Mixing
Generalisation of the following
$begingroup$
Here, the codomain is itself the field of scalars. So, it was easy to write in that form. In Friedberg, we are also asked for an analogous result when the codomain is $mathbb{R}^m$ instead of $mathbb{R}$.
I tried to get a composition of two transformations $T:mathbb{R}^nrightarrowmathbb{R}^m$ and $S:mathbb{R}^mrightarrowmathbb{R}$. So, $S$o$T$ and $S$ is known in the above form. Is this the right approach? What exactly is the analogous statment?
linear-algebra linear-transformations
asked Jan 23 at 15:22
ShanghaikidShanghaikid
548
$endgroup$
add a comment |
$begingroup$
Here, the codomain is itself the field of scalars. So, it was easy to write in that form. In Friedberg, we are also asked for an analogous result when the codomain is $mathbb{R}^m$ instead of $mathbb{R}$.
I tried to get a composition of two transformations $T:mathbb{R}^nrightarrowmathbb{R}^m$ and $S:mathbb{R}^mrightarrowmathbb{R}$. So, $S$o$T$ and $S$ is known in the above form. Is this the right approach? What exactly is the analogous statment?
linear-algebra linear-transformations
asked Jan 23 at 15:22
ShanghaikidShanghaikid
548
$endgroup$
add a comment |
$begingroup$
Here, the codomain is itself the field of scalars. So, it was easy to write in that form. In Friedberg, we are also asked for an analogous result when the codomain is $mathbb{R}^m$ instead of $mathbb{R}$.
I tried to get a composition of two transformations $T:mathbb{R}^nrightarrowmathbb{R}^m$ and $S:mathbb{R}^mrightarrowmathbb{R}$. So, $S$o$T$ and $S$ is known in the above form. Is this the right approach? What exactly is the analogous statment?
linear-algebra linear-transformations
asked Jan 23 at 15:22
ShanghaikidShanghaikid
548
$endgroup$
Here, the codomain is itself the field of scalars. So, it was easy to write in that form. In Friedberg, we are also asked for an analogous result when the codomain is $mathbb{R}^m$ instead of $mathbb{R}$.
I tried to get a composition of two transformations $T:mathbb{R}^nrightarrowmathbb{R}^m$ and $S:mathbb{R}^mrightarrowmathbb{R}$. So, $S$o$T$ and $S$ is known in the above form. Is this the right approach? What exactly is the analogous statment?
linear-algebra linear-transformations
linear-algebra linear-transformations
asked Jan 23 at 15:22
ShanghaikidShanghaikid
548
asked Jan 23 at 15:22
ShanghaikidShanghaikid
548
asked Jan 23 at 15:22
ShanghaikidShanghaikid
548
asked Jan 23 at 15:22
ShanghaikidShanghaikid
548
asked Jan 23 at 15:22
ShanghaikidShanghaikid
548
548
add a comment |
add a comment |
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