Subspaces, Null Spaces and Column Spaces

2 MIN READ · PRACTISE AT YOUR OWN PACE · FREE

A space must survive linear combinations

A nonempty subset of $\mathbb R^n$ is a subspace if it is closed under addition and scalar multiplication. Equivalently, every linear combination of its vectors stays inside it. It must contain zero. Lines and planes through the origin are familiar examples; translated lines and planes generally are not subspaces.

The null space $N(A)=\{\mathbf x:A\mathbf x=0\}$ is a subspace of the input space. The column space $C(A)=\{A\mathbf x\}$ is a subspace of the output space. Linearity proves closure for both.

Visual guide

VISUAL GUIDEA subspace must contain the origin

Subspace y = x
Not a subspace y = x + 1

The line y = x is closed under vector addition and scalar multiplication. The shifted line y = x + 1 misses the zero vector, so it is not a subspace even though it has the same direction.

Worked example: a homogeneous constraint

The set $x+y+z=0$ is a subspace of $\mathbb R^3$ . Setting $y=s,z=t$ gives $(x,y,z)=s(-1,1,0)+t(-1,0,1)$ . These two independent vectors are a basis of the null space of $A=(1\ 1\ 1)$ .

The set $x+y+z=1$ is not a subspace because it excludes zero. If $\mathbf x_p$ is one solution of $A\mathbf x=\mathbf b$ , all solutions are $\mathbf x_p+N(A)$ . This is an affine translation of a subspace.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

Which subset of ℝ² is a subspace?

The circle x²+y²=1

The line y=2x

The line y=2x+1

Hint 1 · Find a starting point

A subspace must contain zero and be closed under linear combinations.

Hint 2 · Take the next step

A line through the origin preserves addition and scaling.

Show the reasoning

Answer: The line y=2x

y=2x is the span of (1,2). The other sets do not even contain the zero vector.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: find a column-space basis

For $A=\begin{pmatrix}1&2&0\\0&0&1\\1&2&1\end{pmatrix}$ , the second column is twice the first. Columns one and three are independent, so they form a basis for $C(A)$ . The output space is the plane $z=x+y$ .

Row reduction identifies pivot column indices, but the basis vectors must be taken from the original matrix. Row operations generally change the column space as a subset of the output space, even though they preserve which columns are independent.

Practice

Is the set $\{(x,y):x\ge0\}$ a subspace?
Find $N(\operatorname{diag}(1,0))$ .
Is every homogeneous system consistent?

Show worked solutions

No. Multiplying $(1,0)$ by $-1$ leaves the set.
$x=0$ and $y$ is free, so the null space is the span of $(0,1)^T$ .
Yes. The zero vector is always a solution, although nonzero solutions may also exist.

MAKE IT YOURS

Pause before the next idea.

Can you explain how the visualization connects to this lesson’s goal? If a step still feels uncertain, put this lesson on your review list and try the check again another day.

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