We consider the Euclidean plane $\mathbb{E}^2$. Through a choice of a coordinate system it becomes a coordinate space $\mathbb{R}^2$ in which each point in the plane is described by two coordinates. The vectors $\vec{e}_1=\cv{1\\ 0}$ and $\vec{e}_2=\cv{1\\ 0}$ are the unit vectors in the chosen coordinate system. Each vector $\vec{v}$ from the origin to a point in the plane, say with coordinates $(v_1,v_2)$, can be uniquely written as a linear combination of the vectors $\vec{e}_1$ and $\vec{e}_2$. If we want to be very precise about our choice of coordinate system, we talk about $e$-coordinates. Vector $\vec{v}=v_1\,\vec{e}_1+v_2\,\vec{e}_2$ we denote in this coordinate system with the $e$-coordinates $\cv{v_1\\v_2}$.

We can look in the chosen coordinate system to the points in the plane in which the first and second coordinates are equal to each other. They form the line $y=x$. Reflection in this line is a linear image $L$ which exchanges the unit vectors $\vec{e}_1$ and $\vec{e}_2$. Herewith corresponds the matrix $A=\matrix{0 & 1\\ 1 & 0}$. In the chosen coordinate space, that is, in $e$-coordinates, the linear mapping $L$ is equal to the matrix multiplication $L_A$: $L_A(\vec{v})=A\,\vec{v}$.

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So far no problem. But the choice of the coordinate system in the Euclidean plane is not unique: We could also have rotated the coordinate system in this example, about the origin on the Euclidean plane by a rotation of 45 degrees. This linear transformation maps $\vec{e}_1$ onto $\vec{f}_{\!1}=\tfrac{1}{2}\sqrt{2}\,\vec{e}_1+\tfrac{1}{2}\sqrt{2}\,\vec{e}_2$ and maps $\vec{e}_2$ onto $\vec{f}_{\!2}=-\tfrac{1}{2}\sqrt{2}\,\vec{e}_1+\tfrac{1}{2}\sqrt{2}\,\vec{e}_2$. We can also use these vectors $\vec{f}_{\!1}$ and $\vec{f}_{\!2}$ to uniquely describe any point in the plane. In this coordinate space, the reflection in the line is a linear mapping $L$ with $L(\vec{f}_{\!1})=\vec{f}_{\!1}$ and $L(\vec{f}_{\!2})=-\vec{f}_{\!2}$. In the new coordinate space, defined in terms of $\vec{f}_{\!1}$ and $\vec{f}_{\!2}$ (in short, in $f$-coordinates), the linear mapping $L$ is the matrix multiplication with the diagonal matrix $\matrix{1 & 0\\ 0 & -1}$.

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In short, it makse a difference for the matrix thar describes a linear mapping in the Euclidean plane which coordinate system you have chosen. In our notation you could not find any information about this choice, but for the rest of this section we will use additional labels to describe a linear transformation in the Euclidean plane as a matrix mapping: instead of the matrix $A$ we write $[A]_e^e$ when we are working with the unit vectors $\vec{e}_1, \vec{e}_2$, and $[A]_f^f$ when we are working with vectors $\vec{f}_{\!1}, \vec{f}_{\!2}$. In our example of a reflection in a line $$[A]_e^e=\matrix{0 & 1\\ 1 & 0}\quad\text{and}\quad [A]_f^f=\matrix{1 & 0\\ 0 & -1}$$

The transition from $f$-coordinates to $e$-coordinates can be described with a matrix that expresses the $f$-s in the $e$-s: the first column is made up of the coefficients of $\vec{f}_{\!1}$ written as a linear combination of $\vec{e}_1$ and $\vec{e}_2$, and the second column, is composed of the coefficients of $\vec{f}_{\!2}$ written as a linear combination of $\vec{e}_1$ and $\vec{e}_2$. In fact you are describing the identity transformation Id with two coordinate systems and this is why we denote the corresponding matrix that expresses the$f$-s in the $e$-s as $[\textit{Id}\,]_e^f$. When a vector $\vec{v}$ is described in $f$-coordinates, then we get the $e$-coordinates by multiplication with this matrix: $[\vec{v}]_e=[\textit{Id}\,]_e^f\cdot [\vec{v}]_f$.

Similarly you can describe the transition from $e$- to $f$-coordinates with a transformation matrix $[\textit{Id}\,]_f^e$ that expresses $e$-s in $f$-s. This is the inverse of the matrix $[\textit{Id}\,]_e^f$. When a vector $\vec{v}$ is described in $e$-coordinates, then we get the $f$-coordinates by multiplication with this matrix: $[\vec{v}]_f=[\textit{Id}\,]_f^e\cdot [\vec{v}]_e$

In our example $$[\textit{Id}\,]_e^f=\matrix{\frac{1}{2}\sqrt{2} & -\frac{1}{2}\sqrt{2}\\ \frac{1}{2}\sqrt{2} & \frac{1}{2}\sqrt{2}}\quad\text{and}\quad [\text{Id}\,]_f^e==\bigl([\textit{Id}\,]_e^f\bigr)^{-1}=\matrix{\frac{1}{2}\sqrt{2} & \frac{1}{2}\sqrt{2}\\ -\frac{1}{2}\sqrt{2} & \frac{1}{2}\sqrt{2}}$$ You can yourself calculate applicable in this case: $$[A]_f^f=[\textit{Id}\,]_f^e\cdot[A]_e^e\cdot[\textit{Id}\,]_e^f$$

By all labels you probably cannot see the wood for the trees. In practice we call $[\textit{Id}]_e^f$ the transformation matrix from $f$- to $e$-coordinates and denote this matrix by the symbol $T$. Then we can write the above equality in a simpler way: $$[A]_f^f=T^{-1}\cdot[A]_e^e\cdot T$$

We will see that this example is a prototype of a coordinate transformation in $\mathbb{R}^n$, and that the last equality defines what is the relationship between the matrices of the same linear mapping in two different coordinate systems.