-2
$\begingroup$

Let $T$ be a linear transformation from $R^3$ to $R^2$ that maps $T(i) = (0,0)$, $T(j)$ = $(1,1)$ and $T(k) = (1,-1)$. In matrix form:

$ T = \left(\begin{array}{ccc}0 & 1 & 1 \\0 & 1 & -1\end{array}\right)$

As I understand it, we say this is the matrix of $T$ relative to the standard basis. If we were asked to find $T$ relative to a new basis, how would you use change of basis because $T$ is not invertible?

$\endgroup$
  • $\begingroup$ what is new basis you have in mind? $\endgroup$ – abel Apr 8 '15 at 19:35
4
+25
$\begingroup$

The most general thing you could ask is this:

Suppose $\mathcal{B}$ is a new basis for $\mathbb{R}^3$, and $\mathcal{C}$ is a new basis for $\mathbb{R}^2$. Then one should be able to find a matrix $[T]_{\mathcal{C} \leftarrow \mathcal{B}}$ whose input is a vector in $\mathcal{B}$ coordinates and whose output is a vector in $\mathcal{C}$ coordinates. In equation form, let $\mathbf{x} \in \mathbb{R}^3$, we want:

$$[T\mathbf{x}]_{\mathcal{C}} = [T]_{\mathcal{C} \leftarrow \mathcal{B}}[\mathbf{x}]_{\mathcal{B}}$$

Let $P_{\mathcal{B}}$ be the 3x3 matrix whose columns are the vectors in $\mathcal{B}$, likewise $P_{\mathcal{C}}$ is the 2x2 matrix whose columns are the vectors in $\mathcal{C}$. Then one has the formula:

$$\boxed{[T]_{\mathcal{C} \leftarrow \mathcal{B}} = P_{\mathcal{C}}^{-1}TP_{\mathcal{B}}}$$

To explain this formula, you should imagine $P_{\mathcal{B}}$ as the 'decoder'. It turns a $\mathcal{B}-$coordinate vector into the standard coordinate vector:

$$P_{\mathcal{B}}[\mathbf{x}]_{\mathcal{B}} = \mathbf{x}$$

And similarly, $P_{\mathcal{C}}^{-1}$ is the 'encoder', which takes a vector in standard coordinates and turns it into $\mathcal{C}-$coordinates.

$\endgroup$

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.