2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

enter image description here

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 $B$:

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.