# Linear transformation has unique standard matrx? [T]=-I

Linear transformation has unique standard matrx?
$\left[T\right]=-I$
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Appohhl
Step 1
The statement is false. Any linear transformation satisfies
$T\left(cv\right)=cT\left(v\right)$
for all scalars c, as a consequence of linearity. In particular, this is true for $c=-1$, so $T\left(-v\right)=-T\left(v\right)$ for all v in ${R}^{n}$
Since all linear transformations satisfy this equation, and there is more than one transformation from ${R}^{n}$ to ${R}^{n}$ (unless $n=0$ or R is a trivial ring), therefore it is not true that there is only one linear transformation satsifying this identity. And not a unique standard matrix neither.
The map with standard matrix $\left[T\right]=-I$ satisfies $T\left(v\right)=-v$. This is sometimes called the antipodal map. That defining equation looks a little like the given equation $T\left(-v\right)=-T\left(v\right)$, which may explain the confusion. There is indeed a unique linear transformation given by that map (a somewhat tautological statement). So if they had written that equation, their answer would have been correct. But the equation as written, as I said, many linear maps satisfy.
###### Not exactly what you’re looking for?
euromillionsna
Step 1
The matrix rel the standard basis for ${\mathbb{R}}^{n}$ is unique, given a linear transformation T.
Then i- th column is given by $T\left({e}_{i}\right)$, expressed in terms of the standard basis, where ei is the i- th standard basis vector.
So, there seems to be an error. Indeed, there is only one $T=-I$.
Of course, every linear transformation T satisfies $T\left(-v\right)=-T\left(v\right)$, however.