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Home Linear Systems Left_Null_Space

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.

Matrix Dimensions

Rows (m) =

Cols (n) =

Maximum size: 6×6

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

A Matrix A (m × n)

The left null space is the null space of A^T

Computing left null space basis...

Load Example:

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

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

Learn About Left_Null_Space

Understanding the concepts behind the calculations.

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)

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

Method 2: From RREF (Advanced)

Compute the RREF of A
Identify the zero rows
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:

Subspace	Notation	Located in	Dimension	Built with
Column Space	Col(A)	ℝᵐ	rank(A)	Columns of A
Left Null Space	Null(Aᵀ)	ℝᵐ	m - rank(A)	Null space of Aᵀ
Row Space	Row(A)	ℝⁿ	rank(A)	Rows of A
Null Space	Null(A)	ℝⁿ	n - 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 Nullity	Geometry in ℝᵐ	Column Space
0	Only the origin (point)	All of ℝᵐ (full dimension)
1	A line through the origin	A hyperplane (dimension m-1)
2	A plane through the origin	(m-2)-dimensional subspace
m	All of ℝᵐ	Only 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

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

Intermediate

Find a basis for the left null space of:

$$ A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \end{bmatrix} $$
For the matrix above, verify that each basis vector is orthogonal to every column of A.

Advanced

Prove that Col(A) and Null(Aᵀ) are orthogonal complements in ℝᵐ.
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:

Enter your matrix A (any size up to 6×6)
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!

Linear Systems Solvers

Solving Systems

Decompositions

Vector Spaces

Need Help?

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

Calculator

A Matrix A (m × n)

Learn About Left_Null_Space

📑 Quick Navigation

What is the Left Null Space?

The Core Concept

Why "Left" Matters

How to Find the Left Null Space

Method 1: Via Transpose (Easiest)

Method 2: From RREF (Advanced)

Complete Example

The Four Fundamental Subspaces

Perfect Orthogonal Relationships

Key Properties

Geometric Interpretation

Real-World Applications

🔍 System Consistency

📊 Least Squares Regression

🔗 Linear Dependencies

🎮 Control Theory

📈 Linear Regression

Frequently Asked Questions

Practice Problems

Beginner

Intermediate

Advanced

Related Linear Algebra Tools

📐 Null Space & Column Space

⚡ System Solvers

Summary

🎯 Key Takeaways

Try It Yourself!