How to Show a Matrix’s Column Space Is Not All of R³

To show that a matrix’s column space is not all of $\mathbb{R}^3$, you need to show that its columns do not span three-dimensional space. The standard test is the matrix’s rank: with three rows, the column space is all of $\mathbb{R}^3$ exactly when the rank is 3.

No matrix is specified here, so a particular rank calculation is not possible. The method below applies to any matrix with three rows, followed by a worked example.

What the column space tells you

If $A$ is a matrix with three rows, each column is a vector in $\mathbb{R}^3$. Its column space, written $\operatorname{Col}(A)$, is the set of all linear combinations of those columns. It is therefore a subspace of $\mathbb{R}^3$.

A subspace of $\mathbb{R}^3$ can have dimension 0, 1, 2, or 3. It equals the whole space only if its dimension is 3. The dimension of the column space is the rank of the matrix, so

$$\operatorname{Col}(A)=\mathbb{R}^3 \quad\text{if and only if}\quad \operatorname{rank}(A)=3.$$

Consequently, to prove the column space is not $\mathbb{R}^3$, it is enough to establish that $\operatorname{rank}(A)<3$.

How to prove the rank is less than 3

There are several useful approaches:

1. Count pivots after row reduction. If the row-echelon form has fewer than three pivot positions, then $\operatorname{rank}(A)<3$.
2. Show the columns are dependent when there are at most three columns. If the columns are linearly dependent, their span has dimension smaller than the number of columns, so with at most three columns its dimension is at most 2. If there are more than three columns, dependence alone does not show that the rank is less than 3. In particular, if there are only two columns, their span has dimension at most 2.
3. Find a nonzero orthogonal vector. If you find $y\ne 0$ such that $y^T A=0$, then $y$ is orthogonal to every column. Every vector in the column space is therefore orthogonal to $y$, so the column space cannot contain every vector in $\mathbb{R}^3$.
4. Exhibit a missing vector. Find one vector in $\mathbb{R}^3$ and show that the equation $Ax=b$ has no solution for it. That directly proves $b\notin\operatorname{Col}(A)$.

Row reduction is usually the quickest way to determine the rank. Although row operations can change the actual columns, they preserve rank; use the pivot count to determine the dimension of the original column space.

Worked example

Consider

$$A=\begin{bmatrix} 1&2\\ 0&1\\ 1&3 \end{bmatrix}.$$

Its columns are $c_1=(1,0,1)^T$ and $c_2=(2,1,3)^T$. Row-reducing gives

$$\begin{bmatrix} 1&2\\ 0&1\\ 1&3 \end{bmatrix} \longrightarrow \begin{bmatrix} 1&0\\ 0&1\\ 0&0 \end{bmatrix}.$$

There are two pivots, so $\operatorname{rank}(A)=2$. Thus $\operatorname{Col}(A)$ is a two-dimensional subspace of $\mathbb{R}^3$, not all of $\mathbb{R}^3$.

You can also verify the claim by finding a vector orthogonal to both columns. Take $y=(-1,-1,1)^T$. Then

$$y^T c_1=0,\qquad y^T c_2=0.$$

Every vector in the column space is a linear combination of $c_1$ and $c_2$, so it is also orthogonal to $y$. But $(0,0,1)^T$ is not orthogonal to $y$, since their dot product is 1. Therefore $(0,0,1)^T$ is not in the column space, giving an explicit witness that the column space is not all of $\mathbb{R}^3$.

Common mistakes

• Ignoring the number of rows. A matrix with three rows has columns in $\mathbb{R}^3$. If it has a different number of rows, its column space is a subspace of a different ambient space.
• Using the number of columns alone. Having three or more columns does not guarantee the columns span $\mathbb{R}^3$; some may be dependent. Check the rank.
• Confusing row space and column space. Row operations preserve rank, not the original column space itself. Use the reduced matrix to count pivots, and use the pivot columns of the original matrix if you need a basis for its column space.
• Claiming a proper subspace without a reason. State the rank or provide a missing vector or orthogonal vector to make the proof explicit.

If you have a particular matrix, row-reduce it and count its pivots. Fewer than three pivots is the exact criterion for its column space to be a proper subspace of $\mathbb{R}^3$.