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

Question

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

Appohhl · Accepted Answer

Step 1
The statement is false. Any linear transformation satisfies

T (c v) = c T (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.

euromillionsna · Accepted Answer

Step 1
The matrix rel the standard basis for

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 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 (- v) = - T (v)

, however.

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

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