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?

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

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.


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