Transpose of a linear transformation

**Sheld** · October 22nd 2011, 11:26 AM

If $F$ is a field and $f$ is the linear functional on $F^2$ s.t. $f(x_1,x_2) = ax_1+bx_2$
For the following linear operators $T$ , let $g = T^{t}f$ and find $g(x_1,x_2)$

lets say $T(x_1,x_2) = (x_1,0)$

So do I find $f$ first, get $f(x_1,x_2) = ax_1+bx_2$

do i now treat " $ax_1$ " and " $bx_2$ " as a vector with two rows and multiply by the transpose of the matrix representation of the linear transformation?

in this case, i get $ax_1$

I read this part of the book, it really isn't clear on what to do. Thank you for your help.

**Deveno** · October 22nd 2011, 03:00 PM

the definition of g doesn't make sense unless $g = f \circ T^t$ , because the domain of $T^t$ is $F^2$ , and the co-domain of f is $F$ .

we're missing some crucial information here. please state the entire problem.

**Sheld** · October 22nd 2011, 04:12 PM

Sorry. To clarify, it is composition. $g = f \; o \; T^{t}$

**Deveno** · October 22nd 2011, 08:36 PM

but you wrote $g = T^tf$ which isn't the same thing. you can't compose two functions unless the co-domain of the first one you do, is the domain of the 2nd one you do.

to make matters even more confusing, some authors means g(f(x)) by gf, and some mean, f(g(x)).

IF $g = f \circ T^t$ , there isn't any problem:

$g(x_1,x_2) = f(T^t(x_1,x_2)) =$

$=f(\begin{bmatrix}t_{11}&t_{21}\\t_{12}&t_{22}\end {bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix})$

$=f(t_{11}x_1+t_{21}x_2,t_{12}x_1+t_{22}x_2)$

$=a(t_{11}x_1+t_{21}x_2) + b(t_{12}x_1+t_{22}x_2)$

where $T = \begin{bmatrix}t_{11}&t_{12}\\t_{21}&t_{22} \end{bmatrix}$ is the matrix for T in the standard basis.

if we use the example you gave, $T(x_1,x_2) = (x_1,0)$ , T has the matrix:

$\begin{bmatrix}1&0\\0&0\end{bmatrix}$ and its transpose is: $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ , which is the same as T, in which case:

$g(x_1,x_2) = f(x_1,0) = ax_1$ .

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