I wanted to make sure I understand how to find the matrix for a linear transformation, but for a non-standard basis. Lets say for example you have the linear operator in $\mathbb{F}^2$ defined in the standard coordinates by T$\begin{pmatrix}x\\y\end{pmatrix}$ = $\begin{pmatrix}x+y\\y-x\end{pmatrix}$ for the basis $(2,3)^T$, and $(1,2)^T$. Would I just the matrix as (T$\begin{pmatrix}2\\3\end{pmatrix}$ | T$\begin{pmatrix}1\\2\end{pmatrix}$)? Which comes out to be $\begin{pmatrix}5&3\\1&1\end{pmatrix}$


If that were true, one would expect the matrix of $T$ in the non-standard basis to take the coordinates of $v$ in that basis to $Tv$ in that basis. However:

$\begin{bmatrix}5&3\\1&1 \end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix} = \begin{bmatrix}5\\1\end{bmatrix}$

So your matrix takes $1(2,3)^T + 0(1,2)^T \to 5(2,3)^T + 1(1,2)^T = (11,17)^T$.

However: $T((2,3)^T) = (5,1)^T$, so this cannot be right.

The matrix for $T$ in the standard basis is:

$A = \begin{bmatrix}1&1\\-1&1\end{bmatrix}$.

Let us denote the coordinates of $(x,y)^T$ in the non-standard basis by:

$[u,v]^T = u(2,3)^T + v(2,1)^T$.

The matrix which sends $[u,v]^T$ to its "standard basis coordinates" is:

$P = \begin{bmatrix}2&1\\3&2\end{bmatrix}$, for example, this sends:

$[1,0]^T = 1(2,3)^T + 0(1,2)^T \to (2,3)^T$

$[0,1]^T = 0(2,3)^T + 1(1,2)^T \to (1,2)^T$, as expected.

Thus $AP([u,v]^T)$ returns the standard-basis coordinates of $T$'s image on non-standard basis coordinate vectors. To get our final answer in non-standard basis coordinates, we must apply:

$P^{-1} = \begin{bmatrix}2&-1\\-3&2\end{bmatrix}$, so that the desired matrix is:

$P^{-1}AP = \begin{bmatrix}9&5\\=13&-7\end{bmatrix}$

and this matrix takes $(2,3)^T = [1,0]^T \to [9,-13]^T = (5,1)^T$ and:

$(1,2)^T = [0,1]^T \to [5,-7]^T = (3,1)^T$, as desired.

  • $\begingroup$ Would it be $P^{-1}AP$? Because Isn't that $P_{SA}A_{SS}P_{AS}$, which doesn't have balance of subscripts, unless I'm not understanding something right. Because A is taking the standard basis to standard basis so it is $A_{SS}$ and P is taking the standard basis to the new basis, so $P_{AS}$ and the inverse is just the opposite. $\endgroup$ – TheStrangeQuark Mar 5 '15 at 3:35
  • $\begingroup$ No, $P$ takes the new basis to the standard basis. $\endgroup$ – David Wheeler Mar 5 '15 at 3:38
  • $\begingroup$ Oh! alright, thank you! $\endgroup$ – TheStrangeQuark Mar 5 '15 at 3:43

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