# Null Space and Column Space query on their separation

I am struggling to relearn some concepts, please help me understand where my logic is incorrect regarding that the Column space and Null Space should only have the zero vector in common.

If we look at Matrix A:

$$A= \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}$$

And consider the phrase "The column space of a matrix consists of all linear combinations of its column vectors".

Since the columns are multiples of each other, the column space can be represented as the span of the first column. $$C(A) = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$$

If I skip the working, the null space can be calculated to be: $$N(A) ={ x2 \begin{bmatrix} -2 \\ 1 \end{bmatrix} }$$

I understand that the null space is all the multiples of the above vector. However, returning to the statement from before: "The column space of a matrix consists of all linear combinations of its column vectors",

Isn't this an appropriate linear combination of the column vectors, i.e why isnt this in the column space?

$$-2 \begin{bmatrix} 1 \\ 2 \end{bmatrix} + 1 \begin{bmatrix} 2 \\ 4 \end{bmatrix} = 0$$

• You are confusing between the coefficients $c_1,c_2$ and the linear combination they produce $c_1\mathbf{v}_1+c_2\mathbf{v}_2$. If $\mathbf{v}_1$ and $\mathbf{v}_2$ are the columns of the matrix $A$, their linear combination $\mathbf{u}=c_1\mathbf{v}_1+c_2\mathbf{v}_2$ will be in the column space of $A$ (not $(c_1,c_2)$). In your case the linear combination that you are producing is the zero vector$(=\mathbf{u})$ and that will be in the column space of $A$. Commented Jan 15 at 20:19
• Thank you Anurag, that helps make sense! Commented Jan 31 at 15:04

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.