Two Matrices with Contrasting Column Spaces
The columns of a matrix determine its column space: it consists of all vectors that can be formed by taking linear combinations of those columns. Two basic matrices make this idea especially concrete. Every column of a zero matrix is the zero vector, so no combination of its columns can produce a nonzero vector. By contrast, the columns of an identity matrix are the standard coordinate vectors, and their combinations can produce every vector in the corresponding coordinate space.
We will work over \(\mathbb{R}\), and take the matrix sizes to be positive integers. For a matrix \(\mathbf{A}\in\mathbb{R}^{m\times n}\) with columns \(\mathbf{a}_1,\dots,\mathbf{a}_n\), its column space is the set of all linear combinations of those columns:
Equivalently, a vector \(\mathbf{b}\in\mathbb{R}^m\) belongs to \(\operatorname{Col}\mathbf{A}\) exactly when there is a coefficient vector \(\mathbf{x}\in\mathbb{R}^n\) such that \(\mathbf{A}\mathbf{x}=\mathbf{b}\). The dimensions matter: the columns of an \(m\times n\) matrix are vectors in \(\mathbb{R}^m\), while the coefficient vector has \(n\) entries.
The Column Space of a Zero Matrix
Proof. Every one of the \(n\) columns of \(\mathbf{O}_{m\times n}\) equals \(\mathbf{0}\in\mathbb{R}^m\). For any real coefficients \(c_1,\dots,c_n\), the corresponding combination is \(c_1\mathbf{0}+\cdots+c_n\mathbf{0}=\mathbf{0}\). Thus every vector in the column space is \(\mathbf{0}\). Conversely, choosing all coefficients to be zero produces \(\mathbf{0}\), so \(\mathbf{0}\) is in the column space. Hence the column space is exactly \(\{\mathbf{0}\}\). A vector \(\mathbf{b}\) belongs to the set \(\{\mathbf{0}\}\) exactly when \(\mathbf{b}=\mathbf{0}\), proving the membership statement as well. \(\square\)
The shape of a zero matrix does not change this conclusion. An \(m\times n\) zero matrix has \(n\) columns, each in \(\mathbb{R}^m\); whether there are two columns or many more, every linear combination remains the zero vector. The number of columns affects how many coefficients are available, but it cannot make a nonzero output when all the columns are zero.
Worked Example: A Rectangular Zero Matrix
Find the column space of \(\mathbf{O}_{2\times3}\), and decide whether \(\mathbf{b}=(0,0)^{\top}\) and \(\mathbf{d}=(1,-2)^{\top}\) belong to it.
Write the matrix by its three columns:
For any coefficient vector \(\mathbf{x}=(x_1,x_2,x_3)^{\top}\), multiplication gives
Therefore \(\operatorname{Col}\mathbf{O}_{2\times3}=\{(0,0)^{\top}\}\). The target \(\mathbf{b}\) belongs: for example, using \(\mathbf{x}=(0,0,0)^{\top}\) gives \(\mathbf{O}_{2\times3}\mathbf{x}=(0,0)^{\top}=\mathbf{b}\). The target \(\mathbf{d}\) does not belong, because every product \(\mathbf{O}_{2\times3}\mathbf{x}\) equals \((0,0)^{\top}\), not \((1,-2)^{\top}\).
Check: the claimed product for \(\mathbf{b}\) is \(\begin{bmatrix}0&0&0\\0&0&0\end{bmatrix}\begin{bmatrix}0\\0\\0\end{bmatrix} =\begin{bmatrix}0\\0\end{bmatrix}\). There can be no coefficient vector producing \(\mathbf{d}\), since the product formula gives zero in both entries for every choice of coefficients.
The Column Space of an Identity Matrix
Proof. The columns of \(\mathbf{I}_n\) are \(\mathbf{e}_1,\dots,\mathbf{e}_n\). First, every vector in \(\operatorname{Col}\mathbf{I}_n\) is a linear combination of these columns and therefore belongs to \(\mathbb{R}^n\). This proves \(\operatorname{Col}\mathbf{I}_n\subseteq\mathbb{R}^n\). For the reverse inclusion, take an arbitrary vector \(\mathbf{b}=(b_1,\dots,b_n)^{\top}\in\mathbb{R}^n\). By the coordinate definition of the standard basis, \(\mathbf{b}=b_1\mathbf{e}_1+\cdots+b_n\mathbf{e}_n\). This is a linear combination of the columns of \(\mathbf{I}_n\), so \(\mathbf{b}\in\operatorname{Col}\mathbf{I}_n\). Thus \(\mathbb{R}^n\subseteq\operatorname{Col}\mathbf{I}_n\) as well, and the two sets are equal. \(\square\)
The same conclusion can be checked through multiplication. For \(\mathbf{x}=(x_1,\dots,x_n)^{\top}\), the \(i\)th entry of \(\mathbf{I}_n\mathbf{x}\) is \(x_i\), since row \(i\) of \(\mathbf{I}_n\) has a \(1\) in position \(i\) and zeros elsewhere. Hence \(\mathbf{I}_n\mathbf{x}=\mathbf{x}\): any desired target can be obtained by choosing that target itself as the coefficient vector.
Worked Example: Membership in the Column Space of \(\mathbf{I}_3\)
Express \(\mathbf{b}=(2,-1,4)^{\top}\) as a combination of the columns of \(\mathbf{I}_3\), and verify the resulting matrix equation.
The identity matrix and its columns are
Using the coordinates of \(\mathbf{b}\) as its coefficients gives
Thus \(\mathbf{b}\in\operatorname{Col}\mathbf{I}_3\). In matrix form, choose \(\mathbf{x}=(2,-1,4)^{\top}\):
This computation also illustrates the whole-space claim: the coordinates \(2,-1,4\) were unrestricted real numbers, and the same construction works for any target in \(\mathbb{R}^3\).
Reading Membership and Spanning Claims Correctly
For a matrix with \(m\) rows, every column and every vector in its column space lies in \(\mathbb{R}^m\). Thus \(\mathbf{O}_{3\times2}\) has two columns in \(\mathbb{R}^3\), and \(\mathbf{I}_3\) has three columns in \(\mathbb{R}^3\). Their outputs are sharply different: the zero matrix produces only \(\mathbf{0}\), whereas the identity produces every vector in \(\mathbb{R}^3\). The target space is determined by the number of rows, not by the number of columns.
Worked Example: Comparing a Zero Matrix and an Identity Matrix
Compare \(\operatorname{Col}\mathbf{O}_{3\times2}\) with \(\operatorname{Col}\mathbf{I}_3\). For each matrix, decide whether \(\mathbf{u}=(0,0,0)^{\top}\) and \(\mathbf{v}=(3,0,-2)^{\top}\) belong to its column space, and decide whether its columns span \(\mathbb{R}^3\).
For the rectangular zero matrix, its columns are both \(\mathbf{0}\in\mathbb{R}^3\):
So \(\operatorname{Col}\mathbf{O}_{3\times2}=\{\mathbf{0}\}\). The target \(\mathbf{u}\) belongs, using \((x_1,x_2)=(0,0)\), but \(\mathbf{v}\) does not: the product is always \((0,0,0)^{\top}\). Since \((3,0,-2)^{\top}\in\mathbb{R}^3\) is not in this column space, the columns do not span \(\mathbb{R}^3\).
For the identity matrix, choose the target itself as the coefficient vector. In particular,
Also, \(\mathbf{I}_3(0,0,0)^{\top}=(0,0,0)^{\top}\). Both targets therefore belong to \(\operatorname{Col}\mathbf{I}_3\), and the identity theorem shows that every vector in \(\mathbb{R}^3\) belongs. Its columns span \(\mathbb{R}^3\).
Check: the two conclusions agree with the general formulas: \(\operatorname{Col}\mathbf{O}_{3\times2}=\{\mathbf{0}\}\) and \(\operatorname{Col}\mathbf{I}_3=\mathbb{R}^3\). The first contains only \(\mathbf{u}\) among the two targets; the second contains both.
Common Pitfall: Counting Columns Instead of Inspecting Them
A tempting but incorrect claim is: “\(\mathbf{O}_{2\times3}\) has three columns, so its columns span \(\mathbb{R}^2\).” The number three counts the available coefficients; it does not show that the combinations can produce every vector in the target space. In fact, if \(\mathbf{x}=(x_1,x_2,x_3)^{\top}\), then
For example, \((1,0)^{\top}\in\mathbb{R}^2\) cannot be obtained, so the columns do not span \(\mathbb{R}^2\). The correction is to use the actual columns and their combinations: here all three columns are zero, and their span is just \(\{\mathbf{0}\}\).
A related dimension check prevents another common error. The product \(\mathbf{O}_{2\times3}\mathbf{x}\) takes a three-entry coefficient vector to a two-entry output. Its column space is a subset of \(\mathbb{R}^2\), not \(\mathbb{R}^3\). Likewise, \(\mathbf{I}_3\) acts on three-entry vectors and has its column space in \(\mathbb{R}^3\). The columns themselves make both the output dimension and the available combinations visible.
| Matrix | Columns | Column space | Do the columns span the target space? |
|---|---|---|---|
| \(\mathbf{O}_{m\times n}\) | \(n\) copies of \(\mathbf{0}\in\mathbb{R}^m\) | \(\{\mathbf{0}\}\) | Only when the target space is \(\{\mathbf{0}\}\); for positive \(m\), they do not span \(\mathbb{R}^m\). |
| \(\mathbf{I}_n\) | \(\mathbf{e}_1,\dots,\mathbf{e}_n\) | \(\mathbb{R}^n\) | Yes. |
These two cases are useful reference points for span-membership questions. A target tested against a zero matrix must be exactly zero; a target in the matching coordinate space always belongs to the column space of an identity matrix. In either case, check the matrix dimensions first, then use the columns to justify the claim.
Check Your Understanding
Use the columns and matrix-vector products to justify each answer.
- What is \(\operatorname{Col}\mathbf{O}_{3\times4}\)? Does it contain \((0,0,0)^{\top}\) and \((1,1,0)^{\top}\)?
- Write \((5,-2)^{\top}\) as a combination of the columns of \(\mathbf{I}_2\), and check the product.
- Do the columns of \(\mathbf{O}_{2\times5}\) span \(\mathbb{R}^2\)? Give a specific vector that demonstrates your conclusion.
- Explain why every vector in \(\mathbb{R}^4\) belongs to \(\operatorname{Col}\mathbf{I}_4\).
- For a matrix with \(m\) rows and \(n\) columns, in which coordinate space do its columns lie, and how many entries does its coefficient vector have?
