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Homogeneous Systems of Equations



Definition
Homogeneous System Notation
When Do Nontrivial Solutions Exist?
The Solution Set Is the Null Space
Finding the Null Space
Parametric Vector Form
The Superposition Principle
Homogeneous vs. Non-Homogeneous
Homogeneous Systems and Linear Independence
The Eigenvalue Connection
Summary: Where Homogeneous Systems Appear
Common Pitfalls
Homogeneous Systems FAQ



Systems with Zero Right-Hand Side

A homogeneous system Ax = 0 always has the trivial solution x = 0. The real question is whether nontrivial solutions exist — and when they do, the solution set is a subspace of Rⁿ whose structure is governed entirely by the rank of the coefficient matrix.



Definition

A homogeneous linear system is one where every equation has zero on the right-hand side:

Definition
Ax=0A\mathbf{x} = \mathbf{0}

The augmented matrix is [A∣0][A \mid \mathbf{0}]. Since the last column is all zeros, row operations on the augmented matrix never produce a contradictory row [0  ⋯  0∣d][0 \; \cdots \; 0 \mid d] with d≠0d \neq 0.

THEOREM
A homogeneous system is always consistent.

The vector x=0\mathbf{x} = \mathbf{0} satisfies every equation — this is the trivial solution. It always exists. The central question for a homogeneous system is never "does a solution exist?" but "does a nontrivial solution exist?"

Homogeneous System Notation

Notation

Homogeneous System Notation

One boldface character carries the whole definition: the right-hand side is a vector of zeros, not the number zero. From that follow a solution that is guaranteed rather than sought, and a vocabulary that names the two cases a homogeneous system can be in. All of it is catalogued among the linear algebra symbols.
The augmented bar [A∣0][A \mid \mathbf{0}] and the row-operation conventions come from Gaussian elimination notation; Nul⁡(A)\operatorname{Nul}(A) from fundamental-space notation; rank⁡\operatorname{rank} and the free-variable count from rank and echelon form notation; det⁡\det from determinant notation.
Ax=0A\mathbf{x} = \mathbf{0}
A x equals the zero vector
The boldface is the definition, not styling: 0\mathbf{0} is a column of mm zeros, while a plain 00 is a single number. Writing Ax=0A\mathbf{x} = 0 equates a vector with a scalar — the shapes do not match, and the equation only typechecks once the right side is bold.
Also written0⃗\vec{0} where the site's arrow convention is in force, and 0m\mathbf{0}_m when the length needs stating; handwriting has no bold, so an underline or an arrow does the work at the board — which is exactly where the scalar/vector slip happens.
Do not confuseThe zero matrix OO. Same idea, different shape: OO is a rectangular array of zeros, 0\mathbf{0} a single column, and matrix operation notation reserves the two letters separately for that reason.
x=0\mathbf{x} = \mathbf{0}
the trivial solution
A named solution rather than a described one: x=0\mathbf{x} = \mathbf{0} satisfies Ax=0A\mathbf{x} = \mathbf{0} for every AA, so it is guaranteed, never discovered. As Definition above puts it, the question for a homogeneous system is never whether a solution exists but whether another one does.
Cases"Nontrivial" is the entire vocabulary for the interesting case, and it is a claim about existence of a second solution, not about complexity — a nontrivial solution can be as simple as (1,0,0)(1, 0, 0). Its presence is equivalent to a free column, to rank⁡(A)<n\operatorname{rank}(A) < n, and for square AA to det⁡(A)=0\det(A) = 0.
Do not confuse"Trivial" as a comment on difficulty. Here it is a technical label for one specific vector, not an editorial judgement — and a system whose only solution is trivial is the more constrained, often harder case.
{x:Ax=0}=Nul⁡(A)\{\mathbf{x} : A\mathbf{x} = \mathbf{0}\} = \operatorname{Nul}(A)
the solution set is the null space of A
An equality between two descriptions of one object: set-builder on the left, a named fundamental space on the right. Only homogeneous systems earn this — the notation records that the solutions form a subspace rather than a scattered set.
CasesThe Superposition Principle below is that subspace closure spelled out: if u\mathbf{u} and v\mathbf{v} solve the system then so does cu+dvc\mathbf{u} + d\mathbf{v}, which is why the general solution can be written as a span rather than a list.
Do not confuseExpecting the same of Ax=bA\mathbf{x} = \mathbf{b}. A non-homogeneous solution set is a translate of Nul⁡(A)\operatorname{Nul}(A), not a subspace — it misses 0\mathbf{0} unless b=0\mathbf{b} = \mathbf{0}, which is precisely the structural difference Homogeneous vs. Non-Homogeneous below turns on.

When Do Nontrivial Solutions Exist?

THEOREM
Nontrivial solutions to Ax=0A\mathbf{x} = \mathbf{0} exist if and only if the rank of AA is less than nn, the number of unknowns.

Fact: Balancing a reaction is a homogeneous system
Balancing a C3H8+b O2→c CO2+d H2Oa\,\text{C}_3\text{H}_8 + b\,\text{O}_2 \to c\,\text{CO}_2 + d\,\text{H}_2\text{O} gives one equation per element and one unknown per molecule. Three elements against four unknowns force a nontrivial solution, and its smallest whole-number point (1,5,3,4)(1, 5, 3, 4) is the balanced equation.

When rank(A)<n\text{rank}(A) < n, at least one free variable appears in the echelon form, and that free variable parametrizes a family of nonzero solutions.

For a square n×nn \times n matrix, nontrivial solutions exist if and only if det⁡(A)=0\det(A) = 0. A nonzero determinant means full rank, which means no free variables, which means only the trivial solution.

One case is automatic: if the system has more unknowns than equations (n>mn > m), nontrivial solutions always exist:

Underdetermined Homogeneous Has Nontrivial
m<n⇒Ax=0 has nontrivial solutionsm < n \Rightarrow A\mathbf{x} = \mathbf{0} \text{ has nontrivial solutions}

Pitfall: Reading the fewer-equations rule backwards
Concluding from m≥nm \ge n that only x=0\mathbf{x} = \mathbf{0} works fails for A=(122436)A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \\ 3 & 6 \end{pmatrix}: three equations, two unknowns, yet rank(A)=1<2\text{rank}(A) = 1 < 2 and x=(−2,1)\mathbf{x} = (-2, 1) solves Ax=0A\mathbf{x} = \mathbf{0}.

The rank of an m×nm \times n matrix cannot exceed mm, and when m<nm < n, the rank is strictly less than nn. This guarantees at least n−mn - m free variables, producing an infinite family of nontrivial solutions. Fewer equations than unknowns always leaves room.
Setting Condition for nontrivial solutions to A x = 0 Equivalent statements
General m × n matrix rank(A) < n nullity(A) > 0; at least one free variable in the echelon form of A
Square n × n matrix det(A) = 0 A is singular; A is not invertible; columns are linearly dependent
Wide matrix (n > m) always — nontrivial solutions guaranteed rank(A) ≤ m < n, so n − rank(A) ≥ n − m ≥ 1 free variables remain
A | b2×412040011→p3×1401+t₁v₁3×1−210
A free column, and with it a family of solutions

The elimination has left a column without a pivot, so one unknown can be chosen freely and the others follow. For a homogeneous system that guarantees solutions beyond the zero one, which is why more unknowns than equations always produces them — there are not enough pivots to go round. Vary the shape and watch the free columns appear on the linear system solutions visualizer.

Counting pivots against unknowns therefore answers the question before any solution is written down.

The Solution Set Is the Null Space

The set of all solutions to Ax=0A\mathbf{x} = \mathbf{0} is the null space of AA:

The Solution Set Is the Null Space
Null(A)={x∈Rn:Ax=0}\text{Null}(A) = \{\mathbf{x} \in \mathbb{R}^n : A\mathbf{x} = \mathbf{0}\}

THEOREM
The null space is a subspace of Rn\mathbb{R}^n.

It contains 0\mathbf{0}, and it is closed under addition and scalar multiplication: if Au=0A\mathbf{u} = \mathbf{0} and Av=0A\mathbf{v} = \mathbf{0}, then A(u+v)=0A(\mathbf{u} + \mathbf{v}) = \mathbf{0} and A(cu)=0A(c\mathbf{u}) = \mathbf{0}.

The dimension of the null space is the nullity:

Homogeneous Solution Space Dimension
dim⁡(Null(A))=n−rank(A)\dim(\text{Null}(A)) = n - \text{rank}(A)

Pitfall: Subtracting the rank from the number of equations
Computing the nullity as m−rank(A)m - \text{rank}(A) gives 3−3=03 - 3 = 0 for a 3×53 \times 5 matrix of rank 33, wrongly ruling out nontrivial solutions. The nullity counts unknowns: n−rank(A)=5−3=2n - \text{rank}(A) = 5 - 3 = 2.

When the nullity is 00, the null space is {0}\{\mathbf{0}\} and only the trivial solution exists. When the nullity is k>0k > 0, the null space is a kk-dimensional subspace, and the solution set contains infinitely many vectors forming a kk-dimensional flat through the origin.
R3×3120001000→A3×3121243364basis of N(A)3×12−10=03×1000
Special solutions, one per free column

Each free variable is set to one in turn while the others are held at zero, and the pivot variables are solved to match. The vectors produced are independent and they span every solution of the system, which is what it means to call the solution set the null space rather than merely to say it resembles one. Generate them for your own matrix on the four fundamental subspaces visualizer.

Because the set is closed under addition and scaling, it is a subspace, and that is what makes a basis for it worth having.

Finding the Null Space

The algorithm is a direct application of Gaussian elimination. Row reduce AA to echelon form (reducing just AA — the zero augmented column adds nothing). Identify the pivot variables and the free variables. For each free variable, set it to 11 with all other free variables at 00, and solve for the pivot variables by back substitution. Each setting produces one basis vector for the null space.

Worked Example


Finding the Null Space
A=(12−10324028−1−2210)A = \begin{pmatrix} 1 & 2 & -1 & 0 & 3 \\ 2 & 4 & 0 & 2 & 8 \\ -1 & -2 & 2 & 1 & 0 \end{pmatrix}

Row reduce:

Finding the Null Space
→R2−2R1,  R3+R1(12−1030022200113)→R3−12R2(12−1030022200002)\xrightarrow{R_2 - 2R_1,\; R_3 + R_1} \begin{pmatrix} 1 & 2 & -1 & 0 & 3 \\ 0 & 0 & 2 & 2 & 2 \\ 0 & 0 & 1 & 1 & 3 \end{pmatrix} \xrightarrow{R_3 - \frac{1}{2}R_2} \begin{pmatrix} 1 & 2 & -1 & 0 & 3 \\ 0 & 0 & 2 & 2 & 2 \\ 0 & 0 & 0 & 0 & 2 \end{pmatrix}

Pivots in columns 11, 33, 55. Free variables: x2=sx_2 = s, x4=tx_4 = t. Row 33: 2x5=0⇒x5=02x_5 = 0 \Rightarrow x_5 = 0. Row 22: 2x3+2t=0⇒x3=−t2x_3 + 2t = 0 \Rightarrow x_3 = -t. Row 11: x1+2s+t+0=0⇒x1=−2s−tx_1 + 2s + t + 0 = 0 \Rightarrow x_1 = -2s - t.

Setting s=1,t=0s = 1, t = 0: v1=(−2,1,0,0,0)\mathbf{v}_1 = (-2, 1, 0, 0, 0). Setting s=0,t=1s = 0, t = 1: v2=(−1,0,−1,1,0)\mathbf{v}_2 = (-1, 0, -1, 1, 0).

Pitfall: Reading null-space vectors from the rows
Taking the rows of the echelon form as null-space vectors fails: A(1,2,−1,0,3)=(15,34,−7)A(1, 2, -1, 0, 3) = (15, 34, -7), not 0\mathbf{0}. Those rows span the row space; the null space comes from solving, here v1=(−2,1,0,0,0)\mathbf{v}_1 = (-2, 1, 0, 0, 0) and v2=(−1,0,−1,1,0)\mathbf{v}_2 = (-1, 0, -1, 1, 0).

The null space is Span{v1,v2}\text{Span}\{\mathbf{v}_1, \mathbf{v}_2\}, a two-dimensional subspace of R5\mathbb{R}^5.

Each basis vector can be tested against the original matrix AA, since row operations change the rows but keep the solutions. For v2=(−1,0,−1,1,0)\mathbf{v}_2 = (-1, 0, -1, 1, 0), the three rows of AA give −1+1=0-1 + 1 = 0, −2+2=0-2 + 2 = 0 and 1−2+1=01 - 2 + 1 = 0. The count matches as well: one basis vector for each free variable, two in all.

Row reduction turns the null space into a recipe: one free variable, one basis vector, and nothing to guess.

Parametric Vector Form

The general solution to Ax=0A\mathbf{x} = \mathbf{0} is a linear combination of the null-space basis vectors:

Parametric Vector Form
x=t1v1+t2v2+⋯+tkvk\mathbf{x} = t_1\mathbf{v}_1 + t_2\mathbf{v}_2 + \cdots + t_k\mathbf{v}_k

Fact: Loops in a circuit are free variables
Kirchhoff's current law is Ax=0A\mathbf{x} = \mathbf{0} with one equation per junction and one unknown current per wire. In a connected circuit the rank is the number of junctions minus one, so the parameters number wires minus junctions plus one: the count of windows in a flat circuit diagram, by Euler's formula.

where v1,…,vk\mathbf{v}_1, \dots, \mathbf{v}_k are the basis vectors found by the algorithm above and t1,…,tkt_1, \dots, t_k are free parameters ranging over all real numbers. The number of parameters k=n−rank(A)k = n - \text{rank}(A) is the nullity.

When k=0k = 0, the only solution is x=0\mathbf{x} = \mathbf{0}. When k=1k = 1, the solutions form a line through the origin in Rn\mathbb{R}^n. When k=2k = 2, a plane through the origin. In general, the solution set is a kk-dimensional subspace passing through the origin.

There is no particular solution xp\mathbf{x}_p to add because the right-hand side is 0\mathbf{0} — the zero vector is itself the particular solution. The entire solution set is the null space, unshifted.

The Superposition Principle

If x1\mathbf{x}_1 and x2\mathbf{x}_2 are solutions to Ax=0A\mathbf{x} = \mathbf{0}, then any linear combination c1x1+c2x2c_1\mathbf{x}_1 + c_2\mathbf{x}_2 is also a solution:

The Superposition Principle
A(c1x1+c2x2)=c1Ax1+c2Ax2=c10+c20=0A(c_1\mathbf{x}_1 + c_2\mathbf{x}_2) = c_1 A\mathbf{x}_1 + c_2 A\mathbf{x}_2 = c_1\mathbf{0} + c_2\mathbf{0} = \mathbf{0}

This is precisely the statement that the solution set is a subspace — it is closed under addition and scalar multiplication. The superposition principle is the reason the general solution is a linear combination of basis vectors, and it is the reason the null space has the clean structure of a vector space rather than an arbitrary collection of points.

RULESuperposition principle
Superposition holds only for homogeneous systems.

For a non-homogeneous system Ax=bA\mathbf{x} = \mathbf{b} with b≠0\mathbf{b} \neq \mathbf{0}, the sum of two solutions is generally not a solution: A(x1+x2)=b+b=2b≠bA(\mathbf{x}_1 + \mathbf{x}_2) = \mathbf{b} + \mathbf{b} = 2\mathbf{b} \neq \mathbf{b}.

Solutions of Ax=0A\mathbf{x} = \mathbf{0} can be added and scaled freely, so a few basis solutions describe every one of them.

Homogeneous vs. Non-Homogeneous

The homogeneous system Ax=0A\mathbf{x} = \mathbf{0} and the non-homogeneous system Ax=bA\mathbf{x} = \mathbf{b} are deeply connected. If xp\mathbf{x}_p is any particular solution to Ax=bA\mathbf{x} = \mathbf{b}, then every solution has the form

Homogeneous vs. Non-Homogeneous
x=xp+xh\mathbf{x} = \mathbf{x}_p + \mathbf{x}_h

where xh∈Null(A)\mathbf{x}_h \in \text{Null}(A) is a solution to the homogeneous system. The particular solution accounts for b\mathbf{b}; the null-space component accounts for the freedom.

This decomposition has two immediate consequences.

COROLLARY
If the null space is trivial (nullity=0\text{nullity} = 0), the non-homogeneous system has at most one solution — either xp\mathbf{x}_p alone or nothing. If the null space is nontrivial (nullity>0\text{nullity} > 0), then either Ax=bA\mathbf{x} = \mathbf{b} has no solution or it has infinitely many — there is no middle ground.

The solution set of Ax=bA\mathbf{x} = \mathbf{b} is therefore a translated copy of the null space: the null space shifted by xp\mathbf{x}_p. The homogeneous system determines the shape and dimension of the solution set; the particular solution determines its position.
The solutions of Ax = b form a line parallel to the null spaceOxₚ = (2, 1)xₕ = (−2, 1)x = xₚ + xₕ = (0, 2)solutions of Ax = bnull space: Ax = 0A = [1 2; 2 4], b = (4, 8): every solution satisfies x + 2y = 4parallel lines: moving along the null space stays among the solutionsevery solution = one particular solution + a null-space vector
Ax = b is Ax = 0, shifted

Both equations of this system reduce to x + 2y = 4, so its solutions fill the amber line; the homogeneous system Ax = 0 has the dashed line x + 2y = 0 through the origin. Pick one solution xp = (2, 1), add any vector of the null space — here xh = (−2, 1) — and you land on the amber line again, at (0, 2). Every solution is reached this way, which is why the two lines are parallel copies of each other.

Solving Ax = b therefore splits into two jobs: find one solution, and find the null space.
Aspect Homogeneous: A x = 0 Non-homogeneous: A x = b, b ≠ 0
Always consistent? yes — the trivial solution x = 0 always works no — depends on whether b ∈ Col(A)
Solution set Null(A) — a subspace passing through the origin xp + Null(A) — an affine flat shifted away from the origin
Superposition closed: any linear combination of solutions is a solution not closed: A(x1 + x2) = 2b ≠ b
When uniquely solvable always uniquely "solvable" — only the trivial solution when rank(A) = n when rank(A) = rank([A | b]) = n
Number of solutions just the trivial one, or infinitely many (never zero, never finite > 1) zero, one, or infinitely many

Homogeneous Systems and Linear Independence

THEOREM
Testing whether vectors v1,…,vk\mathbf{v}_1, \dots, \mathbf{v}_k are linearly independent is equivalent to checking whether the homogeneous system Ac=0A\mathbf{c} = \mathbf{0} has only the trivial solution, where A=[v1  v2  ⋯  vk]A = [\mathbf{v}_1 \; \mathbf{v}_2 \; \cdots \; \mathbf{v}_k].

The equation c1v1+c2v2+⋯+ckvk=0c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k = \mathbf{0} is literally the system Ac=0A\mathbf{c} = \mathbf{0}. If the null space of AA is trivial, the only solution is c=0\mathbf{c} = \mathbf{0} and the vectors are independent. If the null space is nontrivial, some nonzero c\mathbf{c} satisfies the equation, providing an explicit dependence relation — the entries of c\mathbf{c} are the coefficients that express one vector as a combination of the others.

This is the computational link between homogeneous systems and independence. Row reducing AA and checking for free variables is the standard algorithm for deciding independence, and the null-space basis vectors encode the dependence relations when they exist.

The Eigenvalue Connection

The eigenvalue equation Ax=λxA\mathbf{x} = \lambda\mathbf{x} can be rewritten as

The Eigenvalue Connection
(A−λI)x=0(A - \lambda I)\mathbf{x} = \mathbf{0}

Pitfall: Subtracting λ from every entry
Subtracting λ\lambda from all four entries of A=(2112)A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} gives (2−λ)2−(1−λ)2=3−2λ(2 - \lambda)^2 - (1 - \lambda)^2 = 3 - 2\lambda and the false eigenvalue 3/23/2. With λI\lambda I only the diagonal changes: (2−λ)2−1=0(2 - \lambda)^2 - 1 = 0, so λ=1\lambda = 1 and λ=3\lambda = 3.

This is a homogeneous system with coefficient matrix A−λIA - \lambda I. Eigenvectors are precisely the nontrivial solutions. They exist when and only when A−λIA - \lambda I is singular — that is, when det⁡(A−λI)=0\det(A - \lambda I) = 0.

The values of λ\lambda satisfying this determinant condition are the eigenvalues. For each eigenvalue λ\lambda, the set of all solutions to (A−λI)x=0(A - \lambda I)\mathbf{x} = \mathbf{0} is the eigenspace — the null space of A−λIA - \lambda I. The dimension of this eigenspace is the nullity of A−λIA - \lambda I, which equals n−rank(A−λI)n - \text{rank}(A - \lambda I).

This rewriting connects homogeneous systems directly to spectral theory. Every eigenvalue problem is, at its core, a question about when a particular homogeneous system has nontrivial solutions. The machinery of row reduction, null spaces, and rank that governs homogeneous systems is the same machinery that computes eigenvectors and eigenspaces.

Summary: Where Homogeneous Systems Appear

The homogeneous system A·x = 0 shows up far beyond its own statement: every test of linear independence, every eigenvector calculation, every analysis of a non-homogeneous solution set, and every kernel computation reduces to solving one. The table below collects each construction in which homogeneous systems play a role, alongside the specific equation involved and what its solutions represent in that context.
The six settings below are not applications of homogeneous systems so much as the same system wearing different names. In each case something is being asked about a null space, and the only thing that changes is what the trivial solution is taken to mean — no redundancy among vectors, no eigenvector for that scalar, no collapse under the transformation.
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Linear systems · homogeneous

Where Ax=0A\mathbf{x} = \mathbf{0} turns up

Six questions from six different chapters, all of which reduce to solving one homogeneous system. What differs is only what the null space is being called and what its being trivial is taken to mean.

6settings
The system itself2
1
Solution set§ 3
always consistent — x=0\mathbf{x} = \mathbf{0} works
Ax=0A\mathbf{x} = \mathbf{0}
The solution set is Null⁡(A)\operatorname{Null}(A), a subspace rather than merely a set: closed under addition and scaling, which is the superposition principle. One basis vector per free variable, so the dimension is n−rank⁡(A)n - \operatorname{rank}(A).
2
Non-homogeneous solution set§ 7
when Ax=bA\mathbf{x} = \mathbf{b} is consistent
x=xp+Null⁡(A)\mathbf{x} = \mathbf{x}_p + \operatorname{Null}(A)
The homogeneous solution supplies the entire shape and size of the general solution; the particular solution only translates it off the origin. This is why the two systems are never solved independently — the same reduction gives both.
The same equation, asked differently4
3
Linear independence§ 8
independent   ⟺  \iff only c=0\mathbf{c} = \mathbf{0}
Ac=0A\mathbf{c} = \mathbf{0}, columns =vi= \mathbf{v}_i
Independence is the statement that this system has only the trivial solution. When it does not, the nontrivial c\mathbf{c} is not merely evidence of dependence — its entries are the coefficients of an explicit dependence relation among the vectors.
4
Column dependence relations§ 8
one relation per null space basis vector
nontrivial c\mathbf{c} with Ac=0A\mathbf{c} = \mathbf{0}
Read the same solution the other way round: each entry of c\mathbf{c} says how much of the corresponding column enters the relation, so one column is written as a combination of the others. Which columns are redundant is exactly which are non-pivot.
5
Eigenvectors§ 9
λ\lambda already known
(A−λI)x=0(A - \lambda I)\mathbf{x} = \mathbf{0}
The eigenspace for λ\lambda is Null⁡(A−λI)\operatorname{Null}(A - \lambda I), so finding eigenvectors is solving a homogeneous system. A nontrivial solution must exist, which is precisely why det⁡(A−λI)=0\det(A - \lambda I) = 0 is the condition defining λ\lambda in the first place.
6
Kernel of a transformation§ 3
ker⁡(T)=Null⁡(A)\ker(T) = \operatorname{Null}(A) for the matrix of TT
T(x)=0T(\mathbf{x}) = \mathbf{0}
The abstract statement of the same thing. TT is injective exactly when the kernel is trivial — so the size of the null space measures how far a transformation is from being one-to-one.
One computation serves all six: reduce, identify the free variables, read off a basis. That is why the homogeneous case is worth treating first — the non-homogeneous system, the independence test, the eigenvector calculation and the kernel are the same reduction asked about differently.
Where Ax = 0 turns up·/linear-algebra/linear-systems/homogeneousLearn Math Class
Two of these are worth reading together, because they are the same solution used twice. Solving Ac=0A\mathbf{c} = \mathbf{0} answers whether the columns are independent, and when the answer is no, the vector c\mathbf{c} is not merely a witness to dependence — its entries are the coefficients of the relation. The test and the explanation come from one computation, which is why nothing further is needed once the null space basis is in hand.
The eigenvector row is where the ordering matters. (A−λI)x=0(A - \lambda I)\mathbf{x} = \mathbf{0} is only worth solving once λ\lambda is known, and λ\lambda is chosen precisely so that this system has a nontrivial solution — which is what det⁡(A−λI)=0\det(A - \lambda I) = 0 asserts. The characteristic equation exists to make the homogeneous system interesting; the eigenvector calculation that follows is ordinary null space work.

Common Pitfalls

Four mistakes account for many wrong answers about homogeneous systems. Each is marked with a box where it happens in the sections above; here each is worked through in full, with the wrong and the right working side by side.

Reading the fewer-equations rule backwards

The rule says that fewer equations than unknowns (m<nm < n) forces nontrivial solutions. It is read in reverse: with at least as many equations as unknowns, the system is assumed to have only the trivial solution. The rule goes one way only; when m≥nm \ge n, the rank decides.

Wrong working: A=(122436)A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \\ 3 & 6 \end{pmatrix} has 33 equations and 22 unknowns, so only x=0\mathbf{x} = \mathbf{0} is accepted.

Right working: row reduce.
(122436)→R2−2R1,  R3−3R1(120000)\begin{pmatrix} 1 & 2 \\ 2 & 4 \\ 3 & 6 \end{pmatrix} \xrightarrow{R_2 - 2R_1,\; R_3 - 3R_1} \begin{pmatrix} 1 & 2 \\ 0 & 0 \\ 0 & 0 \end{pmatrix}

The rank is 1<21 < 2, so x2=tx_2 = t is free and x1=−2tx_1 = -2t. Setting t=1t = 1 gives the nontrivial solution (−2,1)(-2, 1).

Why it happens: "more equations than unknowns" sounds like too many conditions to leave any freedom. But equations can repeat one another; here rows 22 and 33 are multiples of row 11 and add no condition.

How to check: compute the rank instead of comparing the shape, and substitute the candidate: A(−2,1)=(0,0,0)A(-2, 1) = (0, 0, 0).

Subtracting the rank from the number of equations

The nullity is computed as m−rank(A)m - \text{rank}(A), the number of rows minus the rank, instead of n−rank(A)n - \text{rank}(A). The free variables are counted among the unknowns, that is, among the columns.

Wrong working: the 3×53 \times 5 matrix of the worked example in Finding the Null Space has rank 33, so the nullity is taken as 3−3=03 - 3 = 0 and only the trivial solution is expected.

Right working: the nullity is n−rank(A)=5−3=2n - \text{rank}(A) = 5 - 3 = 2. The null space is two-dimensional, spanned by v1=(−2,1,0,0,0)\mathbf{v}_1 = (-2, 1, 0, 0, 0) and v2=(−1,0,−1,1,0)\mathbf{v}_2 = (-1, 0, -1, 1, 0).

Why it happens: mm and nn are both sizes of the same matrix, and for a square matrix the two formulas agree, so the slip shows only on a non-square one.

How to check: the nullity equals the number of columns without a pivot; here columns 22 and 44. A matrix with m<nm < n can never have nullity 00, since its rank is at most mm.

Reading null-space vectors from the rows

After row reduction, the nonzero rows of the echelon form are taken as a basis of the null space. Those rows span the row space, a different subspace; null-space vectors come only from solving the system.

Wrong working: from the echelon form in Finding the Null Space, the rows (1,2,−1,0,3)(1, 2, -1, 0, 3), (0,0,2,2,2)(0, 0, 2, 2, 2) and (0,0,0,0,2)(0, 0, 0, 0, 2) are listed as the basis, giving a three-dimensional null space.

Right working: set the free variables x2=sx_2 = s, x4=tx_4 = t, solve the pivot variables, and set each free variable to 11 in turn: v1=(−2,1,0,0,0)\mathbf{v}_1 = (-2, 1, 0, 0, 0) and v2=(−1,0,−1,1,0)\mathbf{v}_2 = (-1, 0, -1, 1, 0). Both give Av=0A\mathbf{v} = \mathbf{0}.

Why it happens: the reduced rows are the visible result of the elimination, and the last step, turning the reduced equations into vectors, is skipped.

How to check: multiply each claimed basis vector by the original matrix. A(1,2,−1,0,3)=(15,34,−7)A(1, 2, -1, 0, 3) = (15, 34, -7), not 0\mathbf{0}. The count also gives it away: there must be one vector per free variable, two here, while the echelon form has three nonzero rows.

Subtracting λ from every entry

The matrix A−λIA - \lambda I is formed by subtracting λ\lambda from every entry of AA, instead of only from the diagonal entries. λI\lambda I is the matrix with λ\lambda on the diagonal and 00 elsewhere.

Wrong working: for A=(2112)A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix},
det⁡(2−λ1−λ1−λ2−λ)=(2−λ)2−(1−λ)2=3−2λ=0\det\begin{pmatrix} 2 - \lambda & 1 - \lambda \\ 1 - \lambda & 2 - \lambda \end{pmatrix} = (2 - \lambda)^2 - (1 - \lambda)^2 = 3 - 2\lambda = 0

gives the single value λ=3/2\lambda = 3/2.

Right working:
det⁡(2−λ112−λ)=(2−λ)2−1=λ2−4λ+3=(λ−1)(λ−3)=0\det\begin{pmatrix} 2 - \lambda & 1 \\ 1 & 2 - \lambda \end{pmatrix} = (2 - \lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = (\lambda - 1)(\lambda - 3) = 0

so λ=1\lambda = 1 and λ=3\lambda = 3. The eigenvectors are the null spaces of A−I=(1111)A - I = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}, spanned by (1,−1)(1, -1), and of A−3I=(−111−1)A - 3I = \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}, spanned by (1,1)(1, 1).

Why it happens: rewriting Ax=λxA\mathbf{x} = \lambda\mathbf{x} as (A−λ)x=0(A - \lambda)\mathbf{x} = \mathbf{0} skips the step λx=λIx\lambda\mathbf{x} = \lambda I\mathbf{x}, and A−λA - \lambda is then read as "subtract λ\lambda everywhere".

How to check: test the value found. A−32I=(1/2111/2)A - \frac{3}{2}I = \begin{pmatrix} 1/2 & 1 \\ 1 & 1/2 \end{pmatrix} has determinant 1/4−1=−3/4≠01/4 - 1 = -3/4 \neq 0, so (A−32I)x=0(A - \frac{3}{2}I)\mathbf{x} = \mathbf{0} has only the trivial solution and 3/23/2 is not an eigenvalue. For a 2×22 \times 2 matrix the two eigenvalues must also add up to the trace: 1+3=41 + 3 = 4.

Homogeneous Systems FAQ

Does "trivial solution" mean the easy one?

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No, it is a technical label for one specific vector, the all-zeros solution that every homogeneous system has. Nothing about difficulty is implied, and a system whose only solution is the trivial one is the more constrained case. Similarly, nontrivial claims that a second solution exists, not that it is complicated: (1,0,0)(1, 0, 0) qualifies.Read more →

Is the solution set of Ax=bA\mathbf{x} = \mathbf{b} a subspace?

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Only when b=0\mathbf{b} = \mathbf{0}. A homogeneous solution set is the null space and is closed under addition and scaling, which is why the general solution can be written as a span. A non-homogeneous set is a translate of that null space: it has the same shape but misses the origin, so it fails the first subspace test.Read more →

What is the difference between 0\mathbf{0} and OO?

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Shape. Bold 0\mathbf{0} is the zero vector, a single column, while OO is the zero matrix, a rectangular array. The two letters are reserved separately precisely so an equation can hold both without ambiguity, and the bold face is what distinguishes the vector from the plain scalar zero as well.Read more →