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In these lecture notes: http://www.math.rice.edu/~hassett/teaching/221fall05/linalg5.pdf

the formula (last line on first page) for finding a matrix relative to bases $B'$ and $B$ is:

(1) $$ C_{B'}T C_{B}^{-1}$$, where $B = \{v_1,\dots, v_d\}$ and $B' = \{w_1, \dots, w_k\}$ are ordered bases for $V$ and $W$. $T$ is a linear transformation from $V$ to $W$.

However, I was under the impression that the formula writing a linear transformation as a matrix relative to a bases was:

(2) $$ C^{-1}_{B'}TC_B$$

(as an example, see http://www.math.tamu.edu/~yvorobet/MATH304-503/Lect2-12web.pdf ... the last page of the slides).

Which equation is correct? Or how do I know which equation to use?

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  • $\begingroup$ You ask "which equation", but there are no equations in your question. $\endgroup$ – GFauxPas Apr 16 '15 at 16:50
  • $\begingroup$ The references of (1) or (2), is what I mean by "equations" $\endgroup$ – larry Apr 16 '15 at 17:24
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Suppose $x$ is a vector in the standard basis. Write $[x]_B$ to mean the representation of $x$ in the basis $B$. Let $C$ be the matrix whose columns are the basis vectors in $B$.

Then by parsing the definition of matrix multiplication, we have that $$ x = C [x]_B.$$ Equivalently, $$ C^{-1} x = [x]_B. \tag{1}$$

Now suppose $T_B$ is a transformation given in the basis $B$. That is, $T_B$ acts naturally on $[x]_B$. To understand the action of $T_B$ on $x$, we first put $x$ in $B$ coordinates, operate by $T_B$, and then return to standard coordinates. This is $$ C T_B C^{-1} x. \tag{2}$$

But some people find $(1)$ clunky, and would rather define $C$ to be the inverse of the matrix I wrote above. Your two sources use these different conventions. But I hope walking through it once helps you understand what's really going on.

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Both of the two references have the same formulas. They defined it in an "opposite" way. The $U^{-1}$ in the second reference is the same as the $C_{B}$ in your first reference.

$C_{B}$ is defined as the map that maps a vector to its coordinate vector in terms of the basis $B$, whereas $U$ is the transition matrix from $B$ to the standard basis, so it maps the coordinates of $B$ back to the vector.

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Refering to your example (last page of the slides) you have the linear operator $L$ which can be expressed with the matrix $S$. But this matrix is expressed with the standard basis $B=(e_1,e_2)$ so I will write it as such:

$M_{BB}(L) :=S$

Your goal however is to find the matrix of $L$ under another basis $B'=(v_1,v_2)$, so I call it (instead of $N$)

$M_{B'B'}(L) := N$

Now to solve your problem with the forumla, the rule I learned is the following:

$M_{BB}(L) = C_{BB'} * M_{B'B'}(L) * C_{B'B}$

where $C_{BB'} = C_{B'B}^{-1}$ It only depends on what Basis you chose to be $B$/$B'$!

Keep in mind $C_{BB'}$ is the matrix you get by expressing $B'$ through $B$, so in your example it would be $U$. And from here on you just have to solve the whole thing for the matrix you want: $M_{B'B'}(L) = ...$

I hope that this could help you a little!

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