I am trying to understand what each transformation means and what their differences are but many books that don't state which transformation they are referring to make it a bit confusing to understand which is which.
Also, I want to particularly know how do matrices change when we deal with each transformation, so if anybody could help it would be really appreciated.

Note: I also want to know how is a matrix represented when we change a basis through such transformations


In an active transformation, given a basis, we start from a vector and we find a new vector in the same basis.

In a passive transformation we have a vector expressed in a basis and we express it in a new basis.

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The figure illustrate the action of a matrix $A$ as an active transformation and of $A^{-1}$ as the corresponding passive transformation.

Here we have: $$ A=\begin{bmatrix} 1&2\\ -2&4 \end{bmatrix} \qquad A^{-1}=\frac{1}{8} \begin{bmatrix} 4&-2\\ 2&1 \end{bmatrix} $$

The matrix $A$ acts on a vector $\mathbf{x}$ that in the standard basis $S$ (represented in black) has components $\mathbf{x}=[3,2]_S^T$, and, as active transformation, gives the vector $\mathbf{x'}=A\mathbf{x}=[7,2]_S^T$.

Note that in the new basis $B$ that has as basis vectors the columns of $A$ (represented in blue) this vector has components $\mathbf{x'}=[3,2]_B^T$.

The inverse matrix $A^{-1}$ represents the passive transformation that gives the components of the vector $\mathbf{x}$ in the new basis $B$:

$$ A^{-1}\mathbf{x}= \frac{1}{8} \begin{bmatrix} 4&-2\\ 2&1 \end{bmatrix} \begin{bmatrix} 3\\ 2 \end{bmatrix}= \begin{bmatrix} 1\\ 1 \end{bmatrix} $$

  • $\begingroup$ So, which of the two is what we call a basis change? $\endgroup$ – TheQuantumMan Jul 1 '16 at 21:49
  • $\begingroup$ $A^{-1}$ is the matrix that represents the change of basis form the standard basis to the new basis that has basis vectors the columns of $A$. $\endgroup$ – Emilio Novati Jul 2 '16 at 6:55

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