BYTETOOLS

Null Space (Kernel) Calculator

Solve A·x = 0 to get a basis for the null space, its dimension, and the general solution in parametric vector form — all in exact fractions.

1
Nullity (dim)
2
Rank
3×3
Size
x₃
Free variables

Kernel basis — 1 vector

v1 = (1, -2, 1)from free variable x₃

General solution in parametric vector form

x = t₁·(1, -2, 1)

t₁ can be any real number. Every solution of A·x = 0 is one of these combinations, and no smaller set of vectors spans them.

Reduced row echelon form of A

10-1
012
000

Pivot columns are highlighted; the remaining columns are the free variables. Each basis vector sets one free variable to 1, the others to 0, and reads the pivot values back with the opposite sign.

What is the Null Space (Kernel) Calculator?

The ByteTools Null Space Calculator solves the homogeneous system A·x = 0 and returns a basis for the kernel — the set of all vectors the matrix sends to zero.

  • Exact fraction arithmetic — basis vectors print as clean fractions
  • One basis vector per free variable, each labelled with its variable
  • General solution written in parametric vector form x = t₁v₁ + t₂v₂
  • Trivial kernel stated clearly rather than shown as an empty result
  • Shows the RREF with pivot columns highlighted so you can follow the working
  • Handles matrices up to 10×10, entirely in your browser

How to use the Null Space (Kernel) Calculator

  1. 1

    Enter matrix A with one row per line, values separated by spaces or commas.

  2. 2

    Read the nullity tile — that is how many basis vectors the kernel has.

  3. 3

    Copy the basis vectors, each labelled with the free variable it came from.

  4. 4

    Read the general solution in parametric vector form to describe every solution at once.

  5. 5

    Check the RREF panel at the bottom to see which columns are pivots and which are free.

About the Null Space (Kernel) Calculator

The ByteTools Null Space Calculator solves the homogeneous system A·x = 0 and returns a basis for the kernel — the set of all vectors the matrix sends to zero. It reduces A to reduced row echelon form with exact rational arithmetic, splits the pivot columns from the free columns, and produces one basis vector per free variable by setting that variable to 1 and the other free variables to 0.

You get the nullity (the dimension of the null space), the free variables named, each basis vector written out, and the general solution in parametric vector form — x = t₁v₁ + t₂v₂ and so on. When every column is a pivot column the tool says so plainly: the kernel is trivial, containing only the zero vector, which means the columns of A are linearly independent.

All of it is computed by JavaScript in your browser. Nothing is uploaded to a server, nothing is stored, and it works offline once the page has loaded.

Frequently asked questions

What is the null space of a matrix?

The null space, or kernel, is the set of every vector x for which A·x = 0. It is always a subspace, it always contains the zero vector, and its dimension is called the nullity. Geometrically it is the set of directions the matrix collapses to nothing.

How do you find a basis for the null space?

Row-reduce A to RREF, identify which columns hold pivots and which are free, then build one vector per free variable: set that free variable to 1, the other free variables to 0, and read the pivot variables off the reduced rows. This calculator does exactly that and shows the RREF it used.

What does it mean if the null space is only the zero vector?

It means the nullity is 0 and every column is a pivot column, so the columns of A are linearly independent and A·x = 0 forces x = 0. For a square matrix this is the same as being invertible.

What is the difference between null space and column space?

The null space lives in the input space and holds the vectors sent to zero; the column space lives in the output space and holds everything the matrix can produce. Their dimensions add up neatly through the rank-nullity theorem: nullity + rank = number of columns.

Is the null space basis unique?

No. Any set of independent vectors spanning the same subspace is a valid basis, so different methods can give different-looking answers. The parametric basis this tool produces — one vector per free variable — is the standard textbook form, and the dimension is always the same.

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