Basis For Eigenspace Calculator
An eigenspace basis is the short list of vectors that generates every eigenvector belonging to one eigenvalue. You get it by forming A minus lambda times the identity, row-reducing that matrix to RREF, and reading the basis off the free variables: the eigenspace of A for lambda is exactly the null space of A minus lambda I. This basis for eigenspace calculator does that work for a 2×2 or 3×3 matrix, shows both matrices, and then proves its own answer by multiplying each basis vector back through.
The worked example on this page is the matrix A with rows [4, 1] and [2, 3], at lambda = 5. A minus 5I is the matrix with rows [-1, 1] and [2, -2], its RREF is the matrix with rows [1, -1] and [0, 0] at rank 1, and the basis for the eigenspace is the single vector [1, 1]. Check it by hand: A times [1, 1] is [5, 5], which is 5 times [1, 1]. The calculator is pre-filled with exactly this case.
The zero vector is in every eigenspace but is never part of a basis. A basis lists only the nonzero generators, one per free variable in the RREF.
Quick Answer
Key takeaways
- The eigenspace for lambda is the null space of A minus lambda I. One method, the RREF of that matrix, produces the basis directly from its free variables.
- Each free variable contributes exactly one basis vector: set that variable to 1, the other free variables to 0, and solve for the pivot variables.
- The number of basis vectors is the dimension of the eigenspace, also called the geometric multiplicity of the eigenvalue.
- If A minus lambda I has full rank, the basis is empty because lambda is not an eigenvalue. The calculator says so instead of returning a silent zero.
What an eigenspace basis actually is
An eigenvector of a matrix A is a nonzero vector that A stretches without turning: A v = lambda v for some number lambda, the eigenvalue. Fix one lambda and collect every vector that satisfies that equation, and you get the eigenspace for lambda. It is always a subspace, because it is exactly the set of solutions of (A minus lambda I) v = 0, which makes it the null space of A minus lambda I. A basis for the eigenspace is a minimal generating list for that subspace: independent vectors whose span is the whole eigenspace.
That framing is the standard one in linear algebra courses: the open textbook treatment in the Grand Valley State linear algebra text, chapter on eigenspaces, works an example where the eigenspace for lambda = 1 has basis [1, 1, 0] and [1, 0, 1], a two-dimensional eigenspace that is a plane through the origin. The counting rule behind it is the one Gilbert Strang’s textbook uses for null spaces generally: dimension of the null space equals the number of free variables, which is n minus the rank. That is why this calculator prints the rank of the RREF next to the basis: basis vectors plus rank always add back up to n.
The companion computation lives one page over: a basis for the null space of a matrix itself, with no lambda subtracted, is what the basis for null space calculator finds. An eigenspace basis is that same computation run on A minus lambda I, so if the null space method is clear to you, this page is a single extra subtraction.
The method, step by step
The calculator runs four steps, and each one can be checked against the output it displays.
Form A minus lambda I. Subtract lambda from each diagonal entry of A and leave the off-diagonal entries alone. For the worked example at lambda = 5, [4, 1] and [2, 3] become [-1, 1] and [2, -2].
Row-reduce to RREF. Gauss-Jordan elimination with partial pivoting turns A minus lambda I into reduced row echelon form. The example reduces to [1, -1] and [0, 0], with rank 1: one pivot, one zero row.
Read the free variables. Columns without a pivot are free. Here the second column is free, so x2 is free and the pivot variable is x1 = x2 from the RREF row.
Build one vector per free variable. Set the free variable to 1 and solve for the pivot variables: x2 = 1 gives x1 = 1, so the basis vector is [1, 1]. The tool then re-multiplies each basis vector by A minus lambda I and reports the largest residual, which is 0 for an exact answer.
Worked example: the matrix with rows [4, 1] and [2, 3] at lambda = 5
The calculator is pre-filled with this case, and the table below is what it returns, followed by two more runs on the same matrix that show the range of possible answers.
| Run | A minus lambda I | RREF and rank | Basis |
|---|---|---|---|
| lambda = 5 | [-1, 1] and [2, -2] | [1, -1] and [0, 0], rank 1 | v1 = [1, 1], dimension 1 |
| lambda = 2 | [2, 1] and [2, 1] | [1, 0.5] and [0, 0], rank 1 | v1 = [-0.5, 1], dimension 1 |
| lambda = 7 | [-3, 1] and [2, -4] | Identity, rank 2 | None: 7 is not an eigenvalue |
Each basis vector survives the defining test. At lambda = 5, A times [1, 1] is [4 + 1, 2 + 3] = [5, 5], and 5 times [1, 1] is also [5, 5]. At lambda = 2, A times [-0.5, 1] is [-2 + 1, -1 + 3] = [-1, 2], and 2 times [-0.5, 1] is [-1, 2]. The scaling of a basis vector is arbitrary: [2, 2] spans the same eigenspace as [1, 1], and [-1, 2] is the same answer as [-0.5, 1] scaled by 2. What cannot change is the direction of each vector and the count of vectors.
For a 3×3 case with a wider eigenspace, switch the size to 3 and enter the diagonal matrix with diagonal entries 3, 3, 2 at lambda = 3. A minus 3I is diagonal with entries 0, 0, -1, rank 1, and the basis is [1, 0, 0] and [0, 1, 0]: a two-dimensional eigenspace, geometric multiplicity 2, matching how many times 3 repeats on the diagonal.
Dimension, geometric multiplicity, and the defective case
The dimension of the eigenspace is called the geometric multiplicity of the eigenvalue, and it is always at least 1 for a genuine eigenvalue and never more than the algebraic multiplicity, the number of times lambda appears as a root of the characteristic polynomial. When the two agree for every eigenvalue, the eigenvectors can form a basis for the whole space and the matrix is diagonalizable. When an eigenspace comes up short, the matrix is defective: the classic case is the matrix with rows [2, 1] and [0, 2] at lambda = 2, where A minus 2I is [0, 1] and [0, 0], the basis is the single vector [1, 0], and geometric multiplicity 1 falls short of algebraic multiplicity 2. Run it in the calculator: the rank-1 RREF and the single basis vector are exactly the defect, made visible.
| Situation | Rank of A minus lambda I | Basis size | Meaning |
|---|---|---|---|
| Simple eigenvalue, 2×2 | 1 | 1 | Eigenspace is a line through the origin |
| Repeated eigenvalue, full eigenspace, 3×3 | 1 | 2 | Eigenspace is a plane; geometric multiplicity 2 |
| Repeated eigenvalue, defective, 2×2 | 1 | 1 | Geometric 1 below algebraic 2; not diagonalizable |
| lambda not an eigenvalue | n (full) | 0 | Eigenspace is only the zero vector |
What competing calculators miss
The tools ranking for this search tend to hide the working. dCode returns eigenspaces for matrices of any size, which is genuinely broader than the 2×2 and 3×3 scope here, but it returns an answer with no RREF, no rank, and no check, so a student copying homework gets nothing to learn from and no way to spot an input slip. Eigenvalue-first calculators such as Pearson’s solve a different problem: they find lambda for you and list eigenvectors, which is the wrong shape when your assignment hands you lambda and asks for the eigenspace basis, its dimension, and its multiplicity.
This page is built around the intermediate mathematics instead. A minus lambda I and its RREF are displayed with the rank, the basis is read from the free variables in the open, and every basis vector is verified by multiplication with the residual printed. The not-an-eigenvalue case gets a plain verdict rather than an empty box. For the sizes coursework actually uses, showing the method is worth more than accepting a 6×6 matrix and showing nothing.
One limit to keep in mind: the arithmetic is floating point, rounded to four decimals in the display. For matrices with irrational or complex eigenvalues, a decimal lambda typed in by hand is an approximation, and the basis inherits that approximation. For the rational, integer-heavy matrices of a linear algebra course, the answers here are exact.
Frequently asked questions
Form A minus lambda I by subtracting the eigenvalue from the diagonal, row-reduce it to RREF, and read the basis from the free variables. Each free variable gives one basis vector: set it to 1, set the other free variables to 0, and solve for the pivot variables from the RREF rows.
It is a null space: the eigenspace for lambda is the null space of A minus lambda I, meaning every vector v with (A minus lambda I) v = 0. It is not the null space of A itself, unless lambda is 0. The subtraction is the whole difference.
For the matrix with rows [4, 1] and [2, 3] at lambda = 5, the basis is the single vector [1, 1]. It checks out directly: A times [1, 1] is [5, 5], which is 5 times [1, 1], so [1, 1] is an eigenvector for the eigenvalue 5 and it spans the one-dimensional eigenspace.
It means A minus lambda I has full rank, so its only null vector is the zero vector, which is never part of a basis. The eigenspace is trivial. With the example matrix, lambda = 7 gives a rank-2 A minus 7I, and the calculator reports exactly that instead of returning a basis.
Dimension equals n minus the rank of A minus lambda I, which is also the number of free variables in the RREF and the number of basis vectors. It is called the geometric multiplicity of the eigenvalue. A 3×3 diagonal matrix with entries 3, 3, 2 at lambda = 3 has rank 1 for A minus 3I, so the dimension is 2.
Yes, whenever the eigenvalue has geometric multiplicity above 1. That happens when A minus lambda I has two or more free variables. The diagonal example on this page returns two basis vectors, [1, 0, 0] and [0, 1, 0], and the eigenspace is the plane they span.
A basis must be linearly independent, and any set containing the zero vector is dependent by definition: 1 times the zero vector is zero with a nonzero coefficient. The zero vector belongs to every eigenspace, but it never generates anything, so it is never listed.
No. Any nonzero rescaling of a basis vector, such as [2, 2] instead of [1, 1], is an equally valid basis for a one-dimensional eigenspace, and for wider eigenspaces any independent set spanning the same space works. What is fixed is the space itself, its dimension, and the direction of each generator in the rank-1 case.
Yes. Pick 3 x 3 in the size control, fill all nine entries, and enter lambda. The method is identical: A minus lambda I, RREF, free variables, one basis vector each, with the residual check on every vector.
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References
- Grand Valley State University, “Eigenspaces of a Matrix”, open linear algebra textbook chapter: defines the eigenspace as the null space of A minus lambda I and works a two-dimensional basis example. GVSU: Eigenspaces of a Matrix. Accessed October 2026.
- ProofWiki, “Book: Gilbert Strang, Introduction to Linear Algebra, Fifth Edition”: contents of the standard textbook treatment, including the nullspace and eigenvalues chapters this method comes from. ProofWiki: Strang, Introduction to Linear Algebra. Accessed October 2026.
- MIT OpenCourseWare, “Linear Algebra (18-06SC)”: Gilbert Strang’s course covering eigenvalues, eigenvectors, and diagonalization. MIT OCW: Linear Algebra 18-06SC. Accessed October 2026.
- Silo, “Solutions to Introduction to Linear Algebra by Gilbert Strang”: worked solutions for nullspace basis problems using free variables and special solutions. Solutions to Strang: Introduction to Linear Algebra. Accessed October 2026.
- 123dok, Strang, “The Fundamental Theorem of Linear Algebra”, American Mathematical Monthly 100(9), 1993: the rank and null space relationships behind the dimension count. Strang: The Fundamental Theorem of Linear Algebra. Accessed October 2026.
- dCode, “Eigenspaces of a Matrix”: the definition of an eigenspace as the kernel of M minus lambda I, used as a competitor benchmark for this article. dCode: Eigenspaces of a Matrix. Accessed October 2026.