Linear Transformation: R^n to R^m - Injective?

[georgeh]

Indicate whether each statement is always true, sometimes true, or always false.
IF T: R^n --> R^m is a linear transformation and m > n, then T is 1-1
Not sure to how prove this..

 

[HallsofIvy]

T(x)= 0 for all x is a linear transformation. T(x)= x is a linear transformation.

 

[matt grime]

The second hint doesn't help I'm afraid, Halls, T is a map from R^m to R^n and m>n.

If T were injective, then its image is a subspace of what dimension?

 

[georgeh]

if it is injective and it goes from T^n to T^m, shouldn't the subpace be in the T^m ?

 

[HallsofIvy]

if it is injective and it goes from T^n to T^m, shouldn't the subpace be in the T^m ?

What subspace are you talking about?

If n< m, then you can think of Tⁿ as a subspace of T^{mp/sup]- think of adding 0's to the end of x.}

 

[benorin]

What does it mean for such a linear transformation to be one-to-one?

Let $$T:R^n \rightarrow R^m$$, and suppose $$x_1,x_2\in R^n$$. Then if $$x_1\neq x_2$$ implies that $$Tx_1\neq Tx_2$$, T is 1-1. Since T is linear, we have the last requirement becoming $$x_1\neq x_2\Rightarrow T(x_1-x_2)\neq 0$$

keep going...