Nullspace of non-zero 4x4 matrix

In summary, we discussed the topic of nullspaces and the rank-nullity theorem in relation to a non-zero 4x4 matrix. It was determined that the nullspace of a non-zero 4x4 matrix cannot contain a set of 4 linearly independent vectors, and if the rank of the matrix is zero, the dimension of the nullspace must also be zero. Additionally, the nullity of a matrix is always at least one, as it includes the zero vector.
  • #1
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Hello everyone,
it's final's time next week :cry: , so I will be posting here more often than usual :biggrin:
Here is one problem I came across when doing review:

The nullspace of non-zero 4x4 matrix cannot contain a set of 4 lin. indep. vectors. (T/F)

The way I was thinking is that if I solve a homogeneous s-m with this matrix, and if the dimension of nullspace is 4, that means that there have to be 4 free variables in the homogeneous s-m, but matrix is just 4x4.
And then there is this rank-nullity theorem that n = rank(A) + nullity(A), so in this case rank(A) = 0, is it ever possible? My guess is not.
Does the same hold for dim of nullspace (nullity): it has to have at least one solution (trivial, where everyting = 0, but that does not mean that the nullity is an empty set!) ?
Is it correct?

Thanks in advance!
 
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  • #2
Have you learned what bases are? If you have 4 linearly independent (nonzero) vectors, they form a basis for the space. The action of the matrix on any vector in your space is determined by the action on the 4 basis-vectors.
What does that mean for Av (A = matrix, v=vector), where v is any vector.

Ofcourse, the rank equation works fine too. What can you conclude from A if its rank is zero?
 
  • #3
Galileo said:
Have you learned what bases are? If you have 4 linearly independent (nonzero) vectors, they form a basis for the space.
Yeah, I know about bases. And this part I understand.

The action of the matrix on any vector in your space is determined by the action on the 4 basis-vectors.
What does that mean for Av (A = matrix, v=vector), where v is any vector.
this is a bit blurry.

Ofcourse, the rank equation works fine too. What can you conclude from A if its rank is zero?
I can't say it's non-existent, but if it's rank = 0, dimens. of rowspace = 0!
Even if all the rows/cols are lin. dep. rank would be at least 1.
But if it's zero I dunno, it cannot be zero-vector, since it's one of the conditions. That's why ask.
n = dimesion of nullspace? but then what :confused:
 
  • #4
But if it's zero I dunno, it cannot be zero-vector, since it's one of the conditions. That's why ask.

Precisely. The rank of a non-zero matrix is never zero (the ONLY vector space of dimension zero is the set containing only the zero vector, and its "basis" is the empty set). Thus if you find that null(A)=4 => rank(A)=0, you've answered the question.
 
  • #5
Equivalently: a 4 by 4 matrix maps vectors from R4 to R4- which is 4 dimensional. The null space of any linear operator (or matrix) is a subspace of the domain space (here R4). If it contains 4 linearly independent vectors the its dimension is at least 4. In fact, the only 4 dimensional subspace of R4 is R4 itself. Saying the null space contains 4 independent vectors is simply saying the null space is R4 itself: the matrix maps every vector into the 0 vector and so is the 0 matrix.
 
  • #6
Thanks!
Then a follow-up question: does nullity has to be at least 1? since it always has a solution as zero-vector? Which I assume to be non-empty, so it's 1.

On one hand it makes sense as I described it above, on the other hand, it doesn't: by rank-nullity thm that would mean that matrix contains at least one lin. dep. row/col: n = rank(A) + nullity(A).
But looking at a matrix A with all cols/rows lin. indep: rank(A) = n, which would mean nullity(A) = 0.

Is it a contradiction? Or am I missing something?

Thanks again.
 
  • #7
The nullity of a matrix is the dimension of the nullspace of the matrix. The nullspace of any matrix contains the zero vector. In the case that the nullspace is the zero vector (and only in this case), the nullity of the matrix is 0, because (as I mentioned in my last post), the dimension of the vector space containing only the zero vector is zero (a basis for the vector space has zero elements - ie. the basis is the empty set).
 

What is the nullspace of a non-zero 4x4 matrix?

The nullspace of a non-zero 4x4 matrix is the set of all vectors that, when multiplied by the matrix, result in a zero vector. In other words, it is the set of all solutions to the equation Ax = 0, where A is the 4x4 matrix and x is a vector.

How do you find the nullspace of a non-zero 4x4 matrix?

To find the nullspace of a non-zero 4x4 matrix, you can use the Gaussian elimination method or row reduction to put the matrix in reduced row echelon form. The columns corresponding to the pivot positions in the row echelon form will form a basis for the nullspace.

What is the dimension of the nullspace of a non-zero 4x4 matrix?

The dimension of the nullspace of a non-zero 4x4 matrix is also known as the nullity of the matrix. It is equal to the number of free variables in the solution to the equation Ax = 0. In a 4x4 matrix, the nullity can range from 0 to 4.

Can the nullspace of a non-zero 4x4 matrix be empty?

Yes, it is possible for the nullspace of a non-zero 4x4 matrix to be empty. This means that the only solution to the equation Ax = 0 is the trivial solution, where all components of the vector x are equal to 0. In this case, the nullity of the matrix would be 0.

How is the nullspace of a non-zero 4x4 matrix related to its rank?

The rank of a matrix is equal to the number of pivot positions in its reduced row echelon form. The nullity of a matrix is equal to the number of non-pivot positions in its reduced row echelon form. Therefore, the sum of the rank and nullity of a non-zero 4x4 matrix will always be equal to 4.

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