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Left Null Space Calculator: Find Left Null Space of a Matrix

Find the left null space of a matrix A: all vectors y such that y^T A = 0 (or equivalently A^T y = 0). The left null space is the orthogonal complement of the column space.

Calculator

Enter your matrix below and click "Calculate" to see the step-by-step solution.

Rows (m) =
Cols (n) =
Maximum size: 6×6

The left null space consists of all y such that yTA = 0

A Matrix A (m × n)

The left null space is the null space of AT

Computing left null space basis...

Enter a matrix A to find its left null space (all y such that yTA = 0).

The left null space is the orthogonal complement of the column space.

Learn About Left_Null_Space

Understanding the concepts behind the calculations.

📑 Quick Navigation

What is the Left Null Space?

The left null space of a matrix A is the set of all vectors y such that yᵀA = 0 (equivalently, Aᵀy = 0).

$$ \boxed{\text{Null}(A^T) = \{\mathbf{y} \in \mathbb{R}^m : A^T\mathbf{y} = \mathbf{0}\} = \{\mathbf{y} \in \mathbb{R}^m : \mathbf{y}^T A = \mathbf{0}^T\}} $$

For an m × n matrix A, the left null space lives in ℝᵐ (the same space as the columns of A).

💡 Key Insight: The name "left" comes from the fact that vectors in this space multiply A on the left to give zero: yᵀA = 0ᵀ.

The Core Concept

The left null space contains all vectors that are orthogonal to every column of A.

$$ \mathbf{y}^T A = \mathbf{0}^T \quad \Longleftrightarrow \quad \mathbf{y} \cdot \mathbf{a}_j = 0 \ \text{for every column } \mathbf{a}_j $$

Why "Left" Matters

Right Null Space (Null(A)):

$$A\mathbf{x} = \mathbf{0}$$

Vectors multiply on the right

Left Null Space (Null(Aᵀ)):

$$\mathbf{y}^T A = \mathbf{0}^T$$

Vectors multiply on the left

🎯 The Big Idea: The left null space tells you exactly how rows of A are linearly dependent. Every vector in the left null space gives a linear combination of rows that equals zero!

How to Find the Left Null Space

Method 1: Via Transpose (Easiest)

  1. Compute Aᵀ (the transpose)
  2. Find the null space of Aᵀ using Gaussian elimination
  3. The basis vectors form the left null space

Method 2: From RREF (Advanced)

  1. Compute the RREF of A
  2. Identify the zero rows
  3. Extract coefficients from row operations

💡 Recommendation: Method 1 (via transpose) is straightforward. Use our Null Space Calculator on Aᵀ to find the left null space!

Complete Example

Problem: Find the left null space of

$$ A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \\ 3 & 6 \end{bmatrix} $$

Step 1: Compute Aᵀ

$$ A^T = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \end{bmatrix} $$

Step 2: Set up Aᵀy = 0

$$ \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$

Step 3: Row reduce to REF

$$ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \end{bmatrix} $$

Step 4: Identify free variables

Pivot in column 1 → y₁ is basic. Free variables: y₂, y₃

Step 5: Solve for basic variable

$$ y_1 + 2y_2 + 3y_3 = 0 \implies y_1 = -2y_2 - 3y_3 $$

Step 6: Write general solution

$$ \mathbf{y} = y_2\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} + y_3\begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix} $$

✅ Left Null Space Basis:

$$ \left\{ \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix},\ \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix} \right\} $$

Dimension = 2 = m - rank(A) = 3 - 1 ✓

📐 Interpretation: These basis vectors give the coefficients for linear dependencies among rows:

-2·[1 2] + 1·[2 4] + 0·[3 6] = [0 0] ✓

-3·[1 2] + 0·[2 4] + 1·[3 6] = [0 0] ✓

The Four Fundamental Subspaces

For an m × n matrix A, there are four fundamental subspaces that completely describe the linear transformation:

SubspaceNotationLocated inDimensionBuilt with
Column SpaceCol(A)ℝᵐrank(A)Columns of A
Left Null SpaceNull(Aᵀ)ℝᵐm - rank(A)Null space of Aᵀ
Row SpaceRow(A)ℝⁿrank(A)Rows of A
Null SpaceNull(A)ℝⁿn - rank(A)Solutions to Ax = 0

Perfect Orthogonal Relationships

In ℝᵐ (output space):

$$\text{Col}(A) \perp \text{Null}(A^T)$$

Column Space ⟂ Left Null Space

These are orthogonal complements in ℝᵐ

In ℝⁿ (input space):

$$\text{Row}(A) \perp \text{Null}(A)$$

Row Space ⟂ Null Space

These are orthogonal complements in ℝⁿ

🎯 Key Takeaway: Any vector y ∈ ℝᵐ can be uniquely written as y = y_c + y_n where y_c ∈ Col(A) and y_n ∈ Null(Aᵀ).

Key Properties

  • Dimension formula: dim(Null(Aᵀ)) = m - rank(A)
  • Left nullity = m - rank(A) (the number of zero rows in RREF)
  • Orthogonal to column space: Every left null vector is perpendicular to every column
  • Reveals row dependencies: Coefficients of linear combinations that give zero rows
  • System consistency: For Ax = b to have a solution, b must be ⟂ to Null(Aᵀ)
  • Residuals in least squares: The residual r = b - Ax lies in Null(Aᵀ)
$$ \boxed{\mathbb{R}^m = \text{Col}(A) \oplus \text{Null}(A^T)} $$

Every vector in ℝᵐ decomposes uniquely into a column space component + left null space component.

Geometric Interpretation

The left nullity (m - rank(A)) tells you the "missing dimensions" in the column space:

Left NullityGeometry in ℝᵐColumn Space
0Only the origin (point)All of ℝᵐ (full dimension)
1A line through the originA hyperplane (dimension m-1)
2A plane through the origin(m-2)-dimensional subspace
mAll of ℝᵐOnly the zero vector

Example: For a 3×2 rank-2 matrix:

  • m = 3, rank = 2
  • Left nullity = 3 - 2 = 1
  • Left null space = a line through origin in ℝ³
  • Column space = a plane through origin
  • The line is perpendicular to the plane!

Real-World Applications

🔍 System Consistency

The linear system Ax = b has a solution if and only if b is orthogonal to every vector in Null(Aᵀ).

$$\mathbf{y}^T b = 0 \quad \forall \mathbf{y} \in \text{Null}(A^T)$$

📊 Least Squares Regression

The residual r = b - Ax̂ (the error) lies in Null(Aᵀ), meaning residuals are orthogonal to the column space.

→ Learn more about Least Squares

🔗 Linear Dependencies

Vectors in the left null space give coefficients that combine rows to zero. Perfect for finding redundant equations!

🎮 Control Theory

The left null space determines unobservable states in linear systems—states you cannot detect from outputs.

📈 Linear Regression

In statistics, the residuals are orthogonal to the column space (i.e., lie in the left null space). Essential for least squares estimation.

💡 Key Insight: The left null space is the "error space" of linear regression. The difference between actual and predicted values must lie in the left null space!

Frequently Asked Questions

Q: What's the difference between left null space and regular null space?

A: Regular null space (Null(A)) contains vectors that multiply A on the right to give zero (Ax = 0). Left null space (Null(Aᵀ)) contains vectors that multiply A on the left to give zero (yᵀA = 0ᵀ). They live in different spaces!

Q: Why is it called "left" null space?

A: Because vectors in this space are placed on the left of A: yᵀA = 0ᵀ. Regular null space vectors go on the right: Ax = 0.

Q: How do I find left null space without computing transpose?

A: You can find it from the RREF of A by looking at the zero rows and the row operations used. But the transpose method is much easier! Use our Null Space Calculator on Aᵀ.

Q: What does left null space tell me about row dependencies?

A: Every vector in the left null space gives coefficients for a linear combination of rows that equals zero. If left null space is non-trivial, rows are linearly dependent!

Q: Is the left null space related to the column space?

A: Yes! They are orthogonal complements in ℝᵐ. Every vector in ℝᵐ can be uniquely decomposed into a column space part + left null space part.

Practice Problems

Beginner

  1. Find the dimension of the left null space for a 4×3 matrix with rank 2.
  2. If A is 5×5 with rank 3, what is dim(Null(Aᵀ))?
  3. Can a 2×3 matrix have left null space dimension 0? Why or why not?

Intermediate

  1. Find a basis for the left null space of:

    $$ A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \end{bmatrix} $$

  2. For the matrix above, verify that each basis vector is orthogonal to every column of A.

Advanced

  1. Prove that Col(A) and Null(Aᵀ) are orthogonal complements in ℝᵐ.
  2. Show that the system Ax = b is consistent iff b ⟂ Null(Aᵀ).
Click to reveal solutions

1. dim = m - rank = 4 - 2 = 2

2. dim = m - rank = 5 - 3 = 2

3. No. For 2×3, m = 2. min left nullity = 2 - min(rank) = 2 - 2 = 0? Actually rank ≤ min(m,n) = 2, so nullity = 2 - rank. If rank=2, nullity=0. So YES, possible!

4. Basis: { [-2, 1, 0]ᵀ, [-3, 0, 1]ᵀ }

5. Dot with column 1: (-2)(1) + 1(2) + 0(3) = 0 ✓. Dot with column 2: (-2)(2) + 1(4) + 0(6) = 0 ✓. (And for second vector).

6. Hint: Use the fact that rank(A) + nullity(Aᵀ) = m and dimension arguments with orthogonality.

7. Hint: b must be in Col(A). By orthogonal decomposition, b is orthogonal to any vector in Null(Aᵀ).

Summary

🎯 Key Takeaways

  • Left Null Space = Null(Aᵀ) = {y | yᵀA = 0ᵀ}
  • Dimension: m - rank(A) (the "missing" dimensions in column space)
  • Orthogonal complement of the column space in ℝᵐ
  • Reveals linear dependencies among rows of A
  • Essential for system consistency, least squares, and control theory

💡 Pro Tip: Finding left null space = finding null space of Aᵀ. Use our Null Space Calculator on the transpose!

Try It Yourself!

Use the calculator above to find left null spaces:

  1. Enter your matrix A (any size up to 6×6)
  2. Click "Calculate" to see:
    • The left null space basis vectors
    • Dimension of the left null space
    • Relationship to row dependencies
    • Step-by-step computation via Aᵀ

📐 Try these examples:

  • 3×2 rank-2 matrix: [[1,2],[3,4],[5,6]] → left nullity = 1 (a line)
  • 2×3 rank-1 matrix: [[1,2,3],[2,4,6]] → left nullity = 1
  • Square singular matrix: [[1,2],[2,4]] → left nullity = 1

🔗 Next Steps: Once you've mastered the left null space, explore the Column Space and Null Space to complete the four fundamental subspaces!