linear transformation has unique standard matrix? I have a question for linear algebra - lineartransformation.
The question is
question
In this q, I think all linear transformation satisfy this equation, so not unique standard matrix.
But, In the solution, it said that, there is only one linear transformation that [T] = -I
What is true...
 A: Your answer is correct, and the given solution is incorrect. 
The statement is false. Any linear transformation satisfies $T(cv) = cT(v)$ for all scalars $c$, as a consequence of linearity. In particular, this is true for $c=-1,$ so $T(-v) = -T(v)$ 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 $[T]=-I$ satisfies $T(v)=-v$. This is sometimes called the antipodal map. That defining equation looks a little like the given equation $T(-v) = -T(v)$, 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.
A: The matrix rel the standard basis for $\Bbb R^n$ is unique, given a linear transformation $T$.
Then $i$- th column is given by $T(e_i)$, expressed in terms of the standard basis, where $e_i$ 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 $T(-v)=-T(v)$, however.
