Recipe

Solve a least-squares problem from an existing QR factorization

Use when: you already have $A=QR$ (orthonormal $Q$ , upper-triangular $R$ ) and need the least-squares solution to $A\mathbf x\approx\mathbf b$

Goal: $\hat{\mathbf x}$ , found with one matrix-vector product and one back substitution — no matrix ever multiplied by its own transpose

Taught in Least squares by QR

Quick recipe

Compute $Q^{\mathsf T}\mathbf b$ , one entry at a time
Solve $R\hat{\mathbf x}=Q^{\mathsf T}\mathbf b$ starting at the bottom row
Substitute upward through the remaining rows
Check $A\hat{\mathbf x}$ against $\mathbf b$

1
Dot each column of $Q$ with $\mathbf b$ .

Example

For $Q=\begin{bmatrix}\tfrac23&-\tfrac23\\\tfrac13&\tfrac23\\\tfrac23&\tfrac13\end{bmatrix}$ and $\mathbf b=(5,-2,5)$ : $\mathbf q_1\cdot\mathbf b=\tfrac23(5)+\tfrac13(-2)+\tfrac23(5)=\tfrac{10}{3}-\tfrac23+\tfrac{10}{3}=6$ , and $\mathbf q_2\cdot\mathbf b=-\tfrac23(5)+\tfrac23(-2)+\tfrac13(5)=-\tfrac{10}{3}-\tfrac43+\tfrac53=-3$ . So $Q^{\mathsf T}\mathbf b=(6,-3)$ .

Why: this single matrix-vector product carries every bit of $\mathbf b$ 's information the least-squares solve actually needs — no $A^{\mathsf T}A$ required anywhere.
2
The last row of $R$ has only one nonzero entry, so it already isolates one unknown.

Example

With $R=\begin{bmatrix}3&3\\0&3\end{bmatrix}$ and $Q^{\mathsf T}\mathbf b=(6,-3)$ : the bottom row reads $3\hat x_2=-3$ , so $\hat x_2=-1$ .

Why: $R$ is upper triangular by construction — QR factoring was built column by column precisely so the last row would come out this simple.
3
Plug the values already found into the next row up, solve for its one remaining unknown, and repeat until every entry of $\hat{\mathbf x}$ is known.

Example

The top row reads $3\hat x_1+3\hat x_2=6$ , so $3\hat x_1+3(-1)=6\implies 3\hat x_1=9\implies\hat x_1=3$ . So $\hat{\mathbf x}=(3,-1)$ .

Why: triangularity guarantees each row above the bottom introduces exactly one new unknown once you work upward — the same reason back substitution works on any triangular system.
4
Multiply $A$ by the $\hat{\mathbf x}$ you found; it should equal $\mathbf b$ exactly only if the original system happened to be consistent, and otherwise land as close to $\mathbf b$ as $\operatorname{col}(A)$ allows.

Example

$A=\begin{bmatrix}2&0\\1&3\\2&3\end{bmatrix}$ : $A\hat{\mathbf x}=\begin{pmatrix}2(3)+0(-1)\\1(3)+3(-1)\\2(3)+3(-1)\end{pmatrix}=\begin{pmatrix}6\\0\\3\end{pmatrix}$ , while $\mathbf b=(5,-2,5)$ — a leftover residual of $(-1,-2,2)$ , since $\mathbf b$ wasn't exactly in $\operatorname{col}(A)$ .

Why: this is the same sanity check the normal-equations route uses, just arrived at without ever forming $A^{\mathsf T}A$ — a large or oddly-shaped residual usually means an earlier row or dot product was computed wrong.

Common mistakes

Forming $A^{\mathsf T}A$ out of habit anyway — the entire point of this route is skipping that multiplication.
Using $A^{\mathsf T}\mathbf b$ in place of $Q^{\mathsf T}\mathbf b$ — the two are not the same vector, only the two routes' final $\hat{\mathbf x}$ agree.
Starting back substitution from the top row instead of the bottom — only the bottom row of $R$ is guaranteed to have just one unknown.
Forgetting that $\hat{\mathbf x}$ 's entries correspond to $R$ 's columns in order, and reporting them in the wrong order.