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4.5 The Dimension of a Vector Space

4.4: Isomorphism

A vector space \( V \) with basis \( \mathcal{B} \) containing \( n \) vectors is isomorphic to \( \mathbb{R}^n \).

Example

The standard basis of \( \mathbb{P}_3 \) is \( \mathcal{B} = \{1, t, t^2, t^3\} \).

A typical element in \( \mathbb{P}_3 \) is \( \vec{p}(t) = a_0 + a_1 t + a_2 t^2 + a_3 t^3 \).

It behaves and acts like a vector in \( \mathbb{R}^4 \):

\[ \vec{p} = \begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ a_3 \end{bmatrix} \]
\[ \vec{p}(t) = a_0 + a_1 t + a_2 t^2 + a_3 t^3 \] \[ \vec{q}(t) = b_0 + b_1 t + b_2 t^2 + b_3 t^3 \] \[ \vec{p} + \vec{q} = (a_0 + b_0) + (a_1 + b_1)t + (a_2 + b_2)t^2 + (a_3 + b_3)t^3 \]

Same as:

\[ \vec{p} = \begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ a_3 \end{bmatrix} \quad \vec{q} = \begin{bmatrix} b_0 \\ b_1 \\ b_2 \\ b_3 \end{bmatrix} \] \[ \vec{p} + \vec{q} = \begin{bmatrix} a_0 + b_0 \\ a_1 + b_1 \\ a_2 + b_2 \\ a_3 + b_3 \end{bmatrix} \]
Same for scalar multiplication.
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If a vector space \( V \) has a basis \( B = \{ \vec{b_1}, \dots, \vec{b_n} \} \), then any set in \( V \) containing more than \( n \) vectors must be linearly dependent.

Why?

Each \( \vec{b_i} \) is \( n \times 1 \).

If we have \( m \) vectors, where \( m > n \), then the matrix \( A = [ \vec{b_1}, \dots, \vec{b_m} ] \) has \( n \) rows and \( m \) columns, and \( A \) has \( n \) pivots.

But there are \( m \) columns, so there are free variables in \( A\vec{x} = \vec{0} \rightarrow \) dependent set.

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If a vector space \( V \) has a basis \( B \) of \( n \) vectors, then every basis of \( V \) must also have \( n \) vectors.

Why?

If \( B = \{ \vec{b_1}, \dots, \vec{b_n} \} \) is a basis, then we know the \( \vec{b_i} \) are linearly independent and span \( V \).

If \( C = \{ \vec{c_1}, \dots, \vec{c_m} \} \) is another basis of \( V \). We know \( \vec{c_i} \) must be linearly independent, so knowing there are at most \( n \) linearly independent vectors in \( V \) (from \( B \)), we know \( m \le n \).

But since \( B \), a basis, spans \( V \), we know we need no fewer than \( n \) vectors to span \( V \). So now know \( C \), being a basis, must have no fewer than \( n \) vectors. So \( m \ge n \). So, \( m = n \).

The Basis Theorem

This means ANY other set of \( n \) linearly independent vectors in \( V \) is a basis.

This means ANY other set of \( n \) vectors spanning \( V \) is a basis.

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Dimension of a Vector Space

The number of basis vectors is called the dimension of the vector space. Vector spaces can be finite-dimensional or infinite-dimensional.

Example: all continuous functions (infinite-dimensional).

We already know about \(\dim \text{Col } A\) and \(\dim \text{Nul } A\):

  • \(\dim \text{Col } A\) is the # of basic variables
  • \(\dim \text{Nul } A\) is the # of free variables

Example: Subspace Dimension

How many dimensions does a subspace of all \(\mathbb{P}_{10}\) whose \(5^{\text{th}}\) and \(7^{\text{th}}\) coefficients are the same have?

10. Because \(\mathbb{P}_{10}\) has 11 coefficients, but we can only freely choose 10 of them.

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Example: Chebyshev Polynomials

The first 3 Chebyshev polynomials are \(1, t, 2t^2 - 1\). Can these form a basis for \(\mathbb{P}_2\)?

\(\mathbb{P}_2\) is 3-dimensional, representing polynomials of the form \(a_0 + a_1 t + a_2 t^2\).
The standard basis is: \(\{1, t, t^2\}\).

\(\{1, t, 2t^2 - 1\}\) has 3 vectors, and we know from earlier results if these vectors are linearly independent, then they must form a basis.

Rewrite as vectors in \(\mathbb{R}^3\) using \(\{1, t, t^2\}\) as "coordinates":

\(1\)

\[\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\]

\(t\)

\[\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}\]

\(t^2\)

\[\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\]

For the set \(\{1, t, 2t^2 - 1\}\):

\(1\)

\[\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\]

\(t\)

\[\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}\]

\(2t^2 - 1\)

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

3 pivots, 3 vectors, so linearly independent.

So \( \{ 1, t, 2t^2 - 1 \} \) must be a basis for \( \mathbb{P}_2 \).

Spanning Set Theorem

Spanning Set Theorem allows us to make a basis by throwing out linearly dependent ones: e.g. \( \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \end{bmatrix} \right\} \) is NOT a basis for \( \mathbb{R}^2 \) but \( \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right\} \) is.

Likewise, we can add linearly independent vectors to make a basis for a subspace.

\[ \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix} \right\} \]

add \( \begin{bmatrix} 0 \\ 1 \end{bmatrix} \) (indp from existing ones)

\[ \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right\} \text{ is basis for } \mathbb{R}^2 \]