Find Basis for Ker (T) with S = {1, 0, 1, 0}

In summary, the problem asked to find a basis for the kernel of a linear transformation T that includes the vectors <1, 0, 1, 0> and <0, 1, 0, 2>. However, it was discovered that these two vectors are not in the kernel of T and therefore cannot be included in any basis for the kernel.
  • #1
Luke1294
57
0
Find a basis for Ker T that contains S = [tex]\begin{pmatrix}
1\\
0\\
1\\
0\\
\end{pmatrix}[/tex], [tex]\begin{pmatrix}
0\\
1\\
0\\
2\\
\end{pmatrix}[/tex] where [tex]T : R^4 -> R^4[/tex] is defined by

[tex]T\begin{pmatrix}
1\\
b\\
c\\
d\\
\end{pmatrix} = \begin{pmatrix}
a - b - c\\
a - 2b + c\\
0\\
0\\
\end{pmatrix}[/tex].

Well, I have found a basis 'B' for Ker (T) to be B ={[tex]\begin{pmatrix}
3\\
2\\
1\\
0\\
\end{pmatrix}[/tex], [tex]\begin{pmatrix}
0\\
0\\
0\\
1\\
\end{pmatrix}[/tex]}.

I noticed that the two sets of vectors S and B are linearly independent of one another. Does this mean that there is no basis for Ker(T) that contains S, as a basis must be a minimal spanning set? Or have I gone astray somewhere?
 
Physics news on Phys.org
  • #2
Luke1294 said:
Find a basis for Ker T that contains S = [tex]\begin{pmatrix}
1\\
0\\
1\\
0\\
\end{pmatrix}[/tex], [tex]\begin{pmatrix}
0\\
1\\
0\\
2\\
\end{pmatrix}[/tex] where [tex]T : R^4 -> R^4[/tex] is defined by

[tex]T\begin{pmatrix}
1\\
b\\
c\\
d\\
\end{pmatrix} = \begin{pmatrix}
a - b - c\\
a - 2b + c\\
0\\
0\\
\end{pmatrix}[/tex].

Well, I have found a basis 'B' for Ker (T) to be B ={[tex]\begin{pmatrix}
3\\
2\\
1\\
0\\
\end{pmatrix}[/tex], [tex]\begin{pmatrix}
0\\
0\\
0\\
1\\
\end{pmatrix}[/tex]}.

I noticed that the two sets of vectors S and B are linearly independent of one another. Does this mean that there is no basis for Ker(T) that contains S, as a basis must be a minimal spanning set? Or have I gone astray somewhere?
Perhaps you should reread the problem. "Find a basis for the kernel of T that includes <1, 0, 1, 0> and <0, 1, 0, 2>" makes no sense as it is easy to see that those two vectors are NOT in the kernel of T and so cannot be in any basis for that kernel.
 
  • #3
Alright, that is what I had thought but just wanted to verify it with someone. Thank you very much!
 
  • #4
For future reference: A problem that says "Find a basis for vector space V that include vectors v1 and v2" , assuming v1 and v2 are in V and are independent, means that you are to find a basis that includes those two vectors and possibly more.
 
Last edited by a moderator:

1. What is the definition of "Ker (T)"?

"Ker (T)" refers to the kernel of a linear transformation T, which is the set of all vectors in the domain of T that map to the zero vector in the range of T.

2. How is the basis for Ker (T) found?

The basis for Ker (T) is found by solving the homogeneous equation T(x) = 0, where x represents the vector in the domain of T. The solutions to this equation will form the basis for Ker (T).

3. What does "S = {1, 0, 1, 0}" mean in this context?

"S = {1, 0, 1, 0}" represents a set of vectors in the domain of T. In this case, it means that the vectors (1,0) and (1,0) are in the domain of T.

4. How many vectors are in the basis for Ker (T)?

The number of vectors in the basis for Ker (T) is equal to the nullity of T, which is the dimension of the kernel. In this case, the basis for Ker (T) will have 2 vectors since the nullity of T is 2.

5. Why is finding the basis for Ker (T) important?

Finding the basis for Ker (T) is important because it allows us to understand the behavior of T and its relationship with other linear transformations. It also helps in solving systems of linear equations and in finding solutions to differential equations.

Similar threads

  • Linear and Abstract Algebra
Replies
34
Views
2K
  • Linear and Abstract Algebra
Replies
8
Views
869
  • Linear and Abstract Algebra
Replies
10
Views
977
  • Linear and Abstract Algebra
2
Replies
52
Views
2K
  • Linear and Abstract Algebra
Replies
14
Views
2K
Replies
31
Views
2K
  • Linear and Abstract Algebra
Replies
15
Views
1K
  • Linear and Abstract Algebra
Replies
8
Views
1K
  • Linear and Abstract Algebra
Replies
6
Views
1K
  • Linear and Abstract Algebra
Replies
2
Views
893
Back
Top