The Solutions of a Homogeneous Equation
A homogeneous matrix equation has the form \(\mathbf{A}\mathbf{x}=\mathbf{0}\): its right-hand side is the zero vector. Its solutions have a useful shared property. If you add solutions, or multiply a solution by a real scalar, the result is still a solution. This is not a coincidence tied to a particular matrix. It follows directly from the rules of matrix multiplication, and it holds for every real matrix of compatible size.
Let \(m,n\in\mathbb{N}\) and let \(\mathbf{A}\in\mathbb{R}^{m\times n}\). The unknown \(\mathbf{x}\) is a column in \(\mathbb{R}^n\), so \(\mathbf{A}\mathbf{x}\) is in \(\mathbb{R}^m\). We will use the zero vector in each of these spaces: \(\mathbf{0}\in\mathbb{R}^n\) when discussing an unknown, and \(\mathbf{0}\in\mathbb{R}^m\) on the right-hand side.
To see why solutions behave well under combinations, recall three matrix multiplication identities. For compatible vectors \(\mathbf{u},\mathbf{v}\), scalar \(c\in\mathbb{R}\), and matrix \(\mathbf{A}\), we have \(\mathbf{A}(\mathbf{u}+\mathbf{v})=\mathbf{A}\mathbf{u}+\mathbf{A}\mathbf{v}\), \(\mathbf{A}(c\mathbf{u})=c(\mathbf{A}\mathbf{u})\), and \(\mathbf{A}\mathbf{0}=\mathbf{0}\). The first two express linearity of matrix-vector multiplication; the last follows because every row dotted with a zero vector gives zero.
The Zero Vector Is Always a Solution
The equation is homogeneous precisely because its right-hand side is \(\mathbf{0}\). The zero vector in the domain always produces the zero vector in the codomain when multiplied by \(\mathbf{A}\). Thus every homogeneous equation has at least one solution.
Proof. Each coordinate of \(\mathbf{A}\mathbf{0}\) is the dot product of a row of \(\mathbf{A}\) with the zero vector in \(\mathbb{R}^n\). Every term in each such dot product is zero, so \(\mathbf{A}\mathbf{0}=\mathbf{0}\in\mathbb{R}^m\). By the definition of \(S\), this means \(\mathbf{0}\in S\). \(\square\)
This argument establishes existence of a solution, not that it is the only solution. Other vectors may also be sent to zero by \(\mathbf{A}\); which ones depends on the entries of the matrix. The claims proved here apply regardless of whether the zero vector is the only solution or whether there are other solutions.
Addition, Scaling, and Linear Combinations
Closure under an operation means that applying the operation to members of a set produces another member of that set. For the solution set \(S\), closure under addition says that if \(\mathbf{x},\mathbf{y}\in S\), then \(\mathbf{x}+\mathbf{y}\in S\). Closure under scalar multiplication says that if \(\mathbf{x}\in S\) and \(c\in\mathbb{R}\), then \(c\mathbf{x}\in S\). The conditions on the inputs matter: the claims begin with vectors already known to solve the same homogeneous equation.
Proof. Suppose \(\mathbf{x},\mathbf{y}\in S\). By the definition of \(S\), \(\mathbf{A}\mathbf{x}=\mathbf{0}\) and \(\mathbf{A}\mathbf{y}=\mathbf{0}\). Using distributivity of matrix multiplication over vector addition gives \(\mathbf{A}(\mathbf{x}+\mathbf{y})=\mathbf{A}\mathbf{x}+\mathbf{A}\mathbf{y}=\mathbf{0}+\mathbf{0}=\mathbf{0}\). Therefore \(\mathbf{x}+\mathbf{y}\in S\). Now suppose \(\mathbf{x}\in S\) and \(c\in\mathbb{R}\). Compatibility with scalar multiplication gives \(\mathbf{A}(c\mathbf{x})=c(\mathbf{A}\mathbf{x})=c\mathbf{0}=\mathbf{0}\), so \(c\mathbf{x}\in S\). Finally, let \(p\in\mathbb{N}\), let \(\mathbf{x}_1,\dots,\mathbf{x}_p\in S\), and choose any \(c_1,\dots,c_p\in\mathbb{R}\). Repeated distributivity and compatibility with scalar multiplication yield \(\mathbf{A}(c_1\mathbf{x}_1+\cdots+c_p\mathbf{x}_p)=c_1\mathbf{A}\mathbf{x}_1+\cdots+c_p\mathbf{A}\mathbf{x}_p\). Every \(\mathbf{A}\mathbf{x}_i=\mathbf{0}\), so the right-hand side is \(c_1\mathbf{0}+\cdots+c_p\mathbf{0}=\mathbf{0}\). Hence the linear combination belongs to \(S\). \(\square\)
The conclusion includes choices such as \(c_i=0\), negative scalars, and all coefficients equal to zero. For instance, scaling any solution by \(0\) gives the zero vector, which is indeed a solution. The theorem also concerns one fixed matrix \(\mathbf{A}\): the vectors being combined must all solve the same homogeneous equation.
Worked Checks
Worked Example: Adding Two Solutions
Consider \(\mathbf{A}=\begin{bmatrix}1&2&-1\\2&4&-2\end{bmatrix}\), \(\mathbf{x}=\begin{bmatrix}1\\0\\1\end{bmatrix}\), and \(\mathbf{y}=\begin{bmatrix}0\\1\\2\end{bmatrix}\). Verify that the two vectors solve \(\mathbf{A}\mathbf{z}=\mathbf{0}\), then check their sum.
First multiply \(\mathbf{A}\) by \(\mathbf{x}\), showing each row calculation:
For \(\mathbf{y}\), the row calculations are \(1(0)+2(1)-1(2)=0\) and \(2(0)+4(1)-2(2)=0\). Thus \(\mathbf{A}\mathbf{y}=\begin{bmatrix}0\\0\end{bmatrix}\) as well. Their sum is \(\mathbf{x}+\mathbf{y}=\begin{bmatrix}1\\1\\3\end{bmatrix}\). Direct substitution checks the result:
So the sum is a solution, as the theorem predicts.
Worked Example: Scaling a Solution
Let \(\mathbf{A}=\begin{bmatrix}2&-1\\4&-2\end{bmatrix}\) and \(\mathbf{x}=\begin{bmatrix}1\\2\end{bmatrix}\). Check that \(\mathbf{x}\) is a solution, then check its scalar multiple for \(c=-3\).
For the original vector, the two row products are \(2(1)-1(2)=0\) and \(4(1)-2(2)=0\). Therefore \(\mathbf{A}\mathbf{x}=\begin{bmatrix}0\\0\end{bmatrix}\). Scaling gives \(-3\mathbf{x}=\begin{bmatrix}-3\\-6\end{bmatrix}\). Multiplying this vector by \(\mathbf{A}\) gives
The scaled vector solves the equation. This check also illustrates that a negative scalar causes no problem: each row product remains zero.
Worked Example: A General Linear Combination
Let \(\mathbf{A}=\begin{bmatrix}1&2&3\\2&4&6\end{bmatrix}\) and take \(\mathbf{x}=\begin{bmatrix}-2\\1\\0\end{bmatrix}\) and \(\mathbf{y}=\begin{bmatrix}-3\\0\\1\end{bmatrix}\). Verify that \(2\mathbf{x}-\mathbf{y}\) is a solution.
For \(\mathbf{x}\), the row products are \(1(-2)+2(1)+3(0)=0\) and \(2(-2)+4(1)+6(0)=0\). For \(\mathbf{y}\), they are \(1(-3)+2(0)+3(1)=0\) and \(2(-3)+4(0)+6(1)=0\). Thus both vectors belong to the solution set. The proposed combination is
Substitution verifies both rows: \(1(-1)+2(2)+3(-1)=-1+4-3=0\), and \(2(-1)+4(2)+6(-1)=-2+8-6=0\). Thus \(2\mathbf{x}-\mathbf{y}\) is a solution. This is one concrete instance of the theorem for coefficients \(2\) and \(-1\).
Common Pitfall: Changing the Right-Hand Side
The closure result is specific to a homogeneous equation. It is generally wrong to apply the same conclusion to solutions of \(\mathbf{A}\mathbf{x}=\mathbf{b}\) when \(\mathbf{b}\ne\mathbf{0}\). If \(\mathbf{A}\mathbf{x}=\mathbf{b}\) and \(\mathbf{A}\mathbf{y}=\mathbf{b}\), then distributivity gives \(\mathbf{A}(\mathbf{x}+\mathbf{y})=2\mathbf{b}\), not necessarily \(\mathbf{b}\).
Wrong calculation: take the one-entry matrix \(\mathbf{A}=\begin{bmatrix}1\end{bmatrix}\) and equation \(\mathbf{A}\mathbf{x}=\begin{bmatrix}1\end{bmatrix}\). The vector \(\mathbf{x}=\begin{bmatrix}1\end{bmatrix}\) is a solution, but adding it to itself gives \(\begin{bmatrix}2\end{bmatrix}\), and \(\mathbf{A}\begin{bmatrix}2\end{bmatrix}=\begin{bmatrix}2\end{bmatrix}\ne\begin{bmatrix}1\end{bmatrix}\). Concluding that the sum remains a solution confuses this equation with a homogeneous one. The correction is to check the right-hand side: the closure proof works because each solution is mapped to \(\mathbf{0}\), and \(\mathbf{0}+\mathbf{0}=\mathbf{0}\).
There is a related distinction between the zero vector in the input space and the zero vector on the right-hand side. The input is in \(\mathbb{R}^n\), while the output is in \(\mathbb{R}^m\); their dimensions can differ. The statement \(\mathbf{A}\mathbf{0}=\mathbf{0}\) means that the zero input in \(\mathbb{R}^n\) maps to the zero output in \(\mathbb{R}^m\), not that these two vectors have the same number of coordinates.
Why the Closure Facts Are Useful
Once particular solutions of a homogeneous equation are known, the theorem gives more solutions without requiring a fresh verification of every combination. For example, if \(\mathbf{x}\) and \(\mathbf{y}\) solve the same equation, then \(-\mathbf{x}\) is a solution by choosing \(c=-1\), and \(\mathbf{x}-\mathbf{y}\) is a solution by combining \(\mathbf{x}\) and \(\mathbf{y}\) with coefficients \(1\) and \(-1\). These conclusions are immediate applications of the proved closure properties.
The argument also explains why the zero vector cannot be omitted from the solution set: it is always there, and it is the result of scaling any solution by zero. The theorem does not tell us how to list every solution or how many solutions there are. It establishes the structural facts that any full description must respect: the set contains zero and is closed under addition and scalar multiplication. Describing the solutions themselves requires additional work.
Check Your Understanding
Use matrix multiplication properties and the definition of the solution set to answer each question.
- For \(\mathbf{A}\in\mathbb{R}^{m\times n}\), explain why the input zero vector is in the solution set of \(\mathbf{A}\mathbf{x}=\mathbf{0}\).
- If \(\mathbf{A}\mathbf{x}=\mathbf{0}\) and \(\mathbf{A}\mathbf{y}=\mathbf{0}\), write the calculation proving that \(\mathbf{x}+\mathbf{y}\) is a solution.
- If \(\mathbf{x}\) is a solution and \(c\in\mathbb{R}\), which matrix multiplication identity proves that \(c\mathbf{x}\) is a solution?
- Suppose \(\mathbf{x}_1,\dots,\mathbf{x}_p\) all solve the same homogeneous equation. Why does \(c_1\mathbf{x}_1+\cdots+c_p\mathbf{x}_p\) also solve it?
- For the equation \(\begin{bmatrix}1\end{bmatrix}\mathbf{x}=\begin{bmatrix}2\end{bmatrix}\), check whether the sum of the solution \(\begin{bmatrix}2\end{bmatrix}\) with itself is still a solution. What feature distinguishes this equation from a homogeneous one?
