Worked examples — Orthogonal sets and orthonormal basis

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The scenario matrix

Every problem in this topic is one of these cells. The examples below are labelled with the cell they cover. (Here $W$ is the flat sheet the basis spans, as just defined.)

Cell	Situation	What's tricky	Example
C1	Orthogonal (not unit) basis, $\mathbf{y}$ inside $W$	Must keep the denominator ${\mathbf{v}}_j\cdot {\mathbf{v}}_j$	A
C2	Orthonormal basis	Denominator is $1$ — don't write it	B
C3	Orthogonal set of size $p<n$; $\mathbf{y}$ outside span	Formula gives a projection, not $\mathbf{y}$	C
C4	Degenerate: a zero vector in the set	Independence breaks; length-squared $=0$ divides by zero	D
C5	Sign cases — negative coordinates, mixed-sign components	Bookkeeping of signs in dot products	E
C6	Orthogonal matrix preserves length/angle	Verify $U^{\mathsf{T}}U=I$; check norm	F
C7	Limiting case — near-orthogonal vectors, what "almost" costs	Small off-diagonal error	G
C8	Word problem (signals / forces) mapped to coordinates	Translating words into dot products	H
C9	Exam twist — "orthogonal columns" that are not orthonormal	Naming trap; must normalize first	I

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Example A — Cell C1: orthogonal basis, keep the denominator

Step 1 — check orthogonal. ${\mathbf{v}}_1\cdot {\mathbf{v}}_2=(2)(-1)+(1)(2)=-2+2=0.$ ✓ Why this step? Only then does the shortcut formula $c_j=\frac{\mathbf{y}\cdot {\mathbf{v}}_j}{{\mathbf{v}}_j\cdot {\mathbf{v}}_j}$ apply. (Here the generic ${\mathbf{u}}_j$ of the parent note are our ${\mathbf{v}}_j$ — same formula, matching names.)

Step 2 — the two dot products in the numerator. $\mathbf{y}\cdot {\mathbf{v}}_1=(4)(2)+(3)(1)=11,\mathbf{y}\cdot {\mathbf{v}}_2=(4)(-1)+(3)(2)=2.$ Why this step? The numerator measures "how much of $\mathbf{y}$ points along ${\mathbf{v}}_j$."

Step 3 — the denominators (the length-squared). ${\mathbf{v}}_1\cdot {\mathbf{v}}_1=4+1=5,{\mathbf{v}}_2\cdot {\mathbf{v}}_2=1+4=5.$ Why this step? Because the ${\mathbf{v}}_j$ are not unit vectors, we must divide out their own size. Skipping this is the classic error.

Step 4 — assemble. $c_1=\frac{11}{5},c_2=\frac{2}{5},\mathbf{y}=\frac{11}{5}{\mathbf{v}}_1+\frac{2}{5}{\mathbf{v}}_2.$

Verify: $\frac{11}{5}\left[\begin{array}{c}2\\ 1\end{array}\right]+\frac{2}{5}\left[\begin{array}{c}-1\\ 2\end{array}\right]=\left[\begin{array}{c}(22-2)/5\\ (11+4)/5\end{array}\right]=\left[\begin{array}{c}4\\ 3\end{array}\right]=\mathbf{y}.$ ✓

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Example B — Cell C2: orthonormal, no denominator

Step 1 — confirm unit length. ${\mathbf{q}}_1\cdot {\mathbf{q}}_1=\frac{1}{5}(4+1)=1.$ ✓ Why this step? Once each ${\mathbf{q}}_j\cdot {\mathbf{q}}_j=1$, the coordinate is just $\mathbf{y}\cdot {\mathbf{q}}_j$.

Step 2 — dot products. $c_1=\mathbf{y}\cdot {\mathbf{q}}_1=\frac{1}{\sqrt{5}}(8+3)=\frac{11}{\sqrt{5}},c_2=\mathbf{y}\cdot {\mathbf{q}}_2=\frac{1}{\sqrt{5}}(-4+6)=\frac{2}{\sqrt{5}}.$ Why this step? No denominator needed — orthonormality already did that division for us.

Verify: $c_1{\mathbf{q}}_1+c_2{\mathbf{q}}_2=\frac{11}{\sqrt{5}}\cdot \frac{1}{\sqrt{5}}\left[\begin{array}{c}2\\ 1\end{array}\right]+\frac{2}{\sqrt{5}}\cdot \frac{1}{\sqrt{5}}\left[\begin{array}{c}-1\\ 2\end{array}\right]=\frac{11}{5}\left[\begin{array}{c}2\\ 1\end{array}\right]+\frac{2}{5}\left[\begin{array}{c}-1\\ 2\end{array}\right]=\left[\begin{array}{c}4\\ 3\end{array}\right].$ ✓ Same $\mathbf{y}$, different coordinates — consistent with A.

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Example C — Cell C3: $\mathbf{y}$ outside the span → projection

Step 1 — apply the formula anyway. $c_1=\frac{\mathbf{y}\cdot {\mathbf{u}}_1}{{\mathbf{u}}_1\cdot {\mathbf{u}}_1}=\frac{2}{1}=2,c_2=\frac{\mathbf{y}\cdot {\mathbf{u}}_2}{{\mathbf{u}}_2\cdot {\mathbf{u}}_2}=\frac{3}{1}=3.$ Why this step? The formula never asked whether $\mathbf{y}\in W$; it just projects.

Step 2 — build the result. $\hat{\mathbf{y}}=2{\mathbf{u}}_1+3{\mathbf{u}}_2=\left[\begin{array}{c}2\\ 3\\ 0\end{array}\right].$ Why this step? This $\hat{\mathbf{y}}$ is the point of $W$ closest to $\mathbf{y}$ — the Orthogonal projection onto the plane. The $z=5$ part is dropped because no basis vector points up.

Step 3 — the leftover is orthogonal to $W$. $\mathbf{y}-\hat{\mathbf{y}}=\left[\begin{array}{c}0\\ 0\\ 5\end{array}\right],(\mathbf{y}-\hat{\mathbf{y}})\cdot {\mathbf{u}}_1=0,(\mathbf{y}-\hat{\mathbf{y}})\cdot {\mathbf{u}}_2=0.$ Why this step? This is the defining property of a projection — the error is perpendicular to the subspace, the seed of Least squares.

Verify: $\hat{\mathbf{y}}\ne \mathbf{y}$ (they differ in the third slot), and the residual $[0,0,5]$ dots to $0$ with both basis vectors. ✓ The lesson of C3: the formula always returns the projection; it only equals $\mathbf{y}$ when $\mathbf{y}$ actually lives in the span.

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Example D — Cell C4: a zero vector poisons the set

Step 1 — is it "orthogonal"? ${\mathbf{w}}_1\cdot {\mathbf{w}}_2=0$. Technically yes — the zero vector is orthogonal to everything. Why this step? The definition only bans nonzero distinct pairs from being non-perpendicular; it says nothing about zero, so zero sneaks in.

Step 2 — independence fails. The relation $0\cdot {\mathbf{w}}_1+5\cdot {\mathbf{w}}_2=0$ holds with a nonzero coefficient. So the set is dependent — it is not a basis. Why this step? Section 2 of the parent required nonzero orthogonal vectors. The proof isolates $c_j\Vert {\mathbf{w}}_j{\Vert }^2=0$; when $\Vert {\mathbf{w}}_2\Vert =0$ it cannot conclude $c_2=0$.

Step 3 — the formula divides by zero. $c_2=\frac{\mathbf{y}\cdot {\mathbf{w}}_2}{{\mathbf{w}}_2\cdot {\mathbf{w}}_2}=\frac{0}{0}\text{— undefined.}$ Why this step? The denominator ${\mathbf{w}}_2\cdot {\mathbf{w}}_2=0$ is exactly the length-squared, which is $0$ only for the zero vector.

Verify: ${\mathbf{w}}_2\cdot {\mathbf{w}}_2=0$, confirming the division is illegal. ✓ Rule: always discard zero vectors before calling a set orthogonal.

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Example E — Cell C5: all the signs

Step 1 — orthogonality (three checks). ${\mathbf{a}}_1\cdot {\mathbf{a}}_2=1-1+0=0,{\mathbf{a}}_1\cdot {\mathbf{a}}_3=1+1-2=0,{\mathbf{a}}_2\cdot {\mathbf{a}}_3=1-1+0=0.$ Why this step? Every pair must vanish or the shortcut is invalid.

Step 2 — numerators (watch the signs). $\mathbf{y}\cdot {\mathbf{a}}_1=3-1-2=0,\mathbf{y}\cdot {\mathbf{a}}_2=3+1+0=4,\mathbf{y}\cdot {\mathbf{a}}_3=3-1+4=6.$ Why this step? A zero numerator ($c_1$) is legal and meaningful — $\mathbf{y}$ has no component along ${\mathbf{a}}_1$.

Step 3 — denominators, then divide. ${\mathbf{a}}_1\cdot {\mathbf{a}}_1=3,{\mathbf{a}}_2\cdot {\mathbf{a}}_2=2,{\mathbf{a}}_3\cdot {\mathbf{a}}_3=6.$ $c_1=\frac{0}{3}=0,c_2=\frac{4}{2}=2,c_3=\frac{6}{6}=1.$ Why this step? The ${\mathbf{a}}_j$ are orthogonal but not unit length, so we must divide each numerator by its own length-squared — otherwise the reconstructed vector would come out scaled wrong.

Verify: $0{\mathbf{a}}_1+2{\mathbf{a}}_2+1{\mathbf{a}}_3=\left[\begin{array}{c}0+2+1\\ 0-2+1\\ 0+0-2\end{array}\right]=\left[\begin{array}{c}3\\ -1\\ -2\end{array}\right]=\mathbf{y}.$ ✓ Sign bookkeeping survived intact.

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Example F — Cell C6: rotation preserves length and angle

Step 1 — confirm orthonormal columns. Column 1 length${}^2=\frac{1}{4}+\frac{3}{4}=1$; columns dot to $\frac{1}{2}(-\frac{\sqrt{3}}{2})+\frac{\sqrt{3}}{2}\cdot \frac{1}{2}=0$. So $U^{\mathsf{T}}U=I$ (the clean $1$-diagonal grid). Why this step? Length-preservation follows from $U^{\mathsf{T}}U=I$, not from $U$ being "some" matrix — see QR decomposition where this $U$-with-orthonormal-columns is the $Q$.

Step 2 — apply $U$. $U\mathbf{x}=\left[\begin{array}{c}\frac{1}{2}\cdot 2\\ \frac{\sqrt{3}}{2}\cdot 2\end{array}\right]=\left[\begin{array}{c}1\\ \sqrt{3}\end{array}\right].$ Why this step? We rotate the vector $2$ units long by $60^{\circ }$; the figure shows the arc it sweeps.

Step 3 — compare norms. $\Vert U\mathbf{x}\Vert =\sqrt{1+3}=2=\Vert \mathbf{x}\Vert .$ Why this step? This is the whole point of the example — putting the two lengths side by side is what confirms the geometric claim "rotation doesn't stretch," turning the abstract identity $U^{\mathsf{T}}U=I$ into a concrete number match.

Verify: $\Vert U\mathbf{x}{\Vert }^2=1^2+(\sqrt{3}{)}^2=4$ and $\Vert \mathbf{x}{\Vert }^2=2^2=4$. ✓ Length untouched, exactly as Dot product and norms predicts for orthonormal maps.

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Example G — Cell C7: the limiting / "almost orthogonal" case

Step 1 — the true off-diagonal. ${\mathbf{b}}_1\cdot {\mathbf{b}}_2=ϵ=0.01\ne 0$. Why this step? This nonzero entry is exactly what the shortcut assumes is zero; it is the source of all error.

Step 2 — naive coordinates. $c_1^{\text{naive}}=\mathbf{y}\cdot {\mathbf{b}}_1=1,c_2^{\text{naive}}=\mathbf{y}\cdot {\mathbf{b}}_2=ϵ+1=\mathrm{1.01.}$ Why this step? This is what a careless solver writes, ignoring the non-orthogonality.

Step 3 — reconstruct and measure error. $\hat{\mathbf{y}}=1\cdot {\mathbf{b}}_1+1.01\cdot {\mathbf{b}}_2=\left[\begin{array}{c}1+1.01(0.01)\\ 1.01\end{array}\right]=\left[\begin{array}{c}1.0101\\ 1.01\end{array}\right].$ Error vector $\hat{\mathbf{y}}-\mathbf{y}=\left[\begin{array}{c}0.0101\\ 0.01\end{array}\right]$. Why this step? It quantifies the cost of pretending "almost orthogonal" equals "orthogonal."

Verify: with $ϵ=0.01$, the first error component $=ϵ+{ϵ}^2=0.0101$ and second $=ϵ=0.01$. Both $\to 0$ as $ϵ\to 0$. ✓ Moral: near-orthogonality is not orthogonality — for exactness, run Gram-Schmidt process first.

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Example H — Cell C8: word problem (two microphones)

Step 1 — confirm orthonormal. ${\mathbf{q}}_1\cdot {\mathbf{q}}_1=0.36+0.64=1$; ${\mathbf{q}}_1\cdot {\mathbf{q}}_2=-0.48+0.48=0$. ✓ Why this step? Only orthonormal channels let each reading be a single dot product with no crosstalk.

Step 2 — channel readings = dot products. $c_1=\mathbf{s}\cdot {\mathbf{q}}_1=3+8=11,c_2=\mathbf{s}\cdot {\mathbf{q}}_2=-4+6=2.$ Why this step? Each channel measures the component of $\mathbf{s}$ along its own direction.

Step 3 — energy check (Parseval). $c_1^2+c_2^2=121+4=125,\Vert \mathbf{s}{\Vert }^2=25+100=125.$ Why this step? Orthonormal bases conserve total energy — a physical sanity check the engineer relies on.

Verify: $c_1^2+c_2^2=125=\Vert \mathbf{s}{\Vert }^2$, and $11{\mathbf{q}}_1+2{\mathbf{q}}_2=\left[\begin{array}{c}6.6-1.6\\ 8.8+1.2\end{array}\right]=\left[\begin{array}{c}5\\ 10\end{array}\right]=\mathbf{s}.$ ✓

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Example I — Cell C9: the exam naming trap

Step 1 — columns orthogonal? $\left[\begin{array}{c}3\\ 1\end{array}\right]\cdot \left[\begin{array}{c}-1\\ 3\end{array}\right]=-3+3=0.$ ✓ Perpendicular, yes. Why this step? The claim's premise is genuinely true — that's what makes the trap seductive.

Step 2 — but are they unit vectors? Each column has length${}^2=9+1=10\ne 1$. Why this step? "Orthogonal matrix" (a historical misnomer) demands orthonormal columns. Length matters, so we must test it.

Step 3 — test $M^{\mathsf{T}}M$. $M^{\mathsf{T}}M=\left[\begin{array}{cc}3& 1\\ -1& 3\end{array}\right]\left[\begin{array}{cc}3& -1\\ 1& 3\end{array}\right]=\left[\begin{array}{cc}10& 0\\ 0& 10\end{array}\right]=10I\ne I.$ Why this step? An orthogonal matrix must satisfy $M^{\mathsf{T}}M=I$ (the clean $1$-diagonal grid); here we get $10I$ instead, which shows $M^{\mathsf{T}}\ne M^{-1}$. In fact $M^{-1}=\frac{1}{10}{M}^{\mathsf{T}}$.

Step 4 — the fix: normalize. Divide every column by its length $\sqrt{10}$. Then $Q=\frac{1}{\sqrt{10}}M$ has orthonormal columns, and now $Q^{-1}=Q^{\mathsf{T}}$. Why this step? Normalizing forces each column's length-squared to $1$, which is precisely the missing condition, converting the merely-orthogonal $M$ into a true orthogonal matrix $Q$.

Verify: $M^{\mathsf{T}}M=10I$ (so $M$ is not orthogonal), while $\left(\frac{1}{\sqrt{10}}M{)}^{\mathsf{T}}\right(\frac{1}{\sqrt{10}}M)=\frac{1}{10}(10I)=I$. ✓ The word "orthogonal matrix" secretly means orthonormal + square — the same distinction that matters for Eigenvalues and eigenvectors of symmetric matrices.

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Recall Which cell is this?

A basis has perpendicular columns but lengths $\sqrt{7}$. To get coordinates you... ::: keep the denominator ${\mathbf{v}}_j\cdot {\mathbf{v}}_j=7$ (Cell C1 — orthogonal, not orthonormal). The coordinate formula runs fine but returns a vector $\ne \mathbf{y}$. What happened? ::: $\mathbf{y}$ was outside the span; you got its projection (Cell C3). Your denominator is $0$. Diagnosis? ::: A zero vector is in the set — discard it (Cell C4). "Orthogonal columns ⇒ orthogonal matrix." True? ::: No — needs unit length and square (Cell C9); normalize first.