Columns and the Transposed Equation
For a matrix \(\mathbf{A}\), its columns are vectors in the space where \(\mathbf{A}\) sends its inputs. The null space of \(\mathbf{A}^{\top}\) consists of vectors from that same space. The connection between these objects is encoded directly in the transpose: multiplying a vector by \(\mathbf{A}^{\top}\) records its dot products with the columns of \(\mathbf{A}\). This gives a precise way to recognize vectors orthogonal to every column, and therefore to every linear combination of the columns.
Throughout, let \(\mathbf{A}\in\mathbb{R}^{m\times n}\), with columns \(\mathbf{a}_1,\dots,\mathbf{a}_n\in\mathbb{R}^m\). All vectors are columns, and the inner product on \(\mathbb{R}^m\) is the dot product \(\mathbf{u}\cdot\mathbf{v}=\mathbf{u}^{\top}\mathbf{v}\). The column space \(\operatorname{Col}\mathbf{A}\) is the set of all linear combinations of the columns. The null space \(\operatorname{Nul}\mathbf{A}^{\top}\) is the set of vectors \(\mathbf{y}\in\mathbb{R}^m\) satisfying \(\mathbf{A}^{\top}\mathbf{y}=\mathbf{0}\). Its vectors are sometimes called left-null vectors of \(\mathbf{A}\).
The dimensions matter. Each column \(\mathbf{a}_j\) is in \(\mathbb{R}^m\), and so is every \(\mathbf{y}\in\operatorname{Nul}\mathbf{A}^{\top}\). Their dot product is therefore defined. Also, \(\mathbf{A}^{\top}\) has \(n\) rows and \(m\) columns, so \(\mathbf{A}^{\top}\mathbf{y}\) is a vector in \(\mathbb{R}^n\), with one entry for each column of \(\mathbf{A}\).
Reading the Entries of \(\mathbf{A}^{\top}\mathbf{y}\)
Write \(\mathbf{y}=(y_1,\dots,y_m)^{\top}\). Since the \(j\)th column of \(\mathbf{A}^{\top}\) is the transpose of \(\mathbf{a}_j\), the \(j\)th entry of \(\mathbf{A}^{\top}\mathbf{y}\) is \(\mathbf{a}_j^{\top}\mathbf{y}=\mathbf{a}_j\cdot\mathbf{y}\). Consequently, the equation \(\mathbf{A}^{\top}\mathbf{y}=\mathbf{0}\) is a list of \(n\) equations, one asserting that \(\mathbf{y}\) is orthogonal to each column.
Proof. By matrix multiplication, the \(j\)th entry of \(\mathbf{A}^{\top}\mathbf{y}\) is \(\sum_{i=1}^{m}a_{ij}y_i\). The \(j\)th column of \(\mathbf{A}\) is \(\mathbf{a}_j=(a_{1j},\dots,a_{mj})^{\top}\), so this sum is exactly \(\mathbf{a}_j\cdot\mathbf{y}\). A vector in \(\mathbb{R}^n\) equals \(\mathbf{0}\) precisely when each of its \(n\) entries equals zero. Therefore \(\mathbf{A}^{\top}\mathbf{y}=\mathbf{0}\) holds exactly when \(\mathbf{a}_j\cdot\mathbf{y}=0\) for every \(j\). The dot product is symmetric over \(\mathbb{R}\), so this is equivalent to \(\mathbf{y}\cdot\mathbf{a}_j=0\) for every \(j\). \(\square\)
This equivalence handles both directions: solving the transposed system produces vectors orthogonal to every column, and a vector orthogonal to every column solves the transposed system. It is not necessary to find a basis for the column space before making this check.
Worked Example: Finding Left-Null Vectors
Let \(\mathbf{A}=\begin{bmatrix}1&2\\0&1\\1&3\end{bmatrix}\). Find \(\operatorname{Nul}\mathbf{A}^{\top}\), then check the dot products with the columns.
The columns are \(\mathbf{a}_1=(1,0,1)^{\top}\) and \(\mathbf{a}_2=(2,1,3)^{\top}\), while
For \(\mathbf{y}=(y_1,y_2,y_3)^{\top}\), the equation \(\mathbf{A}^{\top}\mathbf{y}=\mathbf{0}\) gives
Set \(y_1=t\). The first equation gives \(y_3=-t\). Substitution into the second gives \(2t+y_2-3t=0\), so \(y_2=t\). Hence \(\mathbf{y}=t(1,1,-1)^{\top}\), and \(\operatorname{Nul}\mathbf{A}^{\top}=\{t(1,1,-1)^{\top}:t\in\mathbb{R}\}\).
Check: For \(\mathbf{y}=t(1,1,-1)^{\top}\), \(\mathbf{y}\cdot\mathbf{a}_1=t+0-t=0\) and \(\mathbf{y}\cdot\mathbf{a}_2=2t+t-3t=0\). The computed vectors are orthogonal to both columns.
From the Columns to Their Entire Span
The column space includes more than its individual columns: it includes every linear combination of them. Orthogonality to each column is enough to establish orthogonality to all those combinations. Indeed, if \(\mathbf{y}\) has zero dot product with each \(\mathbf{a}_j\), then linearity of the dot product gives \(\mathbf{y}\cdot(c_1\mathbf{a}_1+\cdots+c_n\mathbf{a}_n)=c_1(\mathbf{y}\cdot\mathbf{a}_1)+\cdots+c_n(\mathbf{y}\cdot\mathbf{a}_n)=0\) for any real scalars \(c_1,\dots,c_n\).
Proof. Let \(\mathbf{x}\in\operatorname{Col}\mathbf{A}\). By the definition of column space, there is a vector \(\mathbf{c}=(c_1,\dots,c_n)^{\top}\in\mathbb{R}^n\) such that \(\mathbf{x}=\mathbf{A}\mathbf{c}=\sum_{j=1}^{n}c_j\mathbf{a}_j\). Now take any \(\mathbf{y}\in\operatorname{Nul}\mathbf{A}^{\top}\). The preceding theorem gives \(\mathbf{y}\cdot\mathbf{a}_j=0\) for each \(j\). Distributing the dot product over the sum yields \(\mathbf{y}\cdot\mathbf{x}=\sum_{j=1}^{n}c_j(\mathbf{y}\cdot\mathbf{a}_j)=\sum_{j=1}^{n}c_j\cdot0=0\). This holds for every \(\mathbf{y}\in\operatorname{Nul}\mathbf{A}^{\top}\), so \(\mathbf{x}\in(\operatorname{Nul}\mathbf{A}^{\top})^{\perp}\). Since \(\mathbf{x}\) was arbitrary in \(\operatorname{Col}\mathbf{A}\), the stated inclusion follows. \(\square\)
The inclusion says that every column combination is orthogonal to every left-null vector. It is a statement about containment: by itself, the proof establishes that every vector in the column space belongs to the orthogonal complement. No reverse containment is needed to prove this result.
Worked Example: Several Dependent Columns
Consider \(\mathbf{A}=\begin{bmatrix}1&2&-1\\2&4&-2\end{bmatrix}\). Compute its left null space and verify the inclusion directly from the column combinations.
The columns are \(\mathbf{a}_1=(1,2)^{\top}\), \(\mathbf{a}_2=(2,4)^{\top}\), and \(\mathbf{a}_3=(-1,-2)^{\top}\). For \(\mathbf{y}=(y_1,y_2)^{\top}\), the first equation in \(\mathbf{A}^{\top}\mathbf{y}=\mathbf{0}\) is \(y_1+2y_2=0\). The other two equations are \(2y_1+4y_2=0\) and \(-y_1-2y_2=0\), which are respectively twice and negative once the first equation. Thus \(y_1=-2y_2\), and \(\operatorname{Nul}\mathbf{A}^{\top}=\{t(-2,1)^{\top}:t\in\mathbb{R}\}\).
For any coefficients \(c_1,c_2,c_3\), multiplication gives \(\mathbf{A}\mathbf{c}=(c_1+2c_2-c_3)(1,2)^{\top}\). Write \(q=c_1+2c_2-c_3\), so this column combination is \(q(1,2)^{\top}\). Its dot product with \(t(-2,1)^{\top}\) is \(tq(-2+2)=0\).
Check: The three column dot products with \((-2,1)^{\top}\) are \(-2+2=0\), \(-4+4=0\), and \(2-2=0\). Thus every left-null vector is orthogonal to each column and, as calculated, to every column combination.
A Four-Coordinate Example
Worked Example: Two Independent Left-Null Directions
Let \(\mathbf{A}=\begin{bmatrix}1&0\\0&1\\1&1\\2&-1\end{bmatrix}\). Find all vectors in \(\operatorname{Nul}\mathbf{A}^{\top}\), verify the column dot products, and check a general column combination.
The columns are \(\mathbf{a}_1=(1,0,1,2)^{\top}\) and \(\mathbf{a}_2=(0,1,1,-1)^{\top}\). Write \(\mathbf{y}=(y_1,y_2,y_3,y_4)^{\top}\). Then \(\mathbf{A}^{\top}\mathbf{y}=\mathbf{0}\) is
Let \(y_3=s\) and \(y_4=t\). Solving the two equations gives \(y_1=-s-2t\) and \(y_2=-s+t\). Therefore
For the first displayed vector, the dot products with \(\mathbf{a}_1,\mathbf{a}_2\) are \(-1+1=0\) and \(-1+1=0\). For the second, they are \(-2+2=0\) and \(1-1=0\). Hence every linear combination of these two vectors is orthogonal to both columns. More explicitly, for arbitrary \(c_1,c_2\), the column combination is \(\mathbf{x}=c_1\mathbf{a}_1+c_2\mathbf{a}_2=(c_1,c_2,c_1+c_2,2c_1-c_2)^{\top}\), and direct calculation gives
Check: The coefficients of \(c_1\) in the last expression sum to \(-s-2t+s+2t=0\), and the coefficients of \(c_2\) sum to \(-s+t+s-t=0\). Thus the dot product vanishes for every choice of \(s,t,c_1,c_2\).
Common Pitfall: Using the Wrong Null Space
The left-null space uses \(\mathbf{A}^{\top}\), not \(\mathbf{A}\). Confusing the two can make the matrix product dimensionally invalid, or can produce vectors from the wrong space. For example, with the \(3\times2\) matrix \(\mathbf{A}=\begin{bmatrix}1&2\\0&1\\1&3\end{bmatrix}\), a student might try to test \((1,1,-1)^{\top}\) by writing \(\mathbf{A}\mathbf{y}=\mathbf{0}\). This is not a valid product: \(\mathbf{A}\) has \(2\) columns but \(\mathbf{y}\) has \(3\) entries. The correction is to use \(\mathbf{A}^{\top}\), which has \(3\) columns, and compute \(\mathbf{A}^{\top}(1,1,-1)^{\top}=(1-1,\;2+1-3)^{\top}=(0,0)^{\top}\).
A second error is to check only one column. A vector orthogonal to \(\mathbf{a}_1\) need not be orthogonal to \(\mathbf{a}_2\). Membership in \(\operatorname{Nul}\mathbf{A}^{\top}\) requires all \(n\) dot products to vanish, one for each column. Checking the full product \(\mathbf{A}^{\top}\mathbf{y}\) enforces all of them at once.
Why the Transpose Test Is Useful
The equation \(\mathbf{A}^{\top}\mathbf{y}=\mathbf{0}\) packages several orthogonality conditions into a single matrix equation. It can be solved by ordinary row reduction, after which the resulting vectors can be checked against the columns using dot products. Conversely, once the columns are known to be orthogonal to every vector in the left-null space, linearity immediately handles every vector built from those columns. This is why the orthogonality statement concerns the entire column space rather than only the displayed columns of a particular matrix.
When applying the result, keep the roles of the vectors explicit: \(\mathbf{y}\) is in \(\mathbb{R}^m\), the same space as the columns; \(\mathbf{A}^{\top}\mathbf{y}\) has \(n\) entries; and an arbitrary vector in \(\operatorname{Col}\mathbf{A}\) has the form \(\mathbf{A}\mathbf{c}\) for some \(\mathbf{c}\in\mathbb{R}^n\). These dimensions make the dot-product argument both valid and easy to verify.
Check Your Understanding
Use the transpose equation and dot products to justify each answer.
- For \(\mathbf{A}=\begin{bmatrix}1&0\\1&1\\0&1\end{bmatrix}\), find \(\operatorname{Nul}\mathbf{A}^{\top}\) and check the dot products with both columns.
- Explain why \(\mathbf{A}^{\top}\mathbf{y}=\mathbf{0}\) gives one orthogonality equation for each column of \(\mathbf{A}\).
- For a vector \(\mathbf{x}=\mathbf{A}\mathbf{c}\) and \(\mathbf{y}\in\operatorname{Nul}\mathbf{A}^{\top}\), expand \(\mathbf{y}\cdot\mathbf{x}\) as a sum and show it is zero.
- For \(\mathbf{A}=\begin{bmatrix}1&2\\2&4\end{bmatrix}\), find a nonzero vector in \(\operatorname{Nul}\mathbf{A}^{\top}\) and verify its dot product with each column.
- State the dimensions of \(\mathbf{A}^{\top}\mathbf{y}\) when \(\mathbf{A}\in\mathbb{R}^{m\times n}\) and \(\mathbf{y}\in\mathbb{R}^m\).
