The norm (or magnitude or length) of an element $\bfv=(v_1,\ldots,v_n)$ of $\R^n$ is given by the familiar expression $$ ||\bfv||=\sqrt{\bfv\pd\bfv}=\sqrt{v_1^2+\ldots+v_n^2}. $$ There is a similar notion for any real inner product space $V$. The norm of a vector $\bfv\in V$, denoted by $||\bfv||$, is thus defined by $$ ||\bfv||=\sqrt{\langle\bfv,\bfv\rangle}. $$ A vector with norm $1$ is called a unit vector.

How would we define the distance, $d(\bfu,\bfv)$, between two vectors $\bfu,\bfv\in V$? A natural notion of distance between two vectors should be independent of the order we happen to be viewing them. That is, we want the distance measure to be symmetric: $d(\bfu,\bfv)=d(\bfv,\bfu)$. Again using $\R^n$ as inspiration, we now define the distance between two vectors $\bfu,\bfv\in V$ as $$ d(\bfu,\bfv)=||\bfu-\bfv||. $$

The symmetry $d(\bfu,\bfv)=d(\bfv,\bfu)$ follows from

Note that the notions of norm and distance are relative to the inner product used! For example, with the inner product given in Example 8.8, the norm of a real-valued continuous function on $[-1,1]$ is given as follows:

Note that the notions of norm and distance are relative to the inner product used! For example, with the inner product given in Example 8.8, the norm of a real-valued continuous function on $[-1,1]$ is given as follows:

However, if we define the inner product as \[ \int_{-1}^{1} e^xf(x)g(x) ~dx, \] $$\ds \text{then } \;||\mathbf f|| = \sqrt{\int_{-1}^{1} e^x \big(\,f(x)\big)^2 dx}.$$

As in $\R^n$ with inner product given by the usual dot product, we say that two vectors $\bfu,\bfv\in V$ are orthogonal if $$ \langle\bfu,\bfv\rangle=0. $$ We need a bit of preparation before we can talk more generally about the angle between two vectors, see Section 10.5.

Remark: Notion of orthogonality is also relative to the inner product used.

Let $V$ be a real inner product space, and let $\bfu,\bfv\in V$. Then, $$ |\langle\bfu,\bfv\rangle|\leq||\bfu||\,||\bfv||. $$ Moreover, this inequality is an equality if and only if $\bfu$ or $\bfv$ is a scalar multiple of the other vector.

For $\u = \mathbf 0\;$ ✅ 😃. For $\u \neq \mathbf 0$, let $t\in \R$ and

In this case we must have that $b^2 -4ac \leq 0$. So \[ 4 \langle \u, \v\rangle^2 - 4 \langle \u, \u\rangle\langle \v, \v\rangle \leq 0. \] Or equivalently \[ \langle \u, \v\rangle^2 \leq \langle \u, \u\rangle \langle \v, \v\rangle . \]

Taking square roots of both sides and using the fact that $\langle \u, \u\rangle$ and $\langle \v, \v\rangle $ nonnegative yields

Let $U$ be a subset of the real inner product space $V$. The orthogonal complement of $U$, denoted by $U^\bot$, is the set of all vectors in $V$ that are orthogonal to every vector in $U$. That is, $$ U^\bot=\{\bfv\in V\,|\,\langle\bfv,\bfu\rangle=0\text{ for every } \bfu\in U\}. $$ This is a vector space with addition and scalar multiplication inherited from $V$.

Write $\;\ds A = \begin{pmatrix} \mathbf r_1\\ — \\ \mathbf r_2\\ — \\ \vdots \\ —\\ \mathbf r_m\\ \end{pmatrix}\; $ where $\mathbf r_1, \mathbf r_2,\ldots ,\mathbf r_m$ denote the rows, $$\text{i.e., }\quad\text{Row}(A) = \text{span}\left(\left\{ \mathbf r_1, \mathbf r_2,\ldots ,\mathbf r_n\right\}\right)$$

Let $\mathbf x = \left(x_1,x_2, \ldots, x_n \right)$ be a row vector, then

A nonempty subset $W$ of a vector space $V$ is a subspace of $V$ if it is a vector space under the addition and scalar multiplication defined on $V$. To verify that a subset is a subspace, one checks the following:

• $\mathbf 0 \in W$?
• $\u + \v \in W$? for all $\u,\v\in W$.
• $k\u \in W$? for all $\u \in W, k\in \R$.

Now we prove that $U^\bot$ is an example of a subspace.
📝 Exercise.