{\begin{bmatrix}\vert & \vert & &\vert \\v_1&v_2&\dots&v_k\\\vert&\vert&&\vert\end{bmatrix}}}}\begin{bmatrix}c_1\\c_2\\\vdots\\c_k\end{bmatrix}\)

So

\[\Large\bbox[yellow, 10px]{ \text{C}\quad [\vec a]_B \quad=\quad [\vec a]_{\text{standard}}}\]


The change of basis matrix is the matrix with the basis vectors as columns.


Similarly we can solve for the coordinates of a vector with respect to a basis: If we know that the vector \(\vec d = \begin{bmatrix}8\\-6\\2\end{bmatrix}\) is in the span of the \(\mathbb R^2\) basis \(\vec v_1 = \begin{bmatrix}1\\2\\3\end{bmatrix}\) and \(\vec v_2 = \begin{bmatrix}1\\0\\1\end{bmatrix}\), we can solve \(\text{C}\,\color{red}{ [\vec d]_B}=\vec d,\) as a system of linear equations (expanded matrix).


If \(\text{C}\) is invertible (i.e. it is square and with linearly independent column vectors - a given in a basis), the span of \(\text{C}\) is \(\mathbb R^n\), and:

\[\large \text{C}^{-1} \quad \text{C}\quad [\vec a]_B= \text{C}^{-1} \quad\vec a\]

and hence,

\[\bbox[yellow, 10px]{\Large[\vec a]_B\quad= \quad\text{C}^{-1} \quad[\vec a]_{\text{standard}}}\]

In general,

\[\bbox[yellow, 10px]{\Large[\vec a]_{\text{new coord.}}\quad= \quad\text{C}^{-1} \quad[\vec a]_{\text{old coord}}}\tag{1}\]


\(\large \bf A\vec x\) is a linear transformation (transformation = mapping or function operating on vectors expressed as \(T(\vec x)\)) of \(\vec x\) with \(\vec x\) expressed in the standard basis: \(T(\vec x) = A\vec x.\) But there are multiple different bases. The mapping afforded by the transformation on \([\vec x]_B\) will be \([T(\vec x)]_B.\) We can find a matrix \(\text{D}\), such that \[[T(\vec x)]_B= \text{D} [\vec x]_B\tag{2}\]

\(\text{D}\) is the transformation matrix for \(T\) with respect to \(B\).

Now, \(T(\vec x) = A\vec x\), and replacing in Eq.2:

\[[T(\vec x)]_B=[A\vec x]_B =\text{D} [\vec x]_B\]

From Eq.1, \([A\vec x]_B\) can be expressed as:

\[[A\vec x]_B= \text{C}^{-1} A\vec x\]

at the same time \(\vec x = \text{C}[\vec x]_B.\) Therefore,

\[\Large\bbox[yellow, 10px]{\text{D}\quad [\vec x]_B\quad=\quad [A\vec x]_B \quad= \quad \text{C}^{-1}\quad A\quad \text{C}\quad[\vec x]_B}\]

We can also see that

\[\text{D}\quad\text{C}^{-1}=\text{C}^{-1}\quad A \quad \text{C}\quad\text{C}^{-1}\]

or

\[\text{C}\quad\text{D}\quad\text{C}^{-1}=\text{C}\quad\text{C}^{-1}\quad A\quad \text{C}\quad\text{C}^{-1}\]

Hence,

\[\Large A = \text{C D C}^{-1}\]


Summary diagram:


\(\require{AMScd}\) \[\begin{CD} \text{STD. COORD: }\vec x @>\text{A}>> T(\vec x)\\ @V V \text{C inv} V @VV \text{C inv} V\\ \text{COORD wrt B: }[\vec x]_B @>>\text{D}> T[(\vec x)]_B \end{CD}\]

From this lecture by Norm Wildberger.

Any point on the plane is expressed differently by Emily \(\vec e=\begin{bmatrix}\bf \color{green}{e_1} \\ \bf \color{blue}{e_2 }\end{bmatrix}\) compared to Frank \(\vec f=\begin{bmatrix} \bf \color{orange}{f_1} \\ \bf \color{brown}{f_2} \end{bmatrix}\).

So we can say that Frank’s basis vector \(\vec f_1=\begin{bmatrix}\color{orange}{1_{\text{f}}}\\\color{brown}{0_{\text{f}}}\end{bmatrix}\) is \(\vec f_1 = 5 \, \color{green}{e_1} - 2 \, \color{blue}{e_2}\); and Frank’s basis vector \(\vec f_2=\begin{bmatrix}\color{orange}{0_\text{f}}\\\color{brown}{1_\text{f}}\end{bmatrix}\) is \(\vec f_2 = 1.5 \, \color{green}{e_1} + 2.5 \, \color{blue}{e_2}\) from Emily’s perspective.


Or we can say that the basis vectors in Frank’s system will be seen from Emily’s perspective as:

\[\left(\vec f_1\right)_{\vec e}=\begin{bmatrix}\color{green}{5} & \color{green}{1.5} \\ \color{blue}{-2} & \color{blue}{2.5}\end{bmatrix}\,\begin{bmatrix}\color{orange}{1_{\text{f}}}\\\color{brown}{0_{\text{f}}}\end{bmatrix}=\begin{bmatrix}{\color{green}{5}\\\color{blue}{-2}}\end{bmatrix}\]

and

\[\left(\vec f_2\right)_{\vec e}=\begin{bmatrix}\color{green}{5} & \color{green}{1.5} \\ \color{blue}{-2} & \color{blue}{2.5}\end{bmatrix}\,\begin{bmatrix}\color{orange}{0_{\text{f}}}\\\color{brown}{1_{\text{f}}}\end{bmatrix}=\begin{bmatrix}{\color{green}{1.5}\\\color{blue}{2.5}}\end{bmatrix}\]

So in the columns of \(\bf A =\begin{bmatrix}\color{green}{5} & \color{green}{1.5} \\ \color{blue}{-2} & \color{blue}{2.5}\end{bmatrix}\) we have stored how the basis vectors in Frank’s system are expressed in Emily’s basis:

\[\bf A =\begin{bmatrix}\color{green}{5} & \color{green}{1.5} \\ \color{blue}{-2} & \color{blue}{2.5}\end{bmatrix} = \begin{bmatrix}\vert & \vert \\ (\vec f_1)_{\vec e} & (\vec f_2)_{\vec e} \\ \vert & \vert \end{bmatrix}\]

If we were to express in Emily’s system some other vector initially given in Frank’s system we would just have to perform the operation:

\[\begin{align}\begin{bmatrix}\bf y_{1e}\\\bf{y_{2e}}\end{bmatrix}&= \begin{bmatrix}\color{green}{5} & \color{green}{1.5} \\ \color{blue}{-2} & \color{blue}{2.5}\end{bmatrix}\begin{bmatrix}\bf x_{1f}\\\bf x_{2f}\end{bmatrix}\\[2ex] &=\bf x_{1f}\begin{bmatrix}\color{green}{5}\\\color{blue}{-2}\end{bmatrix} + \bf x_{2f}\begin{bmatrix}\color{green}{1.5}\\\color{blue}{2.5}\end{bmatrix} \end{align}\]

The thing to note is that \(\bf x_{1f}\) will multiply the first column in the matrix to contribute to \(\bf y\) (the first basis vector contribution); and \(\bf x_{2f}\) will multiply the second column vector to contribute to \(\bf y\) (the second basis vector contribution, expressed in Emily’s language).

Clearly we have just made up the equivalence between vector basis in between systems, and any other matrix \(\bf A\) would have served the purpose. Therefore we can say that the operation \(\bf Ax = b\) is the equivalent of a change of basis: \(\begin{bmatrix}{\bf y_{1e} \\ \bf y_{2e}}\end{bmatrix}\) is \(\begin{bmatrix}{\bf x_{1f} \\ \bf x_{2f}}\end{bmatrix}\) viewed from Emily’s perspective!

So is the system of equations:

\[\begin{align} 5{\bf x_{1f}} + 1.5\bf{x_{2f}}&=\bf{y_{1e}}\\ -2{\bf x_{1f}} + 2.5 \bf{x_{2f}} &=\bf{y_{2e}} \end{align}\]

Can we express Emily’s vectors from Frank’s perspective? Yes:

\[\begin{align} \frac{5}{31}{\bf y_{1e}} \frac{-3}{31}\bf{y_{2e}}&=\bf{x_{1f}}\\ \frac{4}{31}{\bf y_{1f}} + \frac{10}{31} \bf{y_{2e}} &=\bf{x_{2f}} \end{align}\]

or

\[\begin{bmatrix}\color{green}{5} & \color{green}{1.5} \\ \color{blue}{-2} & \color{blue}{2.5}\end{bmatrix}^{-1}=\begin{bmatrix}\color{orange}{\frac{5}{31}}&\color{orange}{\frac{-3}{31}}\\\color{brown}{\frac{4}{31}}&\color{brown}{\frac{10}{31}}\end{bmatrix}\]

So now we can look at the Emily’s coordinates vector \(\begin{bmatrix}\bf y_{1e}\\ \bf y_{1e}\end{bmatrix}\) in Frank’s coordinates:

\[\begin{bmatrix}\color{orange}{\frac{5}{31}}&\color{orange}{\frac{-3}{31}}\\\color{brown}{\frac{4}{31}}&\color{brown}{\frac{10}{31}}\end{bmatrix}\begin{bmatrix}{\bf y_{1e} \\ y_{1e}}\end{bmatrix}=\begin{bmatrix}{\bf x_{1f} \\ \bf x_{2f}}\end{bmatrix}\]

Likewise, the basis vectors of Emily with respect to Frank’s are:

\[\left(\vec e_1\right)_{\vec f}=\begin{bmatrix}\color{orange}{\frac{5}{31}}&\color{orange}{\frac{-3}{31}}\\\color{brown}{\frac{4}{31}}&\color{brown}{\frac{10}{31}}\end{bmatrix}\,\begin{bmatrix}\color{green}{1_{\text{e}}}\\\color{blue}{0_{\text{e}}}\end{bmatrix}=\begin{bmatrix}{\color{orange}{\frac{5}{31}}\\\color{brown}{\frac{4}{31}}}\end{bmatrix}\]

and

\[\left(\vec e_2\right)_{\vec f}=\begin{bmatrix}\color{orange}{\frac{5}{31}}&\color{orange}{\frac{-3}{31}}\\\color{brown}{\frac{4}{31}}&\color{brown}{\frac{10}{31}}\end{bmatrix}\,\begin{bmatrix}\color{green}{0_{\text{e}}}\\\color{blue}{1_{\text{e}}}\end{bmatrix}=\begin{bmatrix}{\color{orange}{\frac{-3}{31}}\\\color{brown}{\frac{10}{31}}}\end{bmatrix}\]


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