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2.9 Dimension and Rank (NOT ON EXAM 1)

basis :

the minimum set of vectors needed to span a subspace. These vectors (“bases”) can also be used as the coordinate system in the subspace.

example

The basis of a subspace is

\[ B = \left\{ \begin{bmatrix} 1 \\ 4 \\ -3 \end{bmatrix}, \begin{bmatrix} -2 \\ -7 \\ 5 \end{bmatrix} \right\} \]

plane through origin

a vector in the subspace is \[ \begin{bmatrix} -5 \\ -17 \\ 12 \end{bmatrix} \]

\[ \begin{bmatrix} -5 \\ -17 \\ 12 \end{bmatrix} = (1) \begin{bmatrix} 1 \\ 4 \\ -3 \end{bmatrix} + (3) \begin{bmatrix} -2 \\ -7 \\ 5 \end{bmatrix} \]

but it is also

\[ \begin{bmatrix} -5 \\ -17 \\ 12 \end{bmatrix} = (-5) \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + (-17) \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + (12) \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \]

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Visualizing Subspace Coordinates

The following diagram illustrates a 2D subspace (a plane) within a 3D coordinate system defined by axes \(x_1, x_2, x_3\). The plane is spanned by basis vectors \(v_1\) and \(v_2\), showing how a point \(x\) can be represented as a linear combination.

Figure: 3D coordinate system with axes x1, x2, x3 showing a green plane spanned by vectors v1 and v2. A point x is labeled as 2v1 + 3v2.

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the coordinates in \(\mathbb{R}^3\) is

\[ \begin{bmatrix} -5 \\ -17 \\ 12 \end{bmatrix} \]

but in \(B\) it is

\[ \begin{bmatrix} 1 \\ 3 \end{bmatrix} \Rightarrow \text{coordinates relative to the basis } B \quad \text{or } B\text{-coordinate vector} \]

not really, the basis vectors can simply be looked at as a coord. transformation.

\(B\), in this example, is a subspace of \(\mathbb{R}^3\) and is a plane and it behaves just like \(\mathbb{R}^2\) even though it is not \(\mathbb{R}^2\).

there is a one-to-one correspondence between \(B\) and \(\mathbb{R}^2\)

the subspace preserves linear combination
(looks and acts like \(\mathbb{R}^2\))

\(\Rightarrow\) "isomorphism"

the transformation between \(B\) and \(\mathbb{R}^2\) is both onto and one-to-one

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the basis itself is not unique

but once chosen, every vector can only be described one way

\(\rightarrow\) this is because basis vectors are linearly independent

if \(\{ \vec{v}_1, \vec{v}_2, \vec{v}_3 \}\) is basis set

if we could describe a vector in more than one way,

\[ \begin{aligned} \vec{b} &= c_1 \vec{v}_1 + c_2 \vec{v}_2 + c_3 \vec{v}_3 \\ \vec{b} &= d_1 \vec{v}_1 + d_2 \vec{v}_2 + d_3 \vec{v}_3 \end{aligned} \quad \text{where } c_i \neq d_i \]

then

\[ \vec{0} = (c_1 - d_1) \vec{v}_1 + (c_2 - d_2) \vec{v}_2 + (c_3 - d_3) \vec{v}_3 \]

but this cannot happen because \(\vec{v}_i\) are linearly independent

thus, the \(c_i \neq d_i\) assumption is wrong.

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The number of vectors in a basis set of subspace \(H\) is called the dimension of \(H\). \(\dim H\)

e.g. for \(\mathbb{R}^3\), one possible basis is \(\left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right\} \)

\(\dim \mathbb{R}^3 = 3\)

If \(H\) is an \(n\)-dimensional subspace then any set of \(n\) linearly independent vectors is a basis of \(H\).

\[ \begin{bmatrix} 1 & -1 & 5 \\ 2 & 0 & 7 \\ -3 & -5 & -3 \end{bmatrix} \sim \dots \sim \begin{bmatrix} 1 & -1 & 5 \\ 0 & 2 & -3 \\ 0 & 0 & 0 \end{bmatrix} \]

What is \(\dim \text{Col } A\)?

2 because columns 1 and 2 are pivot columns and are independent, so they form a basis of \(\text{Col } A\).

basic variable

What is \(\dim \text{Nul } A\)?

1 because there is one free variable and all solutions are multiples of one vector.

free variables

all solutions of \(A\vec{x} = \vec{0}\)

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

What is \(\dim \text{Nul } A\) if \(A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)?

\(\text{Nul } A = \{ \vec{0} \} \) we defined \(\dim \text{Nul } A = 0\)

If \(A\) has \(n\) columns, then \(\dim \text{Col } A + \dim \text{Nul } A = n\)

\(\uparrow\)
# of basic

\(\uparrow\)
# of free

\(\uparrow\)
n variables

\(\dim \text{Col } A\) is also called the rank of \(A\)

A \(3 \times 5\) matrix can have at most 3 basic variables and at least 2 free variables.

\[ \begin{bmatrix} \square & \cdot & \cdot & \cdot & \cdot \\ \cdot & \square & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \square & \cdot \end{bmatrix} \]

at most 3 pivots

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Linear Transformations and Null Space

If \( A \) is \( 3 \times 5 \) then \( T: \mathbb{R}^5 \to \mathbb{R}^3 \).

\( \dim \text{Nul } A \) tells us how many axes are "lost" due to the transformation.

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

In the matrix above, the second column \( \begin{bmatrix} 0 \\ 0 \end{bmatrix} \) represents an axis that is lost (is in the null space).

\[ \text{the } A \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x \\ 0 \end{bmatrix} \]