Null Space and Column Space query on their separation

Regarding the statement

$ \text{The Column space and Null Space should only have the zero vector in common.} $

Let,

$ A \in \mathbb{R}^{m \times n} $
$ R(A): $ Range Space/Column Space for matrix $ A $
$ N(A): $ Null Space for matrix $ A $

Now, suppose $ \mathbf{w} \in R(A) $ and $ N(A) $,

$$ A\mathbf{w} = \mathbf{0} \ \because \ \mathbf{w} \in N(A) $$ $$ \mathbf{w} = A\mathbf{u} \ \because \ \mathbf{w} \in R(A), \text{ for some } \mathbf{u} \in \mathbb{R}^n $$

Using these relations: $$ A(A\mathbf{u}) = \mathbf{0} $$ $$ A^2\mathbf{u} = \mathbf{0} \space - (1) $$

So, we have to find the solution of the homogeneous equation (1) to find $ \mathbf{w} ( = A\mathbf{u} )$.

If $ \text{rank}(A^2) = n $, then there will be one solution (trivial solution where $ \mathbf{u} = \mathbf{0} $ and consequently $ \mathbf{w} = \mathbf{0} $).

If $ \text{rank}(A^2) < n $, then there will be an infinite number of solutions for $u$. If for these $u$, $Au = 0$ then also above statement will be true.

Therefore, the given statement is only valid when $ \text{rank}(A^2) = n $. or $ \text{rank}(A^2) < n $ but $ Au=0$ for all $u$ s.t. $A^2u=0$

For the specific matrix $ A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} $: $ A^2 = \begin{bmatrix} 5 & 10 \\ 10 & 20 \end{bmatrix} $

The rank of $ A^2 $ is not 2, as the second row is a multiple of the first row. Thus, $ \text{rank}(A^2) = 1 $.

Moreover, for this matrix, $ \mathbf{u} $ can be expressed as $ \mathbf{u} = \alpha \begin{bmatrix} 2 \\ -1 \end{bmatrix} $, and the corresponding $ \mathbf{w} = A\mathbf{u} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $. Thus, only the zero vector is the common vector between the Column Space and Null Space.