Dimensionality, Rangespace & Nullspace Problem

[e(ho0n3]

[SOLVED] Dimensionality, Rangespace & Nullspace Problem

Homework Statement
Prove (where A is an n x n matrix and so defines a transformation of any n-dimensional space V with respect to B, B where B is a basis of V) that $\dim(R(A) \cap N(A)) = \dim R(A) - \dim R(A^2)$

The attempt at a solution
If I determine the basis of $R(A) \cap N(A)$, I can determine its dimensionality and then compare it with $\dim R(A) - \dim R(A^2)$.

I've been unsuccessful at finding a basis. Also, given that $\dim R(A) = m$, is there a way to determine what $\dim R(A^2)$ is?

 

[pizzasky]

Let $$\{Av_1, \ Av_2, \ ..., \ Av_k\}$$ be a basis for $$R(A)\cap N(A)$$ and $$\{A^2u_1, \ A^2u_2, \ ..., \ A^2u_m\}$$ be a basis for $$R(A^2)$$, with $$v_1, \ v_2, \ ..., \ v_k, \ u_1, \ u_2, \ ..., \ u_m \in V$$.

We claim that X = $$\{Av_1, \ Av_2, \ ..., \ Av_k, \ Au_1, \ Au_2, \ ..., \ Au_m\}$$ is a basis for $$R(A)$$.

Proof:
1) X spans $$R(A)$$.

Take any $$w \in R(A)$$. Since $$Aw \in R(A^2)$$, so $$Aw = b_1 A^2u_1 \ + b_2 A^2u_2 \ + ... \ + b_m A^2u_m$$, for some scalars $$b_1, \ b_2, \ ..., \ b_m$$.

Thus, $$A(w \ - b_1 Au_1 \ - b_2 Au_2 \ - ... \ - b_m Au_m) = \mathbf{0}$$ and so $$w \ - \ b_1 Au_1 \ - \ b_2 Au_2 \ - \ ... \ - \ b_m Au_m \in R(A) \cap N(A)$$ (Why?)
Complete the proof yourself :)

Hence, every element in $$R(A)$$ can be expressed as a linear combination of $$Av_1, \ Av_2, \ ..., \ Av_k, \ Au_1, \ Au_2, \ ..., \ Au_m$$ and so X spans $$R(A)$$.

2) X is a linearly independent set.
We solve the homogeneous equation $$c_1 Av_1 \ + \ c_2 Av_2 \ + \ ... \ + \ c_k Av_k \ + \ c_{k+1} Au_1 \ + \ c_{k+2} Au_2 \ + \ ... \ + \ c_{k+m} Au_m = \mathbf{0}$$, where $$c_1, \ c_2, \ ..., \ c_{k+m}$$ are scalars.

Now, $$A(c_1 Av_1 \ + \ c_2 Av_2 \ + \ ... \ + \ c_k Av_k \ + \ c_{k+1} Au_1 \ + \ c_{k+2} Au_2 \ + \ ... \ + \ c_{k+m} Au_m) = \mathbf{0}$$, and thus...
Complete the proof yourself: I suggest first showing that $$c_{k+1} \ = c_{k+2} \ = ... \ = c_{k+m} \ = 0$$

Since the homogeneous equation has only the trivial solution, X is a linearly independent set.

This proves our claim and the result follows.

 

[e(ho0n3]

Neat. I wish I had your intuition.

Thus, $$A(w \ - b_1 Au_1 \ - b_2 Au_2 \ - ... \ - b_m Au_m) = \mathbf{0}$$ and so $$w \ - \ b_1 Au_1 \ - \ b_2 Au_2 \ - \ ... \ - \ b_m Au_m \in R(A) \cap N(A)$$ (Why?)
Complete the proof yourself :)

w is in R(A) and so is $$- b_1 Au_1 - \cdots - b_m Au_m$$ because it's a linear combination of members of X. Thus $$w - b_1 Au_1 - \cdots - b_m Au_m$$ is in R(A). Since $$A(w - b_1 Au_1 - \cdots - b_m Au_m) = \mathbf{0}$$, $$w - b_1 Au_1 - \cdots - b_m Au_m$$ is also in N(A).

Now, $$A(c_1 Av_1 \ + \ c_2 Av_2 \ + \ ... \ + \ c_k Av_k \ + \ c_{k+1} Au_1 \ + \ c_{k+2} Au_2 \ + \ ... \ + \ c_{k+m} Au_m) = \mathbf{0}$$, and thus...
Complete the proof yourself: I suggest first showing that $$c_{k+1} \ = c_{k+2} \ = ... \ = c_{k+m} \ = 0$$

Let $$\mathfrak{v} = c_1 Av_1 + \cdots + c_k Av_k$$ and let $$\mathfrak{u} = c_{k+1} Au_1 + \cdots + c_{k+m} Au_m$$. $$A(\mathfrak{v} + \mathfrak{u}) = A\mathfrak{v} + A\mathfrak{u} = A\mathfrak{u} = \mathbf{0}$$.

The last equality above plus the fact that $$\{A^2u_1, \ldots, A^2u_m\}$$ is linearly independent means that $$c_{k+1} = \cdots = c_{k+m} = 0$$. Thus $$\mathfrak{v} + \mathfrak{u} = \mathbf{0}$$ reduces to just $$\mathfrak{v} = \mathbf{0}$$ and since $$\{Av_1, \ldots, Av_k\}$$ is a linearly independent set, $$c_1 = \cdots = c_k = 0$$ as well. Ergo, X is linearly independent.