Visual walkthrough — QR decomposition

Before line one, three plain words we will lean on the whole way:

We will use the tiny matrix $A=\left[\begin{array}{cc}1& 1\\ 0& 1\\ 1& 0\end{array}\right]=[{\mathbf{a}}_1\mid {\mathbf{a}}_2],{\mathbf{a}}_1=(1,0,1),{\mathbf{a}}_2=(1,1,0)$ as our running board example. Two crooked arrows in 3D. Let's straighten them.

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Step 1 — Plant the first clean axis

WHAT. We take the first crooked arrow ${\mathbf{a}}_1$ and shrink it to length $1$, keeping its direction. Call the result ${\mathbf{q}}_1$. ${\mathbf{q}}_1=\frac{{\mathbf{a}}_1}{\Vert {\mathbf{a}}_1\Vert },\Vert {\mathbf{a}}_1\Vert =\sqrt{{\mathbf{a}}_1^{\top }{\mathbf{a}}_1}=\sqrt{1^2+0^2+1^2}=\sqrt{2}.$

• ${\mathbf{a}}_1$ — the raw first arrow.
• $\Vert {\mathbf{a}}_1\Vert =\sqrt{2}$ — its length, found by the dot product with itself.
• dividing by $\sqrt{2}$ — the only thing that changes is the length ($\to 1$); the direction is untouched.

So ${\mathbf{q}}_1=\frac{1}{\sqrt{2}}(1,0,1)$.

WHY. Every other arrow will be measured against ${\mathbf{q}}_1$. A ruler must have length $1$ before you measure with it — that's all "normalize" means.

PICTURE. The pale-yellow arrow is ${\mathbf{a}}_1$; the blue arrow is the shortened ${\mathbf{q}}_1$ lying right on top of it, now exactly one unit long.

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Step 2 — Measure the shadow of the second arrow

WHAT. Before we clean ${\mathbf{a}}_2$, we ask: how much of ${\mathbf{a}}_2$ already points along ${\mathbf{q}}_1$? That amount is a single number, $r_{12}={\mathbf{q}}_1^{\top }{\mathbf{a}}_2=\frac{1}{\sqrt{2}}(1\cdot 1+0\cdot 1+1\cdot 0)=\frac{1}{\sqrt{2}}.$

• ${\mathbf{q}}_1^{\top }{\mathbf{a}}_2$ — dot product = "how far ${\mathbf{a}}_2$ reaches in the ${\mathbf{q}}_1$ direction."
• the result $\frac{1}{\sqrt{2}}\approx 0.707$ — the length of the shadow ${\mathbf{a}}_2$ casts on the ${\mathbf{q}}_1$ line.
• we name it $r_{12}$ — it will become an entry of the recipe card $R$ (row $1$, column $2$).

WHY. The dot product is chosen precisely because it reads off a shadow length. This is the projection idea: $r_{12}{\mathbf{q}}_1$ is the piece of ${\mathbf{a}}_2$ that is "already explained" by our first axis.

PICTURE. Drop a dashed perpendicular from the tip of ${\mathbf{a}}_2$ onto the ${\mathbf{q}}_1$ line. Where it lands marks the shadow $r_{12}{\mathbf{q}}_1$ (pink).

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Step 3 — Subtract the shadow to reveal the new direction

WHAT. Erase the explained piece. Whatever remains must be perpendicular to ${\mathbf{q}}_1$: ${\mathbf{u}}_2={\mathbf{a}}_2-r_{12}{\mathbf{q}}_1=(1,1,0)-\frac{1}{\sqrt{2}}\cdot \frac{1}{\sqrt{2}}(1,0,1)=(\frac{1}{2},1,-\frac{1}{2}).$

• ${\mathbf{a}}_2$ — the full second arrow.
• $r_{12}{\mathbf{q}}_1$ — the shadow from Step 2, now scaled back into a real arrow.
• ${\mathbf{u}}_2$ — the leftover, the part of ${\mathbf{a}}_2$ at a right angle to ${\mathbf{q}}_1$.

Check it worked: ${\mathbf{q}}_1^{\top }{\mathbf{u}}_2=\frac{1}{\sqrt{2}}(\frac{1}{2}+0-\frac{1}{2})=0$. ✓ Perpendicular, exactly as promised.

WHY. Subtracting the shadow is the whole trick of the Gram-Schmidt process: remove what's shared, keep what's new. A dot product of $0$ confirms the leftover is a brand-new independent direction.

PICTURE. Vector subtraction as a triangle: ${\mathbf{a}}_2$ minus the pink shadow gives the green ${\mathbf{u}}_2$, standing at $90^{\circ }$ to the ${\mathbf{q}}_1$ line.

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Step 4 — Shrink the leftover to length one

WHAT. Turn ${\mathbf{u}}_2$ into a unit ruler ${\mathbf{q}}_2$, and record its original length: $\Vert {\mathbf{u}}_2\Vert =\sqrt{(\frac{1}{2}{)}^2+1^2+(-\frac{1}{2}{)}^2}=\sqrt{\frac{3}{2}},r_{22}=\Vert {\mathbf{u}}_2\Vert =\sqrt{\frac{3}{2}},{\mathbf{q}}_2=\frac{{\mathbf{u}}_2}{\Vert {\mathbf{u}}_2\Vert }.$

• $\Vert {\mathbf{u}}_2\Vert =\sqrt{3/2}\approx 1.225$ — how long the new direction was before shrinking.
• $r_{22}$ — this length becomes the diagonal recipe entry (row $2$, column $2$).
• ${\mathbf{q}}_2$ — our second clean axis, length $1$, perpendicular to ${\mathbf{q}}_1$.

WHY. We store the length in $r_{22}$ because we will need it to rebuild ${\mathbf{a}}_2$ later. The unit vector is for measuring; the length is for reconstructing. Note $r_{22}>0$ automatically — the leftover was non-zero because our two arrows were genuinely different (linearly independent).

PICTURE. The green ${\mathbf{u}}_2$ (length $\sqrt{3/2}$) shrinks onto the blue unit ${\mathbf{q}}_2$. Now ${\mathbf{q}}_1\perp {\mathbf{q}}_2$: a clean right-angle corner.

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Step 5 — Read the recipe backwards: rebuild the originals

WHAT. We have clean axes ${\mathbf{q}}_1,{\mathbf{q}}_2$. Now flip every equation to express the original arrows as combinations of clean ones: ${\mathbf{a}}_1=\Vert {\mathbf{a}}_1\Vert {\mathbf{q}}_1=r_{11}{\mathbf{q}}_1,{\mathbf{a}}_2=r_{12}{\mathbf{q}}_1+r_{22}{\mathbf{q}}_2.$

• ${\mathbf{a}}_1$ uses only ${\mathbf{q}}_1$ — with $r_{11}=\Vert {\mathbf{a}}_1\Vert =\sqrt{2}$.
• ${\mathbf{a}}_2$ uses ${\mathbf{q}}_1$ (amount $r_{12}$) and ${\mathbf{q}}_2$ (amount $r_{22}$) — but never a later axis, because ${\mathbf{a}}_2$ was cleaned before ${\mathbf{q}}_3$ ever existed.

Collect the recipe numbers into a grid $R$, and the clean axes into $Q$: $Q=\left[\begin{array}{c}{\mathbf{q}}_1\mid {\mathbf{q}}_2\end{array}\right],R=\left[\begin{array}{cc}{r}_{11}& r_{12}\\ 0& r_{22}\end{array}\right]=\left[\begin{array}{cc}\sqrt{2}& \frac{1}{\sqrt{2}}\\ 0& \sqrt{3/2}\end{array}\right].$

WHY the zero (upper-triangular)? ${\mathbf{a}}_1$ was straightened before ${\mathbf{q}}_2$ was born, so ${\mathbf{q}}_2$ cannot appear in ${\mathbf{a}}_1$'s recipe — its coefficient is forced to $0$. That single $0$ in the bottom-left is what "upper-triangular" means: each arrow only ever uses axes that already existed. This is the QR decomposition result, $A=QR$.

PICTURE. $R$ as an instruction card: column $k$ tells you the exact mix of ${\mathbf{q}}_1,\dots ,{\mathbf{q}}_k$ that rebuilds ${\mathbf{a}}_k$. The bottom-left cell is greyed out — the "you can't use a future axis" rule.

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Step 6 — The degenerate case: what if the leftover is zero?

WHAT. Suppose ${\mathbf{a}}_2$ were a copy of ${\mathbf{a}}_1$'s direction, say ${\mathbf{a}}_2=(2,0,2)$. Its shadow onto ${\mathbf{q}}_1$ would be the whole arrow, so the leftover is ${\mathbf{u}}_2={\mathbf{a}}_2-({\mathbf{q}}_1^{\top }{\mathbf{a}}_2){\mathbf{q}}_1=0,\Vert {\mathbf{u}}_2\Vert =0.$

WHY this breaks QR. To make ${\mathbf{q}}_2$ we divide by $\Vert {\mathbf{u}}_2\Vert$ — but here that is $0$, and dividing by zero is undefined. Also $r_{22}=0$, so $R$'s diagonal is no longer strictly positive. Geometrically: there is no new direction to plant an axis in.

The rule this forces. QR (thin, with positive diagonal) needs the columns of $A$ to be linearly independent — no arrow may be a mix of the others. When they are independent, every leftover is non-zero, every $r_{kk}>0$, and the whole construction runs safely.

PICTURE. Two collinear yellow arrows: the shadow eats the entire second arrow, leaving nothing (a red dot at the origin) — the moment the algorithm has to stop.

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Step 7 — Sign convention: why we insist $r_{kk}>0$

WHAT. Nothing stopped us from choosing ${\mathbf{q}}_2$ pointing the opposite way (multiply by $-1$). If we do, then $r_{22}$ flips sign too, and $r_{12}$ shifts to match — so $A=QR$ still holds, but with a different $Q$ and $R$.

WHY. A factorization we cannot pin down is annoying. To make QR unique, we demand $r_{kk}=\Vert {\mathbf{u}}_k\Vert >0\text{for every }k,$ which is guaranteed the moment we always divide by a positive length. That one convention removes the sign ambiguity.

PICTURE. Same corner, two allowed choices of ${\mathbf{q}}_2$ (up-blue vs down-pink); we keep the one giving a positive diagonal.

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The one-picture summary

Everything above, on one board: crooked arrows $A$ on the left flow through "straighten & record" into the clean frame $Q$ plus the triangular recipe $R$, and multiplying them back reproduces $A$.

$\underset{A\text{(crooked)}}{\underbrace{[{\mathbf{a}}_1\mid {\mathbf{a}}_2]}}=\underset{Q\text{(clean frame)}}{\underbrace{[{\mathbf{q}}_1\mid {\mathbf{q}}_2]}}\underset{R\text{(recipe, upper-triangular)}}{\underbrace{\left[\begin{array}{cc}{r}_{11}& r_{12}\\ 0& r_{22}\end{array}\right]}}$

Recall Feynman retelling — say it like a story

I had two bent arrows on the chalkboard. I grabbed the first one and just squeezed it to length one — that's my first clean ruler, ${\mathbf{q}}_1$. Then I looked at the second arrow and asked, "how much of you is just leaning the same way as my ruler?" That amount is one number, $r_{12}$; I wiped that part off. Whatever was left had to be standing at a perfect right angle — I squeezed that down to length one too and called it ${\mathbf{q}}_2$, writing down how long it had been ($r_{22}$). Now I have a tidy square corner of rulers, $Q$. The little card of numbers I wrote down — "${\mathbf{a}}_1$ = so much of ruler 1", "${\mathbf{a}}_2$ = so much of ruler 1 plus so much of ruler 2" — is $R$. Because arrow 1 was straightened before ruler 2 ever existed, ruler 2 can't show up in arrow 1's recipe, so one corner of $R$ is always zero — that's the triangle shape. If ever a "leftover" came out to nothing (the arrows were secretly the same direction), I'd be dividing by zero, so I only play this game with genuinely-different arrows and always keep the recorded lengths positive so the answer is one-of-a-kind.

Recall Quick self-test

• What number is $r_{12}$ geometrically? ::: The length of ${\mathbf{a}}_2$'s shadow on the ${\mathbf{q}}_1$ line, i.e. ${\mathbf{q}}_1^{\top }{\mathbf{a}}_2$.
• Why is the bottom-left of $R$ zero? ::: ${\mathbf{a}}_1$ is built before ${\mathbf{q}}_2$ exists, so ${\mathbf{q}}_2$'s coefficient in ${\mathbf{a}}_1$ must be $0$.
• What goes wrong if columns are dependent? ::: A leftover ${\mathbf{u}}_k=0$, so $\Vert {\mathbf{u}}_k\Vert =0$ and we'd divide by zero.
• Which rule makes QR unique? ::: Every diagonal $r_{kk}>0$.

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Connections

• Gram-Schmidt process — this walkthrough is Gram–Schmidt, drawn.
• Orthonormal bases and projections — Step 2's shadow is a projection.
• Least squares and normal equations — the payoff of having $Q,R$.
• Householder reflections — a rounding-stable route to the same $A=QR$.
• Eigenvalues and the QR algorithm — repeats this factorization.
• LU decomposition — a triangular sibling, without the clean frame.