Another linear algebra true & false +justification.?

Question

1) If the linear transformation T: R^n --> R^n is one-to-one and T(u-v) = 0, then u = v (u and v being vectors and 0 being the zero vector).

2) If the linear transformation R^n --> R^n is onto T(u-v)=0, then u=v (u and v being vectors and 0 being the zero vector).

3) If det(A) = 0, then T is neither onto nor one-to-one.

4) If T: R^n --> R^m is not one-to-one, then ker(T) contains infinitely many vectors.

5) Shears in R^2 are one-to-one linear operators.

Answer 1

1). is true:
Suppose T(u-v) = 0
Then, T is a linear transformation, so
T(u)-T(v) = 0
T(u) = T(v)
Since T is one to one, that means u = v

2). is true:
Let's use the theorem:
dim(range T) + dim(ker T) = n.
(You can find proof of this in any good linear
algebra text. Also remember that range T is
a subspace of R^n.)
Since T is onto, the range of T is all of R^n.
That means dim(range T) = n.
So dim (ker T) = 0
ker(T) = {0},
and that means T is one- to-one.
So, by part 1). u = v.

3). Is true.
If det(A) = 0, rank(A) = dim(range T) < n.
So by the theorem of 2), dim(ker T) > 0.
So there is at least 1 nonzero vector in ker T.
By parts 1) and 2) T is neither 0
nor 1-1.

4). is true. That follows from parts 1) and 2).
If T is not 1-1, there is at least one nonzero
vector x in ker T.
But if x(<>0) is in ker T, so is cx for every
scalar c. So that means ker T contains infinitely
many vectors.

5). is true.
The matrices of the shears are either
1 k
0 1
or
1 0
k 1
Since the determinants of these matrices
are nonzero, dim(range T) = 2
dim(ker T) = 0 and each T is 1-1.
So all these statements are true.

Answer 2

properly it relies upon on the precisely what you mean. greater often than no longer a matrix is diagonalizable if it has different eigenvectors. If the function polynomial of A has complicated roots, then the roots are the complicated conjugate of another. additionally those are the eigenvalues. different eigenvalues have different eigenvectors. you may now answer the question.