Worked examples — QR decomposition

The scenario matrix

Before working examples, let us list every distinct thing a QR problem can throw at you. Each row is a case class; the last column names the example that hits it.

We will now walk them in order. Keep the Gram-Schmidt process steps in mind — every example is that process, just with different inputs.

A one-figure reminder of the geometry

Every example does the same physical move: keep the first arrow, then throw away the shadow of the next arrow so only the perpendicular part survives.

Example 1 — Case 1: clean tall matrix

1. First axis. ${\mathbf{u}}_1={\mathbf{a}}_1=(1,1,0)$, $\Vert {\mathbf{u}}_1\Vert =\sqrt{1^2+1^2+0^2}=\sqrt{2}$. So $r_{11}=\sqrt{2}$ and ${\mathbf{q}}_1=\frac{1}{\sqrt{2}}(1,1,0)$. Why this step? The first clean direction is just the first column, shrunk to length 1 — there is nothing earlier to be perpendicular to.
2. Shadow coefficient. $r_{12}={\mathbf{q}}_1^{\top }{\mathbf{a}}_2=\frac{1}{\sqrt{2}}(1\cdot 2+1\cdot 1+0\cdot 1)=\frac{3}{\sqrt{2}}$. Why this step? $r_{12}$ measures how much of ${\mathbf{a}}_2$ already points along ${\mathbf{q}}_1$ — the part we must remove.
3. Remove the shadow. ${\mathbf{u}}_2={\mathbf{a}}_2-r_{12}{\mathbf{q}}_1=(2,1,1)-\frac{3}{\sqrt{2}}\cdot \frac{1}{\sqrt{2}}(1,1,0)=(2,1,1)-\frac{3}{2}(1,1,0)=(\frac{1}{2},-\frac{1}{2},1)$. Why this step? What remains is the piece of ${\mathbf{a}}_2$ perpendicular to ${\mathbf{q}}_1$ — a brand-new direction.
4. Second axis. $\Vert {\mathbf{u}}_2\Vert =\sqrt{\frac{1}{4}+\frac{1}{4}+1}=\sqrt{\frac{3}{2}}=r_{22}$, and ${\mathbf{q}}_2=\frac{1}{\sqrt{3/2}}(\frac{1}{2},-\frac{1}{2},1)$. Why this step? Normalize so ${\mathbf{q}}_2$ has length 1; now $\{{\mathbf{q}}_1,{\mathbf{q}}_2\}$ is orthonormal.

Verify: ${\mathbf{q}}_1^{\top }{\mathbf{q}}_2=\frac{1}{\sqrt{2}}\cdot \frac{1}{\sqrt{3/2}}(\frac{1}{2}-\frac{1}{2}+0)=0$ ✓ (perpendicular), each $\mathbf{q}$ has length 1, and $QR=A$. All diagonal entries positive → this is the unique QR.

Example 2 — Case 2: a backward-pointing first column (sign convention)

1. First axis. $\Vert {\mathbf{a}}_1\Vert =\sqrt{(-3{)}^2+(-4{)}^2}=\sqrt{25}=5$, so $r_{11}=5$ (a length is always $\ge 0$ — the minus signs squared away). ${\mathbf{q}}_1=\frac{1}{5}(-3,-4)$. Why this step? A norm cannot be negative; the sign lives in the direction ${\mathbf{q}}_1$, not in $r_{11}$. This is exactly why QR fixes $r_{kk}>0$: it forces the sign into $Q$.
2. Shadow coefficient. $r_{12}={\mathbf{q}}_1^{\top }{\mathbf{a}}_2=\frac{1}{5}(-3\cdot 0+-4\cdot 5)=\frac{-20}{5}=-4$. Why this step? $r_{12}$ can be negative — it just means ${\mathbf{a}}_2$ leans against ${\mathbf{q}}_1$. Off-diagonal signs are unconstrained.
3. Remove shadow. ${\mathbf{u}}_2=(0,5)-(-4)\cdot \frac{1}{5}(-3,-4)=(0,5)-(\frac{12}{5},\frac{16}{5})=(-\frac{12}{5},\frac{9}{5})$. Why this step? Subtract the shadow to leave only the part of ${\mathbf{a}}_2$ perpendicular to ${\mathbf{q}}_1$; here the shadow itself has negative sign, so subtracting it adds back — the same rule as always, just with a signed coefficient.
4. Second axis. $\Vert {\mathbf{u}}_2\Vert =\sqrt{\frac{144}{25}+\frac{81}{25}}=\sqrt{\frac{225}{25}}=3=r_{22}$. ${\mathbf{q}}_2=\frac{1}{3}(-\frac{12}{5},\frac{9}{5})=(-\frac{4}{5},\frac{3}{5})$. Why this step? Normalize the perpendicular leftover to length 1 so $\{{\mathbf{q}}_1,{\mathbf{q}}_2\}$ is orthonormal; the diagonal entry $r_{22}=3$ is again a positive length.

Verify: diagonal $5,3>0$ ✓. $Q^{\top }Q=I$ ✓ (compute: $(-\frac{3}{5}{)}^2+(-\frac{4}{5}{)}^2=1$, cross term $(-\frac{3}{5})(-\frac{4}{5})+(-\frac{4}{5})(\frac{3}{5})=\frac{12}{25}-\frac{12}{25}=0$). $QR=A$ ✓. The negativity of the data went entirely into $Q$, never into $R$'s diagonal.

Example 3 — Case 3: an already-orthogonal matrix (limiting case)

1. First axis. $\Vert {\mathbf{a}}_1\Vert =\sqrt{\frac{1}{2}+\frac{1}{2}}=1$, so $r_{11}=1$, ${\mathbf{q}}_1={\mathbf{a}}_1$. Why this step? Column 1 is already length 1 — normalizing changes nothing.
2. Shadow coefficient. $r_{12}={\mathbf{q}}_1^{\top }{\mathbf{a}}_2=\frac{1}{\sqrt{2}}\cdot (-\frac{1}{\sqrt{2}})+\frac{1}{\sqrt{2}}\cdot \frac{1}{\sqrt{2}}=-\frac{1}{2}+\frac{1}{2}=0$. Why this step? Zero shadow means ${\mathbf{a}}_2$ is already perpendicular to ${\mathbf{q}}_1$ — nothing to subtract.
3. Second axis. ${\mathbf{u}}_2={\mathbf{a}}_2-0={\mathbf{a}}_2$, $\Vert {\mathbf{u}}_2\Vert =1=r_{22}$, ${\mathbf{q}}_2={\mathbf{a}}_2$. Why this step? With zero shadow removed, the leftover is ${\mathbf{a}}_2$, and it is already unit length, so ${\mathbf{q}}_2={\mathbf{a}}_2$ with $r_{22}=1$ — Gram–Schmidt confirms the frame was already clean.

Verify: $QR=A\cdot I=A$ ✓. This is the "QR does nothing" limit predicted in the parent note: for an orthogonal $A$, every $r_{kk}=1$ and every off-diagonal $=0$.

Example 4 — Case 4: an already upper-triangular matrix

1. First axis. $\Vert {\mathbf{a}}_1\Vert =\Vert (2,0)\Vert =2=r_{11}$, ${\mathbf{q}}_1=(1,0)$. Why this step? Column 1 lies exactly along the first coordinate axis, so its clean direction is $(1,0)$ and its length is $r_{11}=2$.
2. Shadow coefficient. $r_{12}={\mathbf{q}}_1^{\top }{\mathbf{a}}_2=(1,0)\cdot (5,3)=5$. Why this step? $r_{12}$ measures how far ${\mathbf{a}}_2$ reaches along ${\mathbf{q}}_1$ — the horizontal part $5$ we must strip off.
3. Remove shadow. ${\mathbf{u}}_2=(5,3)-5(1,0)=(0,3)$, $\Vert {\mathbf{u}}_2\Vert =3=r_{22}$, ${\mathbf{q}}_2=(0,1)$. Why this step? Subtracting the horizontal shadow leaves only the vertical part $(0,3)$, which normalizes to the second axis $(0,1)$ — showing the columns already pointed along the coordinate axes.

Verify: $QR=I\cdot A=A$ ✓. Key insight: an already-upper-triangular matrix with positive diagonal is its own $R$ and $Q=I$. (If the diagonal had a negative entry, that column's sign would move into $Q$, exactly as in Example 2 — so $Q$ would be $\pm 1$ on the diagonal, not $I$.)

Example 5 — Case 5: DEGENERATE, dependent columns (QR breaks)

1. First axis. $\Vert {\mathbf{a}}_1\Vert =\sqrt{1+4}=\sqrt{5}=r_{11}$, ${\mathbf{q}}_1=\frac{1}{\sqrt{5}}(1,2)$. Why this step? Same first move as always — normalize the first column to get the first clean direction; nothing is wrong yet.
2. Shadow coefficient. $r_{12}={\mathbf{q}}_1^{\top }{\mathbf{a}}_2=\frac{1}{\sqrt{5}}(1\cdot 2+2\cdot 4)=\frac{10}{\sqrt{5}}=2\sqrt{5}$. Why this step? Measure how much of ${\mathbf{a}}_2$ lies along ${\mathbf{q}}_1$; a large coefficient here is the first hint that ${\mathbf{a}}_2$ points almost entirely along ${\mathbf{q}}_1$.
3. Remove shadow — collapse. ${\mathbf{u}}_2=(2,4)-2\sqrt{5}\cdot \frac{1}{\sqrt{5}}(1,2)=(2,4)-2(1,2)=(0,0)$. Why this matters: ${\mathbf{a}}_2$ was entirely shadow — it lies exactly along ${\mathbf{q}}_1$. After removing the shadow, nothing is left. $\Vert {\mathbf{u}}_2\Vert =0$, so $r_{22}=0$.
4. The break. ${\mathbf{q}}_2={\mathbf{u}}_2/\Vert {\mathbf{u}}_2\Vert =(0,0)/0$ is undefined — division by zero.

Verify: $\det A=1\cdot 4-2\cdot 2=0$, so the columns are dependent (rank 1). The QR theorem in the parent note assumed linearly independent columns precisely so that every $\Vert {\mathbf{u}}_k\Vert >0$. Here that assumption fails, $r_{22}=0$, and thin QR with positive diagonal simply does not exist. (Rank-revealing QR with column pivoting handles this, but the plain algorithm halts.)

Example 6 — Case 6: nearly dependent columns (limiting behaviour)

1. First axis. $\Vert {\mathbf{a}}_1\Vert =1=r_{11}$, ${\mathbf{q}}_1=(1,0)$. Why this step? The first column is already a unit vector, so it is its own clean direction.
2. Shadow coefficient. $r_{12}={\mathbf{q}}_1^{\top }{\mathbf{a}}_2=(1,0)\cdot (1,\epsilon )=1$. Why this step? Measure the along-${\mathbf{q}}_1$ part of ${\mathbf{a}}_2$; almost all of ${\mathbf{a}}_2$ is this shadow.
3. Remove shadow. ${\mathbf{u}}_2=(1,\epsilon )-1\cdot (1,0)=(0,\epsilon )$, so $r_{22}=\Vert {\mathbf{u}}_2\Vert =\epsilon$. Why this step? Strip the shadow; the tiny leftover $(0,\epsilon )$ is all that distinguishes column 2 from column 1.
4. Take the limit. As $\epsilon \to 0^{+}$, $r_{22}=\epsilon \to 0$. Why this step? The leftover perpendicular part shrinks to nothing — the columns are merging into one line, echoing Example 5's collapse.

Verify: $\det A=1\cdot \epsilon -1\cdot 0=\epsilon =r_{11}{r}_{22}$ ✓ (see Example 9 — the product of $R$'s diagonal is the determinant). For $\epsilon =10^{-8}$, $r_{22}=10^{-8}$: any rounding error of size $10^{-8}$ swamps it, so ${\mathbf{q}}_2$ becomes garbage. This is exactly why classical Gram–Schmidt is unstable near dependence, and why practitioners switch to Householder reflections or modified Gram–Schmidt.

Example 7 — Case 7: a WIDE matrix (more columns than rows)

1. First axis. $\Vert {\mathbf{a}}_1\Vert =\Vert (1,0)\Vert =1=r_{11}$, ${\mathbf{q}}_1=(1,0)$. Why this step? Normalize column 1 for the first clean direction — plenty of room so far.
2. Second axis. $r_{12}={\mathbf{q}}_1^{\top }{\mathbf{a}}_2=(1,0)\cdot (0,1)=0$, so ${\mathbf{u}}_2=(0,1)$, $r_{22}=1$, ${\mathbf{q}}_2=(0,1)$. Why this step? Column 2 is already perpendicular to ${\mathbf{q}}_1$; now $\{{\mathbf{q}}_1,{\mathbf{q}}_2\}$ span the entire plane — the room is full.
3. Third axis — collapse. For ${\mathbf{a}}_3=(2,3)$: $r_{13}={\mathbf{q}}_1^{\top }{\mathbf{a}}_3=2$, $r_{23}={\mathbf{q}}_2^{\top }{\mathbf{a}}_3=3$. Then ${\mathbf{u}}_3={\mathbf{a}}_3-r_{13}{\mathbf{q}}_1-r_{23}{\mathbf{q}}_2=(2,3)-2(1,0)-3(0,1)=(0,0)$. Why this matters: $\{{\mathbf{q}}_1,{\mathbf{q}}_2\}$ already span all of ${\mathbb{R}}^2$, so ${\mathbf{a}}_3$ is entirely built from earlier directions. Removing every shadow leaves nothing: $\Vert {\mathbf{u}}_3\Vert =0$, $r_{33}=0$.
4. The break. ${\mathbf{q}}_3=(0,0)/0$ is undefined. Why this step? You cannot create a third direction perpendicular to two that already fill the plane — geometry forbids it.

Verify: the columns of a $2\times 3$ matrix are three vectors in ${\mathbb{R}}^2$, so they are always linearly dependent (you cannot have $3$ independent vectors in $2$ dimensions). Thus $m\ge n$ is required in the QR theorem; a wide matrix ($n>m$) can never have a thin QR with positive diagonal — the very first column that overflows the dimension gives $\Vert {\mathbf{u}}_k\Vert =0$.

Example 8 — Case 8: least-squares word problem

Set up the over-determined system $Ax=b$ with a column of ones (for $x_1$) and a column of $t$ (for $x_2$): $A=\left[\begin{array}{cc}1& 0\\ 1& 1\\ 1& 2\end{array}\right],b=\left[\begin{array}{c}1\\ 2\\ 2\end{array}\right].$

1. QR of $A$. $\Vert {\mathbf{a}}_1\Vert =\sqrt{3}=r_{11}$, ${\mathbf{q}}_1=\frac{1}{\sqrt{3}}(1,1,1)$. $r_{12}={\mathbf{q}}_1^{\top }{\mathbf{a}}_2=\frac{1}{\sqrt{3}}(0+1+2)=\frac{3}{\sqrt{3}}=\sqrt{3}$. ${\mathbf{u}}_2=(0,1,2)-\sqrt{3}\cdot \frac{1}{\sqrt{3}}(1,1,1)=(0,1,2)-(1,1,1)=(-1,0,1)$. $r_{22}=\Vert {\mathbf{u}}_2\Vert =\sqrt{2}$, ${\mathbf{q}}_2=\frac{1}{\sqrt{2}}(-1,0,1)$. Why this step? QR turns the messy least-squares into a triangular solve; see Least squares and normal equations.
2. Form $Q^{\top }b$. ${\mathbf{q}}_1^{\top }b=\frac{1}{\sqrt{3}}(1+2+2)=\frac{5}{\sqrt{3}}$. ${\mathbf{q}}_2^{\top }b=\frac{1}{\sqrt{2}}(-1\cdot 1+0\cdot 2+1\cdot 2)=\frac{1}{\sqrt{2}}$. Why this step? The minimizer of $\Vert Ax-b\Vert$ satisfies $Rx=Q^{\top }b$ (parent note) — $Q^{\top }b$ is $b$ expressed in the clean frame.
3. Back-substitute $\left[\begin{array}{cc}\sqrt{3}& \sqrt{3}\\ 0& \sqrt{2}\end{array}\right]x=\left[\begin{array}{c}5/\sqrt{3}\\ 1/\sqrt{2}\end{array}\right]$. Row 2: $\sqrt{2}{x}_2=\frac{1}{\sqrt{2}}\Rightarrow x_2=\frac{1}{2}$. Row 1: $\sqrt{3}{x}_1+\sqrt{3}{x}_2=\frac{5}{\sqrt{3}}\Rightarrow x_1+x_2=\frac{5}{3}\Rightarrow x_1=\frac{5}{3}-\frac{1}{2}=\frac{7}{6}$. Why this step? Triangular systems solve bottom-up with no matrix inversion — exact and numerically stable, the whole point of using QR.

Verify (units + sanity): intercept $\frac{7}{6}\approx 1.17$ sales at $t=0$ (close to the observed $1$), slope $\frac{1}{2}$ sale per hour. Predictions $(1.17,1.67,2.17)$ vs data $(1,2,2)$ — residuals $(0.17,-0.33,0.17)$ sum to $\approx 0$, as least squares demands (the fit passes through the data's centroid $(\bar{t},\bar{s})=(1,\frac{5}{3})$: $\frac{7}{6}+\frac{1}{2}\cdot 1=\frac{5}{3}$ ✓). Slope $\frac{1}{2}$, matching the forecast option.

Example 9 — Case 9: exam twist, read $|\det A|$ off $R$

1. QR of $A$. ${\mathbf{a}}_1=(0,1)$, $\Vert {\mathbf{a}}_1\Vert =1=r_{11}$, ${\mathbf{q}}_1=(0,1)$. Why this step? Normalize column 1 for the first diagonal entry $r_{11}$ — the only $R$ values we need are these diagonal norms.
2. Shadow + second axis. $r_{12}={\mathbf{q}}_1^{\top }{\mathbf{a}}_2=(0,1)\cdot (1,1)=1$; ${\mathbf{u}}_2=(1,1)-1\cdot (0,1)=(1,0)$; $r_{22}=\Vert {\mathbf{u}}_2\Vert =1$. Why this step? Strip ${\mathbf{a}}_2$'s shadow to get the perpendicular leftover, whose length is the second diagonal entry $r_{22}$.
3. Multiply the diagonal. $\det R=r_{11}{r}_{22}=1\cdot 1=1$. Why this step? For a triangular matrix the determinant is the product down the diagonal — off-diagonal entries never contribute.
4. Use orthogonality. $|\det A|=|\det Q||\det R|=1\cdot 1=1$. Why this step? An orthogonal $Q$ preserves volume, so $|\det Q|=1$; all the "stretch" of $A$ lives in $R$'s diagonal.

Verify: direct expansion $\det \left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]=0\cdot 1-1\cdot 1=-1$, so $|\det A|=1$ ✓. General rule you can quote in an exam: $|\det A|={\prod }_k{r}_{kk}$. (This same trick powers the Eigenvalues and the QR algorithm and mirrors how LU decomposition reads $\det$ off its triangular factor.)

Active recall

Connections

• QR decomposition — the parent this page drills into.
• Gram-Schmidt process — every example is one run of it.
• Orthonormal bases and projections — the shadow ${\mathbf{q}}^{\top }\mathbf{a}$ is a projection.
• Least squares and normal equations — Example 8's engine.
• Householder reflections — the stable fix for Case 6.
• Eigenvalues and the QR algorithm — uses the $|\det |=\prod r_{kk}$ idea.
• LU decomposition — sibling factorization that also reads $\det$ off a triangle.