
# 14.1: Properties of the Standard Basis

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The standard notion of the length of a vector $$x=(x_{1},x_{2},\ldots,x_{n})$$ $$\in$$ $$\mathbb{R}^{n}$$ is

$$||x||=\sqrt{x\cdot x}=\sqrt{(x_{1})^{2}+(x_{2})^{2}+\cdots(x_{n})^{2}}\, .$$

The canonical/standard basis in $$\mathbb{R}^{n}$$

$$e_{1}=\begin{pmatrix}1\\0\\ \vdots \\ 0\end{pmatrix}, e_{2}=\begin{pmatrix}0\\1\\ \vdots \\ 0\end{pmatrix}, \ldots, e_{n}=\begin{pmatrix}0\\0\\ \vdots \\ 1\end{pmatrix}\,$$

has many useful properties with respect to the dot product and lengths:

1. Each of the standard basis vectors has unit length:

$\|e_{i}\|=\sqrt{e_{i}\cdot e_{i}}=\sqrt{e_{i}^{T}e_{i}}=1\, .$

2. The standard basis vectors are $$\textit{orthogonal}$$ (in other words, at right angles or perpendicular):

$e_{i}\cdot e_{j} = e_{i}^{T}e_{j}=0 \textit{ when } i\neq j$

This is summarized by

$e_{i}^{T}e_{j}=\delta_{ij}=\left\{ \begin{array}{cc} 1 & \qquad i=j \\ 0 & \qquad i\neq j \\ \end{array}\right. ,$

where $$\delta_{ij}$$ is the $$\textit{Kronecker delta}$$. Notice that the Kronecker delta gives the entries of the identity matrix.

### Inner and Outer Products

Given column vectors $$v$$ and $$w$$, we have seen that the dot product $$v\cdot w$$ is the same as the matrix multiplication $$v^{T}w$$. This is an inner product on $$\Re^{n}$$. We can also form the outer product $$vw^{T}$$, which gives a square matrix.

The outer product on the standard basis vectors is interesting. Set

$$\Pi_{1} = e_{1}e_{1}^{T} = \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0\end{pmatrix} \begin{pmatrix} 1 & 0 & \cdots & 0\end{pmatrix}$$

$$~~~~~~= \begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & & & \vdots \\ 0 & 0 & \cdots & 0\end{pmatrix}$$

$$\vdots$$

$$\Pi_{n} = e_{n}e_{n}^{T} = \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 1\end{pmatrix} \begin{pmatrix} 0 & 0 & \cdots & 1\end{pmatrix}$$

$$~~~~~~= \begin{pmatrix} 0 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & & & \vdots \\ 0 & 0 & \cdots & 1\end{pmatrix}$$

In short, $$\Pi_{i}$$ is the diagonal square matrix with a $$1$$ in the $$i$$th diagonal position and zeros everywhere else.

Notice that

$\Pi_{i}\Pi_{j}=e_{i}e_{i}^{T}e_{j}e_{j}^{T}=e_{i}\delta_{ij}e_{j}^{T}.$

Then:

$\Pi_{i}\Pi_{j} = \left\{ \begin{array}{cc} \Pi_{i} & \qquad i=j \\ 0 & \qquad i\neq j \\ \end{array}\right. .$

Moreover, for a diagonal matrix $$D$$ with diagonal entries $$\lambda_{1},\ldots, \lambda_{n}$$, we can write

$D= \lambda_{1}\Pi_{1} + \cdots + \lambda_{n}\Pi_{n}.$