Linear Algebra , Linear Transformation

2. ker(T): the kernel of T
If T:V→W is a linear transformation, then the set of vectors in V that T maps into 0.
R (T): the range of T
The set of all vectors in W that are images under T of at least one vector in V.

3. If TA : 𝑅𝑛
→ 𝑅𝑚
is a multiplication by the m×n matrix A, from the discussion preceding
the definition above,
is the nullspace of A
Is the column space of A

4. Let T : V → W be the zero transformation.
Since T maps every vector in V into 0, it follows that ker (T) = V
Moreover, since 0 is the only image under T of vectors in V, we
have R (T) = 0

5. Let T : 𝑅3
→ 𝑅3
be the orthogonal projection on the xy-plane.
The kernel of T is the set of points that T maps into 0 = (0,0,0);
these are the points on the z-axis.

6. Since T maps every points in 𝑅3 into the xy-plane, the range of T
must be some subset of this plane. But every point (X0 , Y0 , 0) in the xy-
plane is the image under T of some point; in fact, it is the image o all points
on the vertical line that passes through (X0 , Y0 , 0) . Thus, R (T) is the entire
xy-plane.

7. Let T : 𝑅2
→ 𝑅2
be the linear operator that rotates each vector in the xy-
plane through the angle 𝜃. Since every vector in the xy-plane can be
obtained by rotating through some vector through angle 𝜃, we have R (T) =
𝑅2
. Moreover, the only vector that rotates into 0 is 0, so ker(T) = 0

8. If T : V → W is linear transformation, then:
Of T is a subspace of V
The
The Of T is a subspace of W

9. If A is any matrix, then the row space and column space of A
have the same dimension

10. The of the matrix A is the dimension of the row space
of A, and is denoted by R(A).
The of a matrix A is the dimension of the null space
of A, and is denoted by N(A). Let A be an m x n matrix. The
null space is the set of solutions to the homogenous system
Ax=0.

11. A =
1 2 4
2 4 8
Step 1. Transform the matrix into REF
A =
1 2 4
2 4 8
R → R2 – 2R1
A =
1 2 4
0 0 0
2 – 2(1) = 0
4 – 2(2) = 0
8 – 2(4) = 0
Rank = r(A) = 1 because only 1 row has non-zero elements
nullity = n(A)
n(A) = n – r(A)
n(A) = 3 – 1
n(A) = 2
r(A) = 1
n(A) = 2

12. A =
1 4 5
0 1 3
0 0 0
r(A) = 2 (because 2 rows have
non-zero elements)
n(A) = n – r(A)
= 3 – 2
= 1
r(A) = 2
n(A) = 1

13. Find the Rank and Nullity
1.) A =
1 4 6
2 8 12
2.) B =
1 4 6
0 1 7
0 0 0
Solutions
1.) A =
1 4 6
2 8 12
R2→2R1
A =
1 4 6
0 0 0
2 – 2(1) = 0
8 – 2(4) = 0
12 -2(6) = 0
r(A) = 1
n(A) = 2
2.) B =
1 4 6
0 1 7
0 0 0
r(B) = 2
n(B) = 1