Lecture 14: Vector Spaces

14.1 Bases, Dimension, and Coordinates

Definition: basis

If the ordered vectors $\xi_1,\xi_2,\ldots,\xi_n$ are linearly independent in $\mathbb R^n$, they form a basis of $\mathbb R^n$. Every vector $\alpha\in\mathbb R^n$ has a unique representation

\alpha=a_1\xi_1+a_2\xi_2+\cdots+a_n\xi_n.

The column vector

x= \begin{bmatrix} a_1\\a_2\\\vdots\\a_n \end{bmatrix}

is the coordinate vector of $\alpha$ relative to the basis $\xi_1,\ldots,\xi_n$. The number $n$ of basis vectors is the dimension of the vector space.

14.2 Changes of Basis and Coordinates

Theorem: change-of-basis

Let

\mathcal B_\xi=(\xi_1,\ldots,\xi_n), \qquad \mathcal B_\eta=(\eta_1,\ldots,\eta_n)

be two bases of $\mathbb R^n$. If

\begin{bmatrix} \eta_1&\eta_2&\cdots&\eta_n \end{bmatrix} = \begin{bmatrix} \xi_1&\xi_2&\cdots&\xi_n \end{bmatrix}C,

then $C$ is the transition matrix from $\mathcal B_\xi$ to $\mathcal B_\eta$. Its $i$th column is the coordinate vector of $\eta_i$ relative to the old basis $\mathcal B_\xi$. Because both ordered sets are bases, $C$ is invertible.

Theorem: coordinate-transformation

If the same vector $\alpha$ has coordinate vectors $x$ and $y$ relative to the two bases, then

\alpha = \begin{bmatrix}\xi_1&\cdots&\xi_n\end{bmatrix}x = \begin{bmatrix}\eta_1&\cdots&\eta_n\end{bmatrix}y.

Using the change-of-basis equation gives

x=Cy, \qquad y=C^{-1}x.

Thus $C$ converts new-basis coordinates $y$ to old-basis coordinates $x$, while $C^{-1}$ performs the reverse conversion.

Two-dimensional example

Let

C= \begin{bmatrix} 2&-1\\ 1&1 \end{bmatrix}, \qquad y= \begin{bmatrix} -1\\2 \end{bmatrix}.

Then

x=Cy =-1 \begin{bmatrix} 2\\1 \end{bmatrix} +2 \begin{bmatrix} -1\\1 \end{bmatrix} = \begin{bmatrix} -4\\1 \end{bmatrix}.

Original illustration:

14.3 Gram-Schmidt Orthogonalization

Theorem: Gram-Schmidt-process

Given linearly independent vectors $\alpha_1,\ldots,\alpha_m$, define

\beta_1=\alpha_1,

and, for $k=2,\ldots,m$,

\beta_k = \alpha_k -\sum_{j=1}^{k-1} \frac{\langle\alpha_k,\beta_j\rangle} {\langle\beta_j,\beta_j\rangle} \beta_j.

The vectors $\beta_1,\ldots,\beta_m$ are mutually orthogonal and span the same subspace as the original vectors. Normalizing them,

e_k=\frac{\beta_k}{\|\beta_k\|},

produces an orthonormal set.