Answer:
Results are (1) True. (2) False. (3) False. (4) True. (5) True. (6) True.
Step-by-step explanation:
Given A is an
matrix. Let
be the corresponding linear transformationover the field F and
be identity vector in V. Now if
.
(1) The kernel of a linear transformation is a vector space : True.
Let
, then,
![T(x+y)=T(x)+T(y)=\theta+\theta=\theta\impies x+y\in Ker( T)](https://tex.z-dn.net/?f=T%28x%2By%29%3DT%28x%29%2BT%28y%29%3D%5Ctheta%2B%5Ctheta%3D%5Ctheta%5Cimpies%20x%2By%5Cin%20Ker%28%20T%29)
hence the kernel is closed under addition.
Let
, then
![T(\lambda x)=\lambda T(x)=\lambda\times \theta=\theta](https://tex.z-dn.net/?f=T%28%5Clambda%20x%29%3D%5Clambda%20T%28x%29%3D%5Clambda%5Ctimes%20%5Ctheta%3D%5Ctheta)
and thus Ket(T) is closed under multiplication
Finally, fore all vectors
,
![T(\theta)=T(\theta+(-\theta))=T(\theta)+T(-\theta)=T(\theta)-T(\theta)=\theta](https://tex.z-dn.net/?f=T%28%5Ctheta%29%3DT%28%5Ctheta%2B%28-%5Ctheta%29%29%3DT%28%5Ctheta%29%2BT%28-%5Ctheta%29%3DT%28%5Ctheta%29-T%28%5Ctheta%29%3D%5Ctheta)
![\implies \theta\in Ker(T)](https://tex.z-dn.net/?f=%5Cimplies%20%5Ctheta%5Cin%20Ker%28T%29)
Thus Ker(T) is a subspace.
(2) If the equation Ax=b is consistent, then Col(A) is
: False
if the equation Ax=b is consistent, then Col(A) must be consistent for all b.
(3) The null space of an mxn matrix is in
: False
The null space that is dimension of solution space of an m x n matrix is always in
.
(4) The column space of A is the range of the mapping
: True.
(5) Col(A) is the set of all vectors that can be written as Ax for some x. : True.
Here Ax will give a linear combination of column of A as a weights of x.
(6) The null space of A is the solution set of the equation Ax=0.
: True