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4.5 Row and Column Spaces

Given a matrix, all possible linear combos form the row space (space spanned by its rows).

\[ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \quad \text{Row}(A) = \text{span} \{ [1 \quad 2], [3 \quad 4] \} \]

\[ B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \quad \text{Row}(B) = \text{span} \{ [1 \quad 0], [0 \quad 0] \} = \text{span} \{ [1 \quad 0] \} \]

What about a basis (minimum spanning set) of row space?

For \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \), \( [1 \quad 2] \) and \( [3 \quad 4] \) are linearly indp so form a basis for \( \text{Row}(A) \) which is \( \mathbb{R}^2 \).

Figure: A 2D coordinate system showing two vectors, [1, 2] and [3, 4], spanning the xy plane.

\( \{ [1 \quad 2], [3 \quad 4] \} \) cover xy plane

So another basis is \( \{ [15 \quad 0], [0 \quad \pi] \} \).

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\[ A = \begin{bmatrix} 1 & -2 & 2 \\ 1 & 4 & 3 \\ 2 & 2 & 5 \end{bmatrix} \quad \begin{matrix} \text{Row}(A)? \\ \text{find a basis} \end{matrix} \]

\( \text{Row}(A) = \text{span} \{ [1 \quad -2 \quad 2], [1 \quad 4 \quad 3], [2 \quad 2 \quad 5] \} \)

But a basis? Knowing a basis helps us see if the \( \text{Row}(A) \) is, for example, \( \mathbb{R}^3 \).

We need to find linearly indp vectors that span the same space (\( \text{Row}(A) \)).

Gaussian elimination produces row equivalent matrices and preserves row space.

For example, \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 2 \\ 0 & -2 \end{bmatrix} = B \)

NOT equal

but are row equivalent

same row space

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

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Row Space and Basis

Back to matrix \( A \):

\[ A = \begin{bmatrix} 1 & -2 & 2 \\ 1 & 4 & 3 \\ 2 & 2 & 5 \end{bmatrix} \]

Performing row reduction:

\[ \rightarrow \dots \rightarrow \begin{bmatrix} 1 & -2 & 2 \\ 0 & 6 & 1 \\ 0 & 0 & 0 \end{bmatrix} = B \]

Row(A) = Row(B)

\[ \text{Row}(B) = \text{span} \{ (1, -2, 2), (0, 6, 1) \} \]

These vectors are linearly independent.

\[ = \text{Row}(A) \]

So, a basis for \( \text{Row}(A) \) is \( \{ (1, -2, 2), (0, 6, 1) \} \).

Procedure for Finding a Basis

To find a basis for row space, perform row reduction and identify the pivot rows (rows w/ pivots) and those pivots form a basis for row space.

# of pivots \( \rightarrow \) rank of the matrix

Matrix \( A \) above is rank two.

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Column Space

Column space: span of columns (all possible linear combinations of columns).

\[ A = \begin{bmatrix} 1 & 2 & 4 \\ 1 & 3 & 2 \\ 3 & 8 & 8 \end{bmatrix} \quad \text{Col}(A) = \text{span} \left\{ \begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix}, \begin{bmatrix} 2 \\ 3 \\ 8 \end{bmatrix}, \begin{bmatrix} 4 \\ 2 \\ 8 \end{bmatrix} \right\} \]

Can we find a basis?

Row ops preserve row space but does NOT, in general, preserve column space.

For example:

\[ C = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 2 & 2 \end{bmatrix} \quad \text{Col}(C) = \text{span} \left\{ \begin{bmatrix} 1 \\ 2 \end{bmatrix} \right\} \]

Figure: A 2D coordinate system showing a line passing through the origin with a positive slope, labeled Col(C).

\[ \rightarrow \begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} = D \quad \text{Col}(D) = \text{span} \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix} \right\} \]

Figure: A 2D coordinate system showing a horizontal line along the x-axis, labeled Col(D).

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Linear Independence and Row Operations

Even though row ops change column space, the linear independency of column is preserved.

For example,

\[ \begin{bmatrix} 1 & 0 & 3 \\ 2 & 1 & 4 \end{bmatrix} \text{ clearly, cols 1, 2 are indp} \]

\(\rightarrow\)

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

column 1, 2 pivot columns so are indp.

Whatever columns that were indp remain indp after row reduction.

Use the reduced matrix to find the columns that were indp in the original matrix.

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Back to Matrix A

\[ A = \begin{bmatrix} 1 & 2 & 4 \\ 1 & 3 & 2 \\ 3 & 8 & 8 \end{bmatrix} \quad \text{Col}(A) = \text{span} \left\{ \begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix}, \begin{bmatrix} 2 \\ 3 \\ 8 \end{bmatrix}, \begin{bmatrix} 4 \\ 2 \\ 8 \end{bmatrix} \right\} \]

Reduce:

\(\rightarrow\)

\[ \begin{bmatrix} 1 & 2 & 4 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{bmatrix} \]

pivot columns: these do NOT form a basis for Col(A)

But this tells us that columns 1 and 2 in A are indp.

So, since cols 1, 2 in A are indp, they form a basis for Col(A):

\[ \left\{ \begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix}, \begin{bmatrix} 2 \\ 3 \\ 8 \end{bmatrix} \right\} \]

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Example: Finding Bases for Row and Column Spaces

Given the matrix:

\[ A = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 3 & 5 & -9 & 10 \\ 1 & 6 & -29 & -11 \end{bmatrix} \]

Find a basis for \( \text{Row}(A) \) and \( \text{Col}(A) \).

Row reducing the matrix:

\[ \to \dots \to \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 2 & -12 & 7 \\ 0 & 0 & 0 & -59 \end{bmatrix} \]

Basis for Row(A)

Pivot rows: all 3, so they form a basis for \( \text{Row}(A) \):

\[ \{ [1, 1, 1, 1], [0, 2, -12, 7], [0, 0, 0, -59] \} \]

Basis for Col(A)

Pivot columns: 1, 2, 4, so we pick columns 1, 2, 4 in the original matrix to form a basis for \( \text{Col}(A) \):

\[ \left\{ \begin{bmatrix} 1 \\ 3 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 5 \\ 6 \end{bmatrix}, \begin{bmatrix} 1 \\ 10 \\ -11 \end{bmatrix} \right\} \]

NOT a basis for Col(A):

\[ \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 7 \\ -59 \end{bmatrix} \right\} \]

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Given a set of vectors, if we want to know which are linearly independent, put them in as columns of a matrix and identify pivot columns which tells us which vectors are independent.

(See previous example pretending the columns are the vectors)

Another way to find basis for Column space:

\[ A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & -9 & 17 \end{bmatrix} \]

\[ A^T = \begin{bmatrix} 1 & 4 \\ 2 & -9 \\ 3 & 17 \end{bmatrix} \]

\[ \text{Row}(A^T) = \text{Col}(A) \]

Reduce \( A^T \):

\[ \to \dots \to \begin{bmatrix} 1 & 4 \\ 0 & -17 \\ 0 & 0 \end{bmatrix} \]

Pivot rows: \( [1, 4], [0, -17] \)
\( \to \) basis for \( \text{Row}(A^T) \)

So, a basis for \( \text{Col}(A) \) is:

\[ \left\{ \begin{bmatrix} 1 \\ 4 \end{bmatrix}, \begin{bmatrix} 0 \\ -17 \end{bmatrix} \right\} \]