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4.6 Rank

If \( A \) is \( m \times n \), \( \dim \text{Col } A = \text{rank } A = \text{# of pivot columns} = \text{# basic variables} \)

\( \dim \text{Nul } A = \text{# free variables} \)

\( \dim \text{Col } A + \dim \text{Nul } A = n \) (The Rank Theorem)

What about the row space of \( A \) (\( \text{Row } A \))?

\[ A = \begin{bmatrix} 1 & -4 & 1 & -7 \\ -1 & 2 & 2 & 1 \\ 2 & 4 & -16 & 22 \end{bmatrix} \]\[ B = \begin{bmatrix} 1 & 0 & -5 & 5 \\ 0 & -2 & 3 & -6 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]

\( A \sim B \)

row space of \( A \): subspace spanned by rows of \( A \)

\( A \) and \( B \) above have the same row space.

We obtained \( B \) by row reductions, so rows of \( B \) are linear combos of rows of \( A \), which means row space of \( B \) is contained in row space of \( A \). But row operations are reversible, so we could have gotten \( A \) from \( B \) by doing ERO's.

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Rows of \( A \) are therefore linear combos of rows of \( B \). So row space of \( A \) is contained in row space of \( B \). But if two spaces are contained within each other, they must be the same.

\[ \text{If } A \sim B, \text{ then } \text{Row } A = \text{Row } B \]

The basis vectors of row space of either are the non zero rows of the echelon matrix. Not from the original matrix because row operations can change linear dependence.

here, the basis of \( \text{Row } A \) or \( \text{Row } B \) is \( \{ (1, 0, -5, 5), (0, -2, 3, -6) \} \)

If we want to know which rows of \( A \) are basis vectors of \( \text{Row } A \), we can find the basis of \( \text{Col } A^T \).

here, the first two rows of \( B \) are linearly independent, but this does NOT mean the first two rows of \( A \) are linearly independent.

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Rank and Subspaces of a Matrix

The above implies that:

\[ \text{rank } A = \dim \text{Col } A = \dim \text{Row } A = \dim \text{Col } A^T \]

\[ A = \begin{bmatrix} 2 & 4 & -2 & 1 \\ -2 & -5 & 7 & 3 \\ 3 & 7 & -8 & 6 \end{bmatrix} \]\[ B = \begin{bmatrix} 1 & 0 & 9 & 0 \\ 0 & 1 & -5 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]

\( A \sim B \)

Is \( \text{Col } A = \mathbb{R}^3 \)?

Yes, 3 pivots, 3 linearly independent columns of 3 elements each, so they must form a basis of \( \mathbb{R}^3 \).

\[ \text{basis for Col } A = \left\{ \begin{bmatrix} 2 \\ -2 \\ 3 \end{bmatrix}, \begin{bmatrix} 4 \\ -5 \\ 7 \end{bmatrix}, \begin{bmatrix} 1 \\ 3 \\ 6 \end{bmatrix} \right\} \]

Does \( A\vec{x} = \vec{b} \) always have a solution for some \( \vec{b} \) in \( \mathbb{R}^3 \)?

Yes, because \( \text{Col } A = \mathbb{R}^3 \) so any \( \vec{b} \) in \( \mathbb{R}^3 \) is a linear combo of columns of \( A \).

\[ \text{basis for Row } A = \{ (1, 0, 9, 0), (0, 1, -5, 0), (0, 0, 0, 1) \} \]

(same dimension as Col A)

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If \( A \) is \( 4 \times 5 \) and \( \dim \text{Nul } A = 2 \). Is \( \text{Col } A = \mathbb{R}^3 \)?

Figure: A 4 by 5 matrix sketch with 3 pivot positions marked with green squares.

2 free variables

3 pivot columns, \( \text{rank } A = 3 \)

No, \( \text{Col } A \neq \mathbb{R}^3 \) because \( \mathbb{R}^3 \) vectors look like \( \begin{bmatrix} a \\ b \\ c \end{bmatrix} \).

Is \( A\vec{x} = \vec{b} \) always consistent?

No, because we need 4 basis vectors/columns to have \( \text{Col } A = \mathbb{R}^4 \) (which guarantees \( A\vec{x} = \vec{b} \) is always consistent) but we only have 3.

Can a \( 6 \times 9 \) matrix have a two-dimensional null space?

Figure: A 6 by 9 matrix sketch with dots representing entries.

2 free variables

7 basic variables

but only 6 rows here, can't place all 7

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What's the relationship between Row A, Col A, Nul A, and Nul A^T?

\[ A = \begin{bmatrix} 3 & -1 & 3 \\ 6 & 0 & 12 \\ 2 & 1 & 7 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \end{bmatrix} \]

Basis for Nul A:

\( x_3 \) free, \( x_2 = -3x_3 \), \( x_1 = -2x_3 \)

\[ \text{Nul } A = \left\{ x_3 \begin{bmatrix} -2 \\ -3 \\ 1 \end{bmatrix} \right\} \quad \text{Nul } A = \text{span } \left\{ \begin{bmatrix} -2 \\ -3 \\ 1 \end{bmatrix} \right\} \]

Column Space and Row Space

\[ \text{Col } A = \text{span } \left\{ \begin{bmatrix} 3 \\ 6 \\ 2 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \right\} \]

\[ \text{Row } A = \text{span } \{ (1, 0, 2), (0, 1, 3) \} \]

Transpose and Left Null Space

\[ A^T = \begin{bmatrix} 3 & 6 & 2 \\ -1 & 0 & 1 \\ 3 & 12 & 7 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 5/6 \\ 0 & 0 & 0 \end{bmatrix} \]

\[ \text{Nul } A^T = \text{span } \left\{ \begin{bmatrix} 1 \\ -5/6 \\ 1 \end{bmatrix} \right\} \]

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Geometric Interpretation of Subspaces

Orthogonality of Row A and Nul A

The relationship between the row space and the null space is one of orthogonality. In \( \mathbb{R}^3 \), the row space is a plane and the null space is a line perpendicular to that plane.

\( \text{Nul } A \perp \text{Row } A \)

Figure: 3D coordinate graph with axes x1, x2, x3 showing a plane for Row A and a perpendicular line for Nul A.

Figure: Abstract diagram showing a plane labeled Row A and a line labeled Nul A intersecting at a right angle.

Orthogonality of Col A and Nul A^T

Similarly, the column space and the left null space are orthogonal complements. The column space forms a plane, while the left null space forms a line perpendicular to it.

Figure: 3D coordinate graph with axes x1, x2, x3 showing a plane for Col A and a perpendicular line for Nul A^T.

Figure: Abstract diagram showing a plane labeled Col A and a line labeled Nul A^T intersecting at a right angle.