Let matrix $T = \left(\begin{array}{cc}1 & 2 \\1 & -1\end{array}\right).$ Let $e_1 = (1,1)^T$ and $e_2 = (3,1)^T$. $T$ is currently relative to the standard basis. If asked to find $T$ relative to bases $e_1,e_2$, do you solve the problem via:

(a) $H$ \ $T$ (note: matlab notation; this solves for $x$ in $Hx = T$).

or (b) $H^{-1} * T * H$?

where $H$ = $\left(\begin{array}{cc}1 & 3 \\-1 & 1\end{array}\right).$

I'm confused because question: Find matrix of linear transformation seems to suggest that (b) is the way to go, but the question Find matrix of linear transformation relative to new bases seems to suggest that a way of representing a matrix relative to a new basis is more like (a) [not exactly the same since $T$ is not invertible, but since $T$ for us is invertible, we can ignore this fact.]

  • $\begingroup$ Should $e_1$ have a $-1$ or should $H$ have a 3 $1$'s? $\endgroup$ – user137731 Mar 6 '15 at 23:39

Basically the way to think about your matrix product is as a series of transformations.

First you transform your vector from the new basis to the standard basis. Notice that $H$ transforms vectors from the new basis to their representations wrt the standard basis. This is exactly what we want. So the first matrix (the one that'll go on the right -- and thus will be multiplied by the vector on the right first) will be $H$.

Now that your vector is in the standard basis, you want to transform your vector with $T$. So that'll be your middle matrix in this product.

Finally, you'd like to transform your vector from it's coordinates with respect to the standard basis to the new basis. This is just the inverse of $H$: i.e. $H^{-1}$.

So your matrix wrt to the new basis is just $T' = H^{-1}TH$.

P.S. I don't see how solving for $x$ in $Hx=T$, which is presumably a column vector, has anything to do with getting the matrix which represents the same linear transformation as $T$ wrt your new basis.

  • $\begingroup$ so I'm confused as to why this works: math.stackexchange.com/questions/1178772/… To get a new representation of $T$, you just do $H\T$, which would give you the representation of $T$ with new bases. $\endgroup$ – larry Mar 7 '15 at 1:27
  • $\begingroup$ The answer given to that one uses the same technique as Bernard uses here. You could also construct it the same way as I did here by finding a matrix $A$ which takes vectors in the standard basis of $\Bbb R^2$ and transforms them into their representations wrt to the basis $\{w_1, w_2\}$ (that matrix would be the inverse of $\pmatrix { 1 & 1 \\ 1 & 2}$). Then the new matrix would be $T'=AT$. Try this method, you'll see that you do indeed get the correct answer (I just tested it). $\endgroup$ – user137731 Mar 7 '15 at 1:33
  • $\begingroup$ My confusion is that I don't see why I shouldn't do this: $T(i)=(1,1)=a_1e_1+b_1e_2$.$T(j)=(2,1)=a_2e_1+b_2e_2$. Solve for the $a$s and b$s. Why doesn't this work? That's what math.stackexchange.com/questions/1178772/… does. $\endgroup$ – larry Mar 7 '15 at 1:46
  • $\begingroup$ If you do that you'll know the coordinates of $T(i)$ and $T(j)$ with respect to your new basis. That will give you $H^{-1}$ from my answer. But your $i$ and $j$ are both given in the standard basis. If you want your matrix $T'$ to take vectors from the new basis to the new basis, you also need to find the coordinates of $i$ and $j$ wrt to the new basis. That algebraic approach is what Bernard did here. $\endgroup$ – user137731 Mar 7 '15 at 1:51
  • 1
    $\begingroup$ Let $A$ be the standard basis and $B =\{e_1, e_2\}$. Also let $[v]_C$ be the vector $v$ written in the basis $C$. You are supposed to find a matrix which does this $[v]_B \mapsto [T(v)]_B$ given only $[v]_A \mapsto [T(v)]_A$ (and implicitly $[v]_B \mapsto [v]_A$). I suggested you do this by $[v]_B \mapsto [v]_A \mapsto [T(v)]_A \mapsto [T(v)]_B$. What you've done algebraically above (assuming you know how to construct a matrix out of $a_1, a_2, b_1, b_2$) is find $[v]_A \mapsto [T(v)]_A \mapsto [T(v)]_B$. You're missing the $[v]_B \mapsto [v]_A$ part. $\endgroup$ – user137731 Mar 7 '15 at 2:04

Method $b)$ is the right one. Btw, $H$ should be $$\begin{bmatrix}1&3\\1&1\end{bmatrix}$$

You also can do all computations by hand, as you have small matrices with small coefficients: let's denote $(c_1, c_2)$ the canonical basis. We have:

$$\begin{cases}e_1=c_1+c_2\\e_2=3c_1+c_2 \end{cases}\enspace\text{whence}\quad\begin{cases}c_1=\frac12(e_2-e_1)\\c_2=\frac12(3e_1-e_2)\end{cases}$$ Now \begin{align*}Te_1&=Tc_1+Tc_2=(c_1+c_2)+(2c_1-c_2)=3 c_1=\tfrac32(e_2-e_1)\\ Te_2&= 3Tc_1+Tc_2=3(c_1+c_2)+(2c_1-c_2)=5c_1+2c_2\\ &=\tfrac52(e_2-e_1)+(3e_1-e_2)=\tfrac12(e_1+3e_2) \end{align*} so that the matrix of T in the basis $(e_1,e_2)$ is: $$T'=\begin{bmatrix}-\frac32&\frac12\\\frac32&\frac32\end{bmatrix}$$


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