Perpendicularity

Inner Product Spaces: Inner product, length, and angle

Perpendicularity

Two vectors having the same direction are at an angle $0$ . If they have opposite direction, then their angle is $\pi$ , which corresponds to $180^\circ$ . The angle $\frac{\pi}{2}$ (or $90^\circ$ ), lying right between these two, has a special significance.

Let $V$ be an inner product space. Two vectors $\vec{a}$ and $\vec{b}$ of $V$ are perpendicular to each other if $\dotprod{\vec{a}}{\vec{b}} = 0$ . In this case, we write $\vec{a}\ \perp\ \vec{b}$ .

In this definition $\vec{a}$ and/or $\vec{b}$ may be the zero vector.

Because $\dotprod{\vec{a}}{\vec{b}} =\dotprod{\vec{b}}{\vec{a}}$ , the vector $\vec{a}$ is perpendicular to $\vec{b}$ if and only if $\vec{b}$ is perpendicular to $\vec{a}$ . In other words: the perpendicularity relation is symmetric.

In $\mathbb{R}^n$ with the standard inner product, for each $i$ and $j$ with $i\neq j$ the vectors $\vec{e}_i$ and $\vec{e}_j$ are perpendicular to each other. In addition, each of these vectors has length $1$ . These properties are called orthonormality and will later play an important role for inner product spaces.

Angle When $\vec{a}$ and $\vec{b}$ are distinct from the zero vector, then it follows from $\vec{a}\perp\vec{b}$ that the angle $\varphi$ between the two vectors satisfies $\cos(\varphi) = 0$ , so $\varphi=\frac{\pi}{2}$ (that is, $90^\circ$ ). This corresponds to the case of the plane which we discussed.

By means of the definition of angles, we can place the Pythagorean theorem in a formal setting.

Let $V$ be an inner product space.

The vectors $\vec{a}$ and $\vec{b}$ are perpendicular to each other if and only if
$\norm{\vec{a}+\vec{b}}^2 = \norm{\vec{a}}^2 + \norm{\vec{b}}^2$
If the vectors $\vec{a}_1 \ldots ,\vec{a}_k$ are mutually perpendicular, that is to say: $\dotprod{\vec{a}_i}{\vec{a}_j }=0$ if $i\neq j$ , then $\norm{\vec{a}_1+\cdots + \vec{a}_k}^2 =\norm{\vec{a}_1}^2 + \cdots + \norm{\vec{a}_k }^2$

In order to prove the first part, we use the properties of the inner product and work out $\norm{\vec{a}+\vec{b}}^2$ :
$\begin{array}{rcl} \norm{\vec{a}+\vec{b}}^2 & =&\dotprod{(\vec{a}+\vec{b})}{( \vec{a} + \vec{b})}\\&=&\dotprod{\vec{a} }{\vec{a}} + \dotprod{\vec{a}}{ \vec{b}}+ \dotprod{\vec{b}}{\vec{a}}+\dotprod{\vec{b}}{\vec{b}}\\ & =&\norm{\vec{a}}^2 + \norm{\vec{b}}^2 +2(\dotprod{\vec{a}}{\vec{b}}) \end{array}$ Since $\vec{a}$ and $\vec{b}$ are perpendicular by assumption, we have $\dotprod{\vec{a} }{\vec{b}}=0$ . When we use this in the above expression, we find $\norm{\vec{a}+\vec{b}}^2 =\norm{\vec{a}}^2 +\norm{\vec{b}}^2$ .

In order to prove the other implication we assume that $\norm{\vec{a}+\vec{b}}^2 =\norm{\vec{a}}^2 + \norm{\vec{b}}^2$ . This implies that $2(\dotprod{\vec{a} }{\vec{b}})=0$ , so $\dotprod{\vec{a}}{\vec{b}}=0$ and consequently $\vec{a}\perp\vec{b}$ . This proves the first statement.

The second statement follows by induction from the first. The full proof is left to the reader.

If we look at the vector space $V=\mathbb{R}^2$ with the standard inner product, we get the Pythagorean theorem as we know it. If the vectors $\vec{a}$ and $\vec{b}$ are perpendicular to each other, then, according to the theorem, $\norm{\vec{a}+\vec{b}}^2=\norm{\vec{a}}^2+\norm{\vec{b}}^2$ . The vector $\vec{a}+\vec{b}$ is of course the vector $\vec{c}$ of the Pythagorean theorem.

Counterexample

The fact that the second statement contains no 'if and only if', can be understood by looking at the following counterexample. Take the vectors $\vec{a}_1=\rv{1,0}$ , $\vec{a}_2=\rv{0,1}$ , and $\vec{a}_3=\rv{1,-1}$ . We have $\norm{\vec{a}_1+\vec{a}_2+\vec{a}_3}^2 = \norm{\rv{2,0}}^2=4$ and $\norm{\vec{a}_1}^2 + \norm{\vec{a}_2}^2 + \norm{\vec{a}_3}^2= 1+1+2 = 4$ so the equality of the statement holds. However, the inner product $\dotprod{\vec{a}_1}{\vec{a}_3}$ is equal to $1$ and so the vectors $\vec{a}_1$ and $\vec{a}_3$ are not perpendicular!

Provide a vector of the inner product space $\mathbb{R}^3$ (with the standard inner product), which is distinct from the zero vector and perpendicular to the vector $\rv{8,9,5}$

$\rv{0,1,-{{9}\over{5}}}$

The answer is not unique.

A vector $\rv{x,y,z}$ is perpendicular to $\rv{8,9,5}$ if and only if $\dotprod{\rv{8,9,5}}{\rv{x,y,z}}=0$ . This means that we must have $x\cdot 8+y\cdot 9 + z\cdot 5=0$ This is a single linear equation with three unknowns, so we can simply choose values for $x$ and $y$ (not both equal to $0$ so as to avoid the zero vector for an answer) and solve the resulting linear equation in $z$ .

We take $x=0$ and $y=1$ . This gives the equation $9+z\cdot 5=0$ , which has solution $z=-{{9}\over{5}}$ . We then find the vector
$\rv{0,1,-{{9}\over{5}}}$

The solution is not unique; each solution $\rv{x,y,z}$ of $8 x+9 y+5 z=0$ distinct from the zero vector is a good answer.

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