Linear Algebra Four fundamental subspaces small proof.

[wjv4]

Homework Statement

Given A$\in$ $M$_nxn and A = A², show that C(A) +N(A) = ℝⁿ.

note: C(A) means the column space of A.
N(A) means the null space of A

Homework Equations

These equations were proved in earlier parts of the problem...

C(A) = {$\vec{x}$$\in$ ℝⁿ such that $\vec{x}$ = $\vec{u}$-A$\vec{u}$ for some $\vec{u}$ $\in$ℝⁿ}

N(A) = {$\vec{x}$$\in$ℝⁿ such that $\vec{x}$ = A$\vec{x}$}

The Attempt at a Solution

I feel that my attempt is logical and it works, but I'm not sure if the last step I took works, but if anyone could prove me wrong, confirm that I am right, or offer an alternative, that would be cool! My solⁿ is attached as a picture.

 

[mighty2000]

Your attempt is a good start, but there are a few errors. First, when you wrote "Let x be any vector in ℝn", you should specify that x is in the column space of A, since you are showing that C(A) + N(A) = ℝn. Also, when you wrote "x = u - Au for some u in ℝn", this is not necessarily true. x is in the column space of A, so it can be written as a linear combination of the columns of A, but u may not be in the column space of A.

Instead, let x be any vector in C(A). Then, by definition, x = u - Au for some u in ℝn. Now, since A = A^2, we have A(x) = A(u - Au) = Au - A^2u = Au - Au = 0. This means that x is also in N(A). Therefore, C(A) is a subset of N(A), and we can conclude that C(A) + N(A) = N(A).

To show that N(A) = ℝn, we can use the fact that A is an nxn matrix. This means that the null space of A has dimension n, since n is the number of columns in A. Since the dimension of ℝn is also n, we can conclude that N(A) = ℝn.

Therefore, we have shown that C(A) + N(A) = ℝn, as desired.