Linear transformation has unique standard matrix

linear algebralinear-transformations

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…

Best Answer

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.

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