Solution. To show that vectors $\boldsymbol{v}$ and $\boldsymbol{w}$ form an orthogonal set, we must check that they are perpendicular to each other. We know from theoremlink that if the dot product of two vectors is zero, then they are perpendicular. So let's compute the dot product between them:

Because $\boldsymbol{v}$ and $\boldsymbol{w}$ are orthogonal to each other, we conclude that they form an orthogonal set. Let's also visually confirm this:

Indeed, they are perpendicular!

Solution. Let's first check that they are perpendicular to each other:

Because their dot product is zero, they are perpendicular. We can also easily tell that these vectors have both length one, and so $\boldsymbol{e}_1$ and $\boldsymbol{e}_2$ form an orthonormal set by definition!

We can visually confirm this as well:

Proof. We know from theoremlink that:

• the vector $\boldsymbol{u}$ in formula \eqref{eq:WDgE7iuxsGj6txyl1ss} has a length of one.

• the vector $\boldsymbol{u}$ points in the same direction as $\boldsymbol{v}$.

Since the direction of every vector in the orthogonal set is preserved, the new unit vectors will also be perpendicular to each other. This means that the new set of vectors is orthonormal. This completes the proof.

Solution. To convert an orthogonal set of vectors into an orthonormal set, we simply have to convert them into unit vectors. Let's convert $\boldsymbol{v}$ into a unit vector first:

Next, let's convert $\boldsymbol{w}$ into a unit vector:

Visually, $\boldsymbol{u}_1$ and $\boldsymbol{u}_2$ look as follows:

Proof. Suppose $S$ is an orthogonal set with the following vectors:

We can use the definitionlink of linear independence to show that $S$ is linearly independent:

If the only way for this equality to hold is $c_1=c_2=\cdots=c_n$, then $S$ is linearly independent.

Let's perform a dot product of both sides of \eqref{eq:bdBBKeLSAfSp81Aw4Eb} with $\boldsymbol{v}_i$ where $i$ is any value between $1$ and $n$ to get:

Because every pair of vectors in $S$ is orthogonal, we have that $\boldsymbol{v}_j\cdot{\boldsymbol{v}_i}=0$ if $j\ne{i}$. Therefore, \eqref{eq:eKibp1fyLn3VM48eUi7} reduces to:

Since vectors in $S$ are assumed to be non-zero, the only way for \eqref{eq:JXXpECsUlFoSOLVadhk} to hold is if $c_i=0$. This means all the scalar values must be zero for \eqref{eq:bdBBKeLSAfSp81Aw4Eb} to hold. Therefore, we conclude that $S$ is linearly independent. This completes the proof.

Solution. We already know from this examplelink that these vectors are orthonormal, that is, they are perpendicular to each other and both have a length of one. Since $\boldsymbol{e}_1$ and $\boldsymbol{e}_2$ are the standard basis vectors for $\mathbb{R}^2$, we conclude that $\boldsymbol{e}_1$ and $\boldsymbol{e}_2$ form an orthonormal basis for $\mathbb{R}^2$.

Proof. By definitionlink of basis, we know that $V$ is spanned by $S=\{\boldsymbol{v}_1,\boldsymbol{v}_2,\cdots,\boldsymbol{v}_n\}$. Therefore, any vector $\boldsymbol{x}$ in $V$ can be expressed as a linear combination of vectors in $S$ like so:

Where $c_1$, $c_2$, $\cdots$, $c_n$ are some scalars. The dot product of both sides of \eqref{eq:QoehhocPZoAdSMKFRA8} with any vector $\boldsymbol{v}_i$ in $S$ gives:

Since $S$ is an orthogonal set, we have that $\boldsymbol{v}_j\cdot{\boldsymbol{v}_i}=0$ if $j\ne{i}$. Therefore, \eqref{eq:W75Aq4NFrJZKwLENRhG} becomes:

Note that the last step holds because of theoremlink. Rearranging \eqref{eq:jRRm2HwsYMyAVjhpGih} to make to $c_i$ the subject gives:

We know that \eqref{eq:FdyfpqsrgKlsrW5fpKk} holds for $i=1$, $2$, $\cdots$, $n$. We can substitute \eqref{eq:FdyfpqsrgKlsrW5fpKk} into \eqref{eq:QoehhocPZoAdSMKFRA8} to get:

This completes the proof.

Proof. We have just proven in theoremlink that if $S_*=\{\boldsymbol{v}_1,\boldsymbol{v}_2,\cdots,\boldsymbol{v}_n\}$ is an orthogonal basis for $V$, then any vector $\boldsymbol{x}$ in $V$ can be expressed as:

If $S=\{\boldsymbol{u}_1,\boldsymbol{u}_2,\cdots,\boldsymbol{u}_n\}$ is an orthonormal basis, then each vector in $S$ is a unit vector by definition. Therefore, \eqref{eq:LUhuxjlcwyV5z0S9cns} simplifies to:

This completes the proof.

Solution. A simple orthonormal basis for $\mathbb{R}^2$ is:

We know from theoremlink that $\boldsymbol{v}$ can be expressed as:

The next example demonstrates how we could extend a given set of orthogonal vectors into an orthogonal basis of some vector space. The technique demonstrated here will be useful for future proofs.

Solution. By theoremlink, we need $3$ basis vectors for $\mathbb{R}^3$. We must find $\boldsymbol{q}_3$ such that $\boldsymbol{q}_3$ is perpendicular to both $\boldsymbol{q}_1$ and $\boldsymbol{q}_2$. Let's represent $\boldsymbol{q}_3$ as:

By propertylink of dot product, for $\boldsymbol{q}_3$ to be perpendicular to $\boldsymbol{q}_1$ and $\boldsymbol{q}_2$, the following must hold:

We can explicitly write the dot products as:

Notice that this is a system of linear equationslink! Solving this system will give us all the components of $\boldsymbol{q}_3$. Let's use Gaussian elimination to solve the system:

We have that $x_3$ is a free variablelink so let $x_3=t$ where $t$ is any scalar. The second row gives us:

The first row gives us:

Therefore, $\boldsymbol{q}_3$ can be expressed as:

For simplicity, setting $t=5$ gives us:

Let's just check that $\boldsymbol{q}_3$ is orthogonal to both $\boldsymbol{q}_1$ and $\boldsymbol{q}_2$ like so:

Great, this is exactly what we would expect. Therefore, an orthogonal basis for $\mathbb{R}^3$ is: