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🧮 Linear Algebra Kingdom · Linear Algebra

Kernel, Range & Rank–Nullity

What a transformation loses and what it can reach: decide whether a vector is sent to zero, find the dimension of the kernel and a vector that spans it, decide whether a target is in the range, and use rank plus nullity to account for every dimension of the input.

In short

  • The kernel is everything sent to the zero vector; it always holds that vector, so it is never empty, and its dimension is the number of columns without a pivot.
  • The range is the column space — everything the columns can build between them — and its dimension is the rank.
  • b is in the range exactly when the augmented matrix has no row reading 0 = something not zero, which is the same as the two ranks agreeing.
  • Rank plus nullity is the number of columns, so any two of the three give the third — and a nullity of 0 is exactly what one-to-one means.

What a map loses: the kernel

Walk into one of the black patches on the isle and every direction takes you to the same point. That is what a transformation does when it has a kernel: a whole set of inputs, all flattened onto the origin.

The kernel of T is every vector x with T(x) = 0. If T is given by a matrix, that is the solution set of Ax = 0 — the same equation as always, with a right-hand side of zeros that stays zero the whole way through the row reduction. So you never write the augmented column down at all; you reduce A and read the answer off it.

The kernel is never empty. T(0) = 0 for every linear map, so the zero vector is always in there. When nothing else joins it, the kernel is a single point — the origin — and its dimension is 0. That is a perfectly ordinary answer, and the phrase to avoid is "the kernel is empty", which is never true of anything.

To find its dimension, reduce A and look at the columns. Every column with a pivot pins its unknown to a value; every column without one leaves an unknown free to be anything, and each free unknown adds a dimension:

dimension of the kernel = number of columns without a pivot

That number is also called the nullity of A, and the kernel itself is also called the null space.

Reading a spanning vector off the free column

When exactly one column comes out free, the kernel is a single line through the origin, and one vector is enough to describe all of it. Here is how to read that vector off the reduced form.

Take A = [[1, 2, 3], [0, 1, 1]]. Reducing it clears the 2 above the second pivot and leaves [[1, 0, 1], [0, 1, 1]]. Pivots sit in columns 1 and 2, so the third unknown is the free one.

Now read each reduced row as an equation and move the free unknown across, changing sign as it goes:

x1 + x3 = 0 so x1 = -x3 x2 + x3 = 0 so x2 = -x3

Choose x3 = 1 and the vector is (-1, -1, 1). Every multiple of it is in the kernel too, so there are infinitely many correct-looking answers — which is why a question always asks for one particular one: whole entries, no common factor, and the first non-zero entry positive. Multiplying by -1 gives (1, 1, -1), and that is the one to write down.

Two habits keep this clean. First, always check at the end: A times your vector really should be the zero vector, and it takes ten seconds. Second, if the reduced form leaves a fraction, choose the free unknown to be the denominator instead of 1, so every entry lands whole before you scale it down.

What a map can reach: the range

The range of T is everything it can hand back — every vector of the form Ax. Since Ax is a combination of the columns of A with the entries of x as weights, the range is exactly the set of combinations of the columns, which is why it is also called the column space of A.

So "is b in the range?" is the same question as "can Ax = b be solved?", and row reduction answers it. Build the augmented matrix with b in the last column and reduce. There are two things it can say:

  • a row reading 0 = something that is not zero means the system has no solution, so b is outside the range
  • no such row means b is in the range, and then the free columns decide whether there is exactly one solution or infinitely many

The tidy way to say the same thing is with ranks. The rank of A is its number of pivots, and b is in the range exactly when the rank of A and the rank of the augmented matrix agree — a pivot appearing in that last column is precisely the contradictory row.

The rank is also the dimension of the range: the pivot columns are the ones that genuinely add something new, and the rest are combinations of them. So the range of a map into R3 can be a plane, or a line, or just the origin — it does not have to fill the space it lives in, and it fills it exactly when every row has a pivot.

Rank and nullity add up

Here is the accounting rule that ties the two halves together. For a matrix A with n columns:

rank + dimension of the kernel = n

Every column either carries a pivot or it does not. The ones that do build the range; the ones that do not leave an unknown free and build the kernel. Nothing is counted twice and nothing is left out, so the two dimensions have to add up to the number of columns. This is the rank-nullity relation, and it means that of the three numbers, any two give you the third.

Two things about it are worth saying out loud, because both are slipped on constantly.

It counts columns, not rows. The number n is the dimension of the domain — the space the vectors come from. The number of rows is the dimension of the space the images land in, and it does not appear in the relation at all.

Rank and nullity are not interchangeable. The rank measures what survives; the nullity measures what is lost. If a map from R5 has rank 3, its kernel has dimension 2, not 3.

Finally, the relation explains why one number decides so much. A kernel of dimension 0 means no free columns, which means every column has a pivot, which means T is one-to-one — and a rank equal to the number of rows means every row has a pivot, which means T is onto. One reduction, and every question in this craft is answered.

Worked examples

Example 1

A = [[1, 2, 3], [0, 1, 1]]. Find the dimension of the kernel of T(x) = Ax, and a vector that spans it.

  1. Reduce A. Subtracting 2 times the second row from the first gives [[1, 0, 1], [0, 1, 1]], with pivots in columns 1 and 2.
  2. A has 3 columns and 2 pivots, so 3 - 2 = 1 column is free and the kernel has dimension 1: it is a single line.
  3. The free unknown is the third. The rows say x1 = -x3 and x2 = -x3, so choosing x3 = 1 gives (-1, -1, 1).
  4. Scaled the way a question asks — no common factor, first non-zero entry positive — that is (1, 1, -1). Check: A times (1, 1, -1) is (1 + 2 - 3, 0 + 1 - 1) = (0, 0).

Example 2

A = [[1, 0], [0, 1], [1, 1]] and b = (1, 2, 5). Is b in the range of T(x) = Ax?

  1. Build the augmented matrix [[1, 0 | 1], [0, 1 | 2], [1, 1 | 5]] and reduce it.
  2. Subtracting the first row and then the second row from the third leaves a bottom row of [0, 0 | 2].
  3. That row reads 0 = 2, which nothing can satisfy, so Ax = b has no solution.
  4. Equivalently: A has rank 2 while the augmented matrix has rank 3, and the extra pivot sits in the last column. So b is not in the range — the range here is a plane in R3, and b is off it.

Example 3

T is a linear transformation from R5 to R3 with rank 3. What is the dimension of its kernel, and is T onto?

  1. Rank-nullity says rank + dimension of the kernel = the number of columns, which is 5 because the domain is R5.
  2. So 3 + dimension of the kernel = 5, giving a kernel of dimension 2.
  3. The rank, 3, is also the dimension of the range, and the images live in R3, so the range fills the whole codomain and T is onto.
  4. T is not one-to-one, though: the kernel holds more than the zero vector, so different inputs share images. Note that the codomain being R3 never entered the rank-nullity line — only the 5 columns did.

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

T(x) = Ax is a linear transformation with A = [[1, 4], [4, 0]] and v = [0, 2]. Is v in the kernel of T? Pick the statement that is true.

  1. No: Av = [0, 0], which is not the zero vector.
  2. No: Av = [8, 0], which is not the zero vector.
  3. No: Av = [9, 0], which is not the zero vector.
  4. Yes: Av is the zero vector, so v is in the kernel of T.

Answer: B. No: Av = [8, 0], which is not the zero vector.

  1. Row 1: (1)(0) + (4)(2) = 8
  2. Row 2: (4)(0) + (0)(2) = 0
  3. So Av = [8, 0].
  4. At least one entry is not 0, so Av is not the zero vector and v is outside the kernel of T.

Problem 2

Difficulty 3 of 5

T(x) = Ax is a linear transformation from R2 to R3, with A = [[-5, 5], [0, 1], [3, -1]] What is the dimension of the kernel of T?

Answer: 0

  1. Row reducing A leaves 2 pivots, in columns 1 and 2, so the rank of A is 2.
  2. A has 2 columns, so 2 - 2 of them carry no pivot and their unknowns are free.
  3. The dimension of the kernel is therefore 0.

Problem 3

Difficulty 4 of 5

T(x) = Ax is a linear transformation from R4 to R3, with A = [[0, 1, -1, 1], [2, -1, 1, 0], [4, -1, -1, 1]] The kernel of T is a single line through the origin. Find a vector that spans it. Every multiple of that vector spans the same line, so give the one with whole entries, no common factor, and a positive first non-zero entry. Write the vector as its entries separated by commas, like 1,-2,3.

Answer: [1,2,0,-2]

  1. Row reducing A puts pivots in columns 1, 2 and 3, leaving the 4th column free.
  2. Each reduced row reads "pinned unknown = -(entry) times the free unknown", so choosing the free unknown to clear the fractions gives [1, 2, 0, -2].
  3. Checking: A times [1, 2, 0, -2] is the zero vector, and the entries have no common factor with a positive first non-zero entry, so this is the vector the question asks for.

Common mistakes

  • Saying the kernel is empty when nothing but the zero vector is sent to zero. The zero vector is always in the kernel; the right word is dimension 0.
  • Counting rank-nullity against the rows of the matrix. The relation adds up to the number of columns, because that is the dimension of the domain.
  • Swapping rank and nullity, and reporting the number of pivots as the dimension of the kernel. The pivots measure what survives; the free columns measure what is lost.
  • Giving a spanning vector of the null space without scaling it as asked, or with the sign of the free unknown left the other way round.

What you should be able to do

  • Decide whether a vector lies in the kernel of a transformation, and find a vector that spans the kernel.
  • Find the dimension of the kernel and of the column space of a matrix.
  • Decide whether a vector lies in the range of a transformation.
  • Apply the rank-nullity relation, and connect a kernel of only the zero vector to a one-to-one map.

Where this fits in the curriculum

Common Core

  • HSN-VM.C.11

    High school — Multiply a vector (regarded as a matrix with one column) by a matrix of suitable dimensions to produce another vector; work with matrices as transformations of vectors.

    HSN-VM.C.11 is a (+) standard — beyond the college- and career-ready threshold, i.e. precalculus rather than Algebra II. "Matrices as transformations of vectors" is as far as the Common Core takes this. What a transformation sends to zero, what it can reach, and the rank-nullity relation that ties the two together are named in no published framework.

  • HSN-VM.C.12

    High school — Work with 2 × 2 matrices as transformations of the plane, and interpret the absolute value of the determinant in terms of area.

    HSN-VM.C.12 is a (+) standard — beyond the college- and career-ready threshold, i.e. precalculus rather than Algebra II. A determinant of zero is the area collapsing, which is the plane's version of a kernel larger than the zero vector.

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