From Linear Equations to a Subspace
A homogeneous linear system imposes linear conditions on the coordinates of a vector. Its solutions are not merely a collection of vectors that happen to satisfy the same equations: they have a useful structure. Adding solutions, or multiplying a solution by a real scalar, preserves every homogeneous constraint. This makes the solution set a subspace of its ambient coordinate space.
In this tutorial, the ambient space is \(\mathbb{R}^n\), with vectors written as columns. A homogeneous linear equation in the coordinates \(x_1,\dots,x_n\) has the form \(a_1x_1+\cdots+a_nx_n=0\), where the coefficients \(a_1,\dots,a_n\) are fixed real numbers. A finite homogeneous linear system consists of one or more such equations, all required to hold at once.
The distinction between homogeneous and nonhomogeneous equations is the right-hand side: in a homogeneous system every equation has right-hand side zero. In particular, the zero vector satisfies every homogeneous linear equation. A vector is a solution only if it satisfies all equations in the system, not just one of them.
Writing the Constraints as a Matrix Equation
Suppose a system has \(m\) equations in \(n\) unknowns. Let \(a_{ij}\) be the coefficient of \(x_j\) in equation \(i\), and place these coefficients in the matrix \(\mathbf{A}\in\mathbb{R}^{m\times n}\). With \(\mathbf{x}=(x_1,\dots,x_n)^{\top}\), the whole system can be written as \(\mathbf{A}\mathbf{x}=\mathbf{0}\). The \(i\)th coordinate of \(\mathbf{A}\mathbf{x}\) is \(a_{i1}x_1+\cdots+a_{in}x_n\), exactly the left-hand side of equation \(i\).
The solution set is also called the null space of \(\mathbf{A}\), written \(\operatorname{Nul}\mathbf{A}\). In this setting, that notation means precisely \(\{\mathbf{x}\in\mathbb{R}^n:\mathbf{A}\mathbf{x}=\mathbf{0}\}\). The matrix packages the constraints; the vectors in its null space are the columns that pass every constraint.
Proof. Let \(\mathbf{x}_1,\dots,\mathbf{x}_k\in\operatorname{Nul}\mathbf{A}\), so \(\mathbf{A}\mathbf{x}_j=\mathbf{0}\) for every \(j\). Choose any real scalars \(c_1,\dots,c_k\). Matrix multiplication distributes over addition and scalar multiplication, so
Thus the linear combination is a solution. Taking the combination with all coefficients zero shows that \(\mathbf{0}\) belongs to \(\operatorname{Nul}\mathbf{A}\), so the set is nonempty. Taking two solutions and coefficients \(1,1\) proves closure under addition; taking one solution and any real coefficient proves closure under scalar multiplication. These are exactly the requirements for a subspace. \(\square\)
The same reasoning can be read one equation at a time. If \(\mathbf{x}\) and \(\mathbf{y}\) satisfy the \(i\)th equation, then its left-hand side evaluated at \(c\mathbf{x}+d\mathbf{y}\) is \(c\) times its value at \(\mathbf{x}\), plus \(d\) times its value at \(\mathbf{y}\). Both values are zero, so the resulting value is zero as well. Since this holds for every equation, the combined vector satisfies the whole system.
Worked Examples: Membership and Closure
Worked Example: One Constraint in \(\mathbb{R}^3\)
Let \(S\) be the set of columns \(\mathbf{x}=(x,y,z)^{\top}\in\mathbb{R}^3\) satisfying \(x+2y-z=0\). Decide whether \((1,0,1)^{\top}\) and \((0,1,0)^{\top}\) belong to \(S\), and verify the subspace property directly.
For \(\mathbf{x}=(1,0,1)^{\top}\), the left-hand side is \(1+2(0)-1=0\), so this vector belongs to \(S\). For \(\mathbf{y}=(0,1,0)^{\top}\), it is \(0+2(1)-0=2\ne0\), so \(\mathbf{y}\notin S\).
To check closure, take arbitrary \(\mathbf{u}=(u_1,u_2,u_3)^{\top}\) and \(\mathbf{v}=(v_1,v_2,v_3)^{\top}\) in \(S\). Their equations are \(u_1+2u_2-u_3=0\) and \(v_1+2v_2-v_3=0\). For any real \(c,d\), the constraint evaluated at \(c\mathbf{u}+d\mathbf{v}\) is
Also, the zero column gives \(0+2(0)-0=0\). Hence \(S\) contains zero and is closed under linear combinations, so it is a subspace. As another membership check, \((2,-1,0)^{\top}\in S\) because \(2+2(-1)-0=0\).
Worked Example: Two Constraints and a Matrix Reduction
Find the solutions in \(\mathbb{R}^3\) to \(x-y+z=0\) and \(2x+y-z=0\), then test whether \((0,3,3)^{\top}\) is one of them.
Put the coefficients into a matrix and reduce. First apply \(R_2\to R_2-2R_1\), then \(R_2\to \tfrac13R_2\), and finally \(R_1\to R_1+R_2\):
The final matrix says \(x=0\) and \(y-z=0\). Set \(z=t\), where \(t\in\mathbb{R}\); then \(y=t\), so every solution has the form \((0,t,t)^{\top}\). Conversely, substituting \((0,t,t)^{\top}\) into the original equations gives \(0-t+t=0\) and \(0+t-t=0\), so every column of this form really is a solution.
For the proposed vector, \(x-y+z=0-3+3=0\), and \(2x+y-z=2(0)+3-3=0\). It belongs to the solution set; it is the case \(t=3\). The matrix equation check agrees: the two rows of the original coefficient matrix give \(0\) and \(0\).
Worked Example: Three Constraints in \(\mathbb{R}^4\)
Describe the solutions to \(x_1+x_2-x_3=0\), \(2x_1+2x_2-2x_3=0\), and \(x_3+x_4=0\). Check one proposed solution and reject another.
The second equation is twice the first, so it adds no new condition: if \(x_1+x_2-x_3=0\), then \(2x_1+2x_2-2x_3=2(0)=0\). From the third equation, \(x_3=-x_4\). Substituting into the first gives \(x_1+x_2+x_4=0\), or \(x_2=-x_1-x_4\). Let \(x_1=s\) and \(x_4=t\), with \(s,t\in\mathbb{R}\). The solutions are therefore exactly
For verification, the first left-hand side is \(s+(-s-t)-(-t)=s-s-t+t=0\); the second is twice that value, also \(0\); and the third is \(-t+t=0\). Taking \(s=1,t=2\) gives \((1,-3,-2,2)^{\top}\), which satisfies all three equations. In contrast, \((1,0,0,0)^{\top}\) fails the first equation, since \(1+0-0=1\ne0\), and is not a solution.
Why Homogeneity Matters
The closure argument depends on the right-hand side being zero. For a homogeneous equation, multiplying a solution by \(c\) multiplies the left-hand side \(0\) by \(c\), which remains \(0\). For a system with nonzero right-hand sides, that reasoning does not establish closure under scalar multiplication. In particular, the zero vector satisfies every homogeneous system, a useful first check when deciding whether a proposed solution set could be a subspace.
There is also a practical difference between describing solutions and checking membership. A parameter description gives a way to generate solutions, but a proposed column should still be tested in the original equations. When there are multiple constraints, passing one equation is not enough. Likewise, when reducing a matrix, each row represents a condition; the solution must satisfy all rows, including any condition that might not be visible from checking only one of them.
Common Mistake: Checking Only One Equation
Consider the homogeneous system \(x+y=0\) and \(x-y=0\). A tempting but incorrect membership check claims that \((1,-1)^{\top}\) is a solution because its first equation gives \(1+(-1)=0\). But the second equation gives \(1-(-1)=2\ne0\). Therefore this vector is not in the solution set. Solving both equations confirms the correction: adding them gives \(2x=0\), so \(x=0\); then \(x+y=0\) gives \(y=0\). The solution set is just \(\{\mathbf{0}\}\), which is a subspace.
A related error is to treat any collection of linear-looking constraints as homogeneous. The equation \(x+y=1\), for example, is not homogeneous: the zero vector gives \(0+0\ne1\). A system must have zero on the right-hand side of every equation to use the theorem above. Always identify the full system and test each of its equations before claiming membership or applying the subspace conclusion.
Check Your Understanding
For each system, work from its displayed equations and check proposed vectors in every constraint.
- Write \(2x-y+3z=0\) as a matrix equation in \(\mathbb{R}^3\). Is \((1,5,1)^{\top}\) a solution?
- For \(x+y=0\) and \(y-z=0\), find all solutions using one real parameter.
- Check whether \((1,-1,0)^{\top}\) satisfies both equations in question 2.
- Explain why the solution set of \(\mathbf{A}\mathbf{x}=\mathbf{0}\) contains \(\mathbf{0}\) for every real matrix \(\mathbf{A}\) of compatible size.
- Give a specific nonzero right-hand-side equation for which the zero vector is not a solution, and show the failed substitution.
