Rank of a matrix using elementary row/column operations

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Rank of a Matrix – Linear Algebra

Rank of a Matrix Using Elementary Row and Column Operations

😊 Hello, students! The rank of a matrix is one of the most fundamental concepts in linear algebra. It tells us the “true size” of a matrix — how many independent equations or directions it really contains. Today, we’ll explore rank deeply using elementary row operations, connect it to linear systems, and see why it’s so powerful.

1. What Is the Rank of a Matrix?

The rank of a matrix \( A \), denoted \( \text{rank}(A) \), is the number of non-zero rows in any row echelon form (REF) of \( A \).

Equivalently, it is the maximum number of linearly independent rows (or columns) of \( A \).

💡 Key insight: Rank is invariant under elementary row operations. That’s why we can reduce \( A \) to REF or RREF to find it!

2. Why Elementary Row Operations Don’t Change Rank

Each elementary row operation is reversible and corresponds to multiplying by an invertible matrix. Hence:

  • Row replacement: doesn’t change the span of the rows.
  • Row interchange: just reorders rows.
  • Row scaling (by \( k \ne 0 \)): doesn’t change linear dependence.

Therefore, the row space (and its dimension) remains unchanged.

🤔 Remark: While row operations preserve row rank, column operations preserve column rank. A deep theorem says: row rank = column rank. So we just say “rank”!

3. How to Compute Rank: Step-by-Step

Example: Find the rank of \[ A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 1 \end{bmatrix} \]

Step 1: \( R_2 \to R_2 – 2R_1 \), \( R_3 \to R_3 – R_1 \)

\[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & -1 & -2 \end{bmatrix} \]

Step 2: Swap \( R_2 \leftrightarrow R_3 \)

\[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & -1 & -2 \\ 0 & 0 & 0 \end{bmatrix} \]

✅ This is in REF. There are 2 non-zero rows ⇒ \( \text{rank}(A) = 2 \).

😠 Common mistake: Counting zero rows or forgetting to fully reduce. Always go to REF (or better, RREF) before counting!

4. Rank and Consistency of Linear Systems

Consider a system \( A\mathbf{x} = \mathbf{b} \) with augmented matrix \( [A \mid \mathbf{b}] \).

The system is consistent if and only if

\[ \text{rank}(A) = \text{rank}([A \mid \mathbf{b}]) \]

If consistent:

  • Unique solution ⇔ \( \text{rank}(A) = n \) (number of variables)
  • Infinitely many solutions ⇔ \( \text{rank}(A) < n \)

System: \[ \begin{cases} x + y = 2 \\ 2x + 2y = 4 \\ x + y = 1 \end{cases} \]

Augmented matrix:

\[ \left[\begin{array}{cc|c} 1 & 1 & 2 \\ 2 & 2 & 4 \\ 1 & 1 & 1 \end{array}\right] \xrightarrow{\text{row reduce}} \left[\begin{array}{cc|c} 1 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{array}\right] \]

\( \text{rank}(A) = 1 \), \( \text{rank}([A \mid \mathbf{b}]) = 2 \) ⇒ inconsistent.

5. Key Properties of Matrix Rank

  • \( 0 \leq \text{rank}(A) \leq \min(m, n) \) for \( A \in \mathbb{R}^{m \times n} \)
  • \( \text{rank}(A) = \text{rank}(A^T) \)
  • \( \text{rank}(A) = n \) ⇔ columns of \( A \) are linearly independent
  • \( \text{rank}(A) = m \) ⇔ rows of \( A \) span \( \mathbb{R}^n \)
  • If \( A \) is \( n \times n \), then \( A \) is invertible ⇔ \( \text{rank}(A) = n \)
  • \( \text{rank}(AB) \leq \min(\text{rank}(A), \text{rank}(B)) \)

6. Full Example: Rank via RREF

Let \[ A = \begin{bmatrix} 2 & 4 & 6 & 8 \\ 1 & 2 & 1 & 3 \\ 3 & 6 & 4 & 9 \end{bmatrix} \]

Reduce to RREF:

\[ \begin{aligned} &\xrightarrow{R_1 \leftrightarrow R_2} \begin{bmatrix} 1 & 2 & 1 & 3 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 4 & 9 \end{bmatrix} \xrightarrow{R_2 – 2R_1,\ R_3 – 3R_1} \begin{bmatrix} 1 & 2 & 1 & 3 \\ 0 & 0 & 4 & 2 \\ 0 & 0 & 1 & 0 \end{bmatrix} \\ &\xrightarrow{R_2 \leftrightarrow R_3} \begin{bmatrix} 1 & 2 & 1 & 3 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 4 & 2 \end{bmatrix} \xrightarrow{R_3 – 4R_2} \begin{bmatrix} 1 & 2 & 1 & 3 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{bmatrix} \\ &\xrightarrow{\text{scale and eliminate}} \begin{bmatrix} 1 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \end{aligned} \]

✅ RREF has 3 non-zero rows ⇒ \( \text{rank}(A) = 3 \).

7. Can We Use Column Operations?

Yes! Column operations preserve column rank. Since row rank = column rank, you can use column operations to count independent columns.

However, **for linear systems**, only row operations preserve solution sets. So in practice, we almost always use row reduction.

8. Summary Table

Matrix REF / RREF Rank
\( \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \) \( \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix} \) 1
\( \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix} \) (already in RREF) 2
\( I_n \) \( I_n \) \( n \)

9. Self-Check: Test Your Understanding

  1. What is the rank of a \( 4 \times 5 \) matrix with 3 pivot columns?
  2. Can a \( 3 \times 3 \) matrix have rank 4? Why or why not?
  3. Find the rank of \( \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} \).

Answers: (1) 3, (2) No — max rank is 3, (3) 1

Rank of a Matrix – Questions & Full Solutions

Rank of a Matrix – Questions & Full Solutions

😊 Hello, students! Below are all the rank-related questions from your documents, each with a complete solution using elementary row operations. Use this to master rank, REF, and consistency!

Q1. Find the rank of \[ A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 1 \end{bmatrix} \] using elementary row operations.

Solution:

Row reduce to REF:

  • \( R_2 \to R_2 – 2R_1 \)
  • \( R_3 \to R_3 – R_1 \)
\[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & -1 & -2 \end{bmatrix} \xrightarrow{R_2 \leftrightarrow R_3} \begin{bmatrix} 1 & 2 & 3 \\ 0 & -1 & -2 \\ 0 & 0 & 0 \end{bmatrix} \]

Two non-zero rows ⇒ \( \text{rank}(A) = 2 \).

Q2. The following matrix is in row echelon form. What is its rank?

\[ \begin{bmatrix} 1 & 0 & 2 & -1 \\ 0 & 1 & -1 & 3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]

Solution:

Count non-zero rows: **2**.

✅ Rank = 2.

Q3. Reduce to RREF and find the rank of

\[ A = \begin{bmatrix} 2 & 4 & 6 & 8 \\ 1 & 2 & 1 & 3 \\ 3 & 6 & 4 & 9 \end{bmatrix} \]

Solution:

  • Swap \( R_1 \leftrightarrow R_2 \)
  • \( R_2 \to R_2 – 2R_1 \), \( R_3 \to R_3 – 3R_1 \)
  • \( R_3 \to R_3 – 4R_2 \), normalize pivots
\[ \xrightarrow{\text{RREF}} \begin{bmatrix} 1 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]

✅ Three non-zero rows ⇒ \( \text{rank}(A) = 3 \).

Q4. Consider the system with augmented matrix

\[ \left[\begin{array}{ccc|c} 1 & 1 & 1 & 2 \\ 2 & 2 & 2 & 5 \\ 3 & 3 & 3 & 6 \end{array}\right] \]

Find: (a) rank of coefficient matrix \( A \), (b) rank of augmented matrix \( [A \mid \mathbf{b}] \), (c) Is the system consistent?

Solution:

Row reduce:

\[ \xrightarrow{R_2 – 2R_1,\ R_3 – 3R_1} \left[\begin{array}{ccc|c} 1 & 1 & 1 & 2 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] \]
  • Coefficient matrix after reduction has 1 non-zero row ⇒ \( \text{rank}(A) = 1 \)
  • Augmented matrix has 2 non-zero rows ⇒ \( \text{rank}([A \mid \mathbf{b}]) = 2 \)
  • Since ranks differ ⇒ inconsistent (no solution).

Q5. Find the rank of

\[ A = \begin{bmatrix} 1 & 3 & 2 \\ 2 & 6 & 4 \\ -1 & -3 & -2 \end{bmatrix} \]

Solution:

All rows are multiples of row 1.

\[ \xrightarrow{R_2 – 2R_1,\ R_3 + R_1} \begin{bmatrix} 1 & 3 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \]

✅ Only 1 non-zero row ⇒ \( \text{rank}(A) = 1 \).

Q6. What is the maximum possible rank of a \( 4 \times 5 \) matrix? A \( 3 \times 3 \) matrix?

Solution:

Maximum rank = \( \min(m, n) \).

  • For \( 4 \times 5 \): \( \min(4,5) = 4 \)
  • For \( 3 \times 3 \): \( \min(3,3) = 3 \)

✅ Max rank = 4 and 3, respectively.

Q7. If a \( 5 \times 5 \) matrix \( A \) is invertible, what is \( \text{rank}(A) \)?

Solution:

A square matrix is invertible ⇔ \( \det(A) \ne 0 \) ⇔ rows are linearly independent ⇔ full rank.

✅ So \( \text{rank}(A) = 5 \).

Q8. What is the rank of the \( 3 \times 4 \) zero matrix?

Solution:

All rows are zero ⇒ no non-zero rows in REF.

✅ Rank = 0.

Q9. Can you use elementary column operations to find the rank of a matrix? Explain.

Solution:

✅ Yes! Column operations preserve **column rank**, and since **row rank = column rank**, they preserve the (overall) rank.

However, for solving linear systems, we use **row operations** because column ops change variable roles.

Q10. (From document) Find the rank of:

\[ A = \begin{bmatrix} 1 & 3 & 5 & 7 \\ 2 & 4 & 6 & 8 \\ 3 & 5 & 7 & 9 \end{bmatrix} \]

Solution:

  • \( R_2 \to R_2 – 2R_1 \), \( R_3 \to R_3 – 3R_1 \)
  • \( R_3 \to R_3 – R_2 \)
\[ \begin{bmatrix} 1 & 3 & 5 & 7 \\ 0 & -2 & -4 & -6 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]

✅ Two non-zero rows ⇒ \( \text{rank}(A) = 2 \).

📌 Key Takeaways:

  • Rank = number of pivots = non-zero rows in REF/RREF.
  • Elementary row operations preserve rank.
  • System \( A\mathbf{x} = \mathbf{b} \) is consistent ⇔ \( \text{rank}(A) = \text{rank}([A \mid \mathbf{b}]) \).
  • Max rank of \( m \times n \) matrix is \( \min(m,n) \).

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