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

[descendents]

Homework Statement

Given A is nxn matrix. Aⁿ=0 and A^n-1 $$\neq$$ 0.
Find dim(null (A))

Homework Equations

The Attempt at a Solution

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

column space of A^n-1$$\subseteq$$null space of A

because if we think that

A^n-1=[v1 v2 v3 v4 ...] (v_i is column vector of A^n-1),

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

Then, dim(null(A))$$\geq$$dim(Col(A^{n-1})).

2.dim(null(Aⁿ))=n since Aⁿ=0.

3.dim(null(A^n-1)) $$\leq$$ dim(null(Aⁿ))
Since if we let v be in null space of A^n-1, we can get Aⁿv=0.
Then we conclude that v is in null space of Aⁿ

 

[lanedance]

could you show that null(A^(n-1)) is a subset of null(A^n)? and consider a form of induction back to A?

 

[matematikeren]

Just to be sure aren't we going to say

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

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??

 

[lanedance]

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