How do you find the dimension of the null space of a matrix A?

  • #1
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Homework Statement



Given A is nxn matrix. An=0 and An-1 [tex]\neq[/tex] 0.
Find dim(null (A))

Homework Equations





The Attempt at a Solution


1.
I split An= A An-1=0.
From this, I conclude that

column space of An-1[tex]\subseteq[/tex]null space of A

because if we think that

An-1=[v1 v2 v3 v4 ...] (vi is column vector of An-1),

we can get
An=A An-1 =[Av1 Av2 Av3 ...] = 0.

Then, dim(null(A))[tex]\geq[/tex]dim(Col(A^{n-1})).

2.dim(null(An))=n since An=0.

3.dim(null(An-1)) [tex]\leq[/tex] dim(null(An))
Since if we let v be in null space of An-1, we can get Anv=0.
Then we conclude that v is in null space of An
 
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  • #2
could you show that null(A^(n-1)) is a subset of null(A^n)? and consider a form of induction back to A?
 
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  • #3
Just to be sure aren't we going to say

[tex]\begin{pmatrix}a_{11} & a_{12} & \ldots & a_{1n} \\ \vdots &\vdots & \vdots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix}^n = \begin{pmatrix} 0 \\ \vdots \\ 0_n \end{pmatrix}[/tex]

When the vectors which spans the corresponding Vectorspace V then this has the dimension n if and only n > 1? Hence the second condition?

Doesn't this result in the that as long as n > 1 that the dimension of the nullspace meaning the set of solutions for Ax = 0 will have dimension n?? Because V contains and infinite number of vectors??
 
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  • #4
i was more thinking clearly thinking the nullspace of A^n is all of R^n..

Clearly the nullspace of A^n-1, must be less than that and so is a proper subset of null(A^n)

If you could show that for similar i, i-1, then you would inductively get to the point where dim(null(A))=1, though i haven't worked through the details and there may be another way to get there
 
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