Download original notes

PAGE 1

2.8 Subspaces of \(\mathbb{R}^n\)

Subspace: collection of vectors in some subset of \(\mathbb{R}^n\)

for example, vectors of the form \(\begin{bmatrix} x \\ y \\ 0 \end{bmatrix}\) are in a subspace of \(\mathbb{R}^3\)

here, the space is \(\mathbb{R}^2\)

A subspace must have these properties:

a)
It must contain the zero vector \((\vec{0})\)
b)

It must be closed under addition

for each \(\vec{u}\) and \(\vec{v}\) in the set, \(\vec{u} + \vec{v}\) is also in the set.
c)

It must be closed under multiplication

for each \(\vec{u}\) in the set, \(c\vec{u}\) must also be in the set.

PAGE 2

all vectors of the form \(\begin{bmatrix} x \\ y \\ 0 \end{bmatrix}\). Are the requirements met?

a) is \(\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\) in the set? Yes

Figure: 3D coordinate system showing the xy-plane as a shaded parallelogram with x, y, and z axes.

b) closed under addition?

Figure: Vector addition diagram on a plane showing vectors u and v and their sum u+v staying within the plane.

Yes, ANY vectors \(\begin{bmatrix} x_1 \\ y_1 \\ 0 \end{bmatrix}\) and \(\begin{bmatrix} x_2 \\ y_2 \\ 0 \end{bmatrix}\) always have a sum \(\begin{bmatrix} x_1 + x_2 \\ y_1 + y_2 \\ 0 \end{bmatrix}\) that's on xy plane.

c) closed under multiplication?

\[ c \begin{bmatrix} x \\ y \\ 0 \end{bmatrix} = \begin{bmatrix} cx \\ cy \\ 0 \end{bmatrix} \text{ still on xy plane} \]

so, yes.

therefore, \(\mathbb{R}^2\) (xy plane) is a subspace.

PAGE 3

Example: Is this a subspace?

All vectors \((x, y)\) (or \(\begin{bmatrix} x \\ y \end{bmatrix}\)) such that \(x \ge 0, y \ge 0\) and \(x \le 0, y \le 0\).

All \((x, y)\) such that \(x\) and \(y\) have same sign or one or both zero.

Analysis

Is \(\vec{0}\) in this set? Yes
Closed under addition?
No, because \(\vec{u} + \vec{w}\) is in 2nd quadrant.

\[ \begin{bmatrix} 1 \\ 2 \end{bmatrix} + \begin{bmatrix} -3 \\ -1 \end{bmatrix} = \begin{bmatrix} -2 \\ 1 \end{bmatrix} \]
Closed under multiplication? Yes

Therefore, this set is NOT a subspace.

Figure: Graph showing vectors u and v in the first quadrant and their sum u+v also in the first quadrant.

Figure: Graph showing vector u in the first quadrant and w in the third quadrant, with their sum u+w in the second quadrant.

PAGE 4

Some notable subspaces connected to a matrix \(A\)

column space — all linear combos of columns of \(A\)
null space — all solutions of \(A\vec{x} = \vec{0}\)

Example

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

Column space of \(A\): \(\text{Col } A = \text{span} \left\{ \begin{bmatrix} 1 \\ 4 \\ 0 \end{bmatrix}, \begin{bmatrix} 2 \\ 5 \\ -1 \end{bmatrix}, \begin{bmatrix} 3 \\ 6 \\ -2 \end{bmatrix} \right\}

Is \(\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\) in \(\text{Col } A\)? \(\Rightarrow\) Solve \(A\vec{x} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\)

\[ \begin{bmatrix} 1 & 2 & 3 & | & 0 \\ 4 & 5 & 6 & | & 0 \\ 0 & -1 & -2 & | & 1 \end{bmatrix} \sim \dots \sim \begin{bmatrix} 1 & 0 & -1 & | & 0 \\ 0 & 1 & 2 & | & 0 \\ 0 & 0 & 0 & | & 1 \end{bmatrix} \]

No solution, so no \(x_1, x_2, x_3\) exist such that

\[ x_1 \begin{bmatrix} 1 \\ 4 \\ 0 \end{bmatrix} + x_2 \begin{bmatrix} 2 \\ 5 \\ -1 \end{bmatrix} + x_3 \begin{bmatrix} 3 \\ 6 \\ -2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \]

Therefore, \(\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\) is NOT in \(\text{Col } A\).

PAGE 5

Null Space of A: \(\text{Nul } A\)

Solutions of \(A\vec{x} = \vec{0}\)

\[\begin{bmatrix} 1 & 2 & 3 & 0 \\ 4 & 5 & 6 & 0 \\ 0 & -1 & -2 & 0 \end{bmatrix} \sim \dots \sim \begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}\]

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

so \(\vec{x} = \begin{bmatrix} x_3 \\ -2x_3 \\ x_3 \end{bmatrix} = x_3 \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}\)

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

is \(\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}\) in \(\text{Nul } A\)? No

equivalent question: is \(A \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \vec{0}\)?

PAGE 6

\[A = \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \end{bmatrix}\]

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

the last one, being a linear combo of first 3, is not needed to describe \(\text{Col } A\).

the minimum set of vectors needed to describe a subspace is called a basis

the vectors in a basis (called bases) are linearly independent.

PAGE 7

Example: Finding a Basis for the Null Space

Given the matrix:

\[ A = \begin{bmatrix} 3 & 6 & 21 & -24 \\ 5 & 9 & 29 & -34 \\ 2 & 3 & 8 & -10 \end{bmatrix} \]

Find a basis for \( \text{Nul } A \)

So reduce the augmented matrix \( [A \mid \vec{0}] \):

\[ [A \mid \vec{0}] = \begin{bmatrix} 3 & 6 & 21 & -24 & 0 \\ 5 & 9 & 29 & -34 & 0 \\ 2 & 3 & 8 & -10 & 0 \end{bmatrix} \]

Performing row reduction:

\[ \sim \dots \sim \begin{bmatrix} 1 & 3 & 13 & -14 & 0 \\ 0 & 1 & 6 & -6 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \]

From the reduced row echelon form, we see that \( x_3 \) and \( x_4 \) are free variables.

The equations are:

\( x_2 = -6x_3 + 6x_4 \)

\( x_1 = -3x_2 - 13x_3 + 14x_4 \)

\( x_1 = -3(-6x_3 + 6x_4) - 13x_3 + 14x_4 \)

\( x_1 = 18x_3 - 18x_4 - 13x_3 + 14x_4 \)

\( x_1 = 5x_3 - 4x_4 \)

The general solution vector \( \vec{x} \) is:

\[ \vec{x} = \begin{bmatrix} 5x_3 - 4x_4 \\ -6x_3 + 6x_4 \\ x_3 \\ x_4 \end{bmatrix} = x_3 \begin{bmatrix} 5 \\ -6 \\ 1 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -4 \\ 6 \\ 0 \\ 1 \end{bmatrix} \]

PAGE 8

The basis of \( \text{Nul } A \) is:

\[ \left\{ \begin{bmatrix} 5 \\ -6 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -4 \\ 6 \\ 0 \\ 1 \end{bmatrix} \right\} \]

Basis of \( \text{Col } A \)?

\[ A = \begin{bmatrix} 3 & 6 & 21 & -24 \\ 5 & 9 & 29 & -34 \\ 2 & 3 & 8 & -10 \end{bmatrix} \sim \dots \sim \begin{bmatrix} 1 & 3 & 13 & -14 \\ 0 & 1 & 6 & -6 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]

The columns in \( A \) that correspond to the pivot columns are basis vectors of \( \text{Col } A \).

\( \text{Col } A \) has basis:

\[ \left\{ \begin{bmatrix} 3 \\ 5 \\ 2 \end{bmatrix}, \begin{bmatrix} 6 \\ 9 \\ 3 \end{bmatrix} \right\} \]

Why?

Basis vectors are linearly independent.
Solving \( [A \mid \vec{0}] \) identifies the dependence relationship (and thus which are linearly independent).
This method works because Elementary Row Operations (EROs) do not change the dependency of columns.

PAGE 9

Basis of Column Space and Pivots

The basis of \(\text{Col } A\) is related to pivots.

If an \(n \times n\) matrix has \(n\) pivots, then the columns of \(A\) are basis vectors of \(\text{Col } A\).

→ Columns span \(\mathbb{R}^n\), \(\text{Col } A = \mathbb{R}^n\)
→ \(A\) is invertible
→ \(A\vec{x} = \vec{0}\) only has the trivial solution