Question bank — QR decomposition

True — that is the defining property, the columns are orthonormal by construction, so the Gram matrix of $Q$ is the identity.

False for thin QR — $Q{Q}^{\top }$ is the $m\times m$ projection onto the column space of $A$, not the identity, unless $m=n$ (full/square case).

True in a weak sense: $A=QR$ exists for any $A$, but uniqueness with positive diagonal needs linearly independent columns; with dependent columns some $r_{kk}=0$ and $Q$ isn't fully determined.

False — it is forced: ${\mathbf{a}}_k$ is built only from ${\mathbf{q}}_1,\dots ,{\mathbf{q}}_k$, so any coefficient $r_{ik}$ with $i>k$ must be zero, which is upper-triangularity.

False — independence means ${\mathbf{u}}_k\ne 0$ at every step, so $r_{kk}=\Vert {\mathbf{u}}_k\Vert >0$; a zero would signal a linear dependence.

True — Gram–Schmidt finds each column already unit-length and perpendicular, so every $r_{kk}=1$ and all off-diagonals vanish, giving $Q=A$, $R=I$.

True — scaling ${\mathbf{a}}_1$ only rescales ${\mathbf{u}}_1$; after normalizing, ${\mathbf{q}}_1$ is identical, though $r_{11}$ (and the first row of $R$) doubles.

False — LU is two triangular factors; QR pairs an orthogonal factor with a triangular one, so they coincide only in trivial cases, never generically.

True — it is triangular with all diagonal entries $>0$, and a triangular matrix's determinant is the product of the diagonal, which is nonzero here.

False — reordering columns changes who gets orthogonalized against whom, so both $Q$ and $R$ can change entirely; QR is order-dependent.

Error — $Q^{-1}$ needs $Q$ square; a tall $m\times n$ $Q$ has no two-sided inverse. Only $Q^{\top }$ acts as a left inverse ($Q^{\top }Q=I$), not $Q{Q}^{\top }$.

Error in spirit — QR avoids forming $A^{\top }A$, whose condition number is squared; solving $Rx=Q^{\top }b$ keeps the original conditioning, which is the whole numerical point.

Error of practice — you never invert $R$; being triangular you back-substitute from the bottom row up, which is faster and more stable than forming $R^{-1}$.

Error — up to sign conventions they produce the same $Q,R$; they differ only in numerical accuracy, not in the exact answer.

Error — a negative $r_{kk}$ corresponds to flipping the sign of ${\mathbf{q}}_k$; it's still a valid QR, but you lose uniqueness, so a positive-diagonal convention is fixed.

Error — you must multiply by $Q^{\top }$, giving $Rx=Q^{\top }b$, not $Rx=b$; $Q^{\top }b$ generally differs from $b$ (and $b$ may not even lie in $Q$'s column space).

Error — the projection vector is $({\mathbf{q}}_i^{\top }{\mathbf{a}}_k){\mathbf{q}}_i$; you scale the unit direction ${\mathbf{q}}_i$, not ${\mathbf{a}}_k$. This scalar ${\mathbf{q}}_i^{\top }{\mathbf{a}}_k$ is exactly the entry $r_{ik}$.

So that $Q$'s columns are unit length, making $Q^{\top }Q=I$; without it the diagonal of $R$ would be tangled into $Q$ and lengths wouldn't be preserved.

Because ${\mathbf{q}}_i$ is a unit vector, the dot product ${\mathbf{q}}_i^{\top }{\mathbf{a}}_k$ is precisely the signed length of ${\mathbf{a}}_k$'s shadow along ${\mathbf{q}}_i$ — the amount of ${\mathbf{q}}_i$ used to rebuild ${\mathbf{a}}_k$.

The normal equations use $A^{\top }A$, squaring the condition number and amplifying rounding; QR works with $A$ directly through the orthogonal $Q$, which never inflates errors.

A dependent column produces ${\mathbf{u}}_k=0$, so ${\mathbf{q}}_k$ is undefined (dividing by zero) and $r_{kk}=0$ — the recipe breaks and the direction is arbitrary.

Repeatedly setting $A_{k+1}=R_k{Q}_k$ is an orthogonal similarity transform that preserves eigenvalues while pushing the matrix toward triangular form, revealing eigenvalues on the diagonal.

Negating ${\mathbf{q}}_k$ and simultaneously negating row $k$ of $R$ leaves $QR$ unchanged, since the two sign flips cancel in the product — so validity survives but uniqueness needs a sign rule.

Because $\Vert Q\mathbf{x}{\Vert }^2={\mathbf{x}}^{\top }{Q}^{\top }Q\mathbf{x}={\mathbf{x}}^{\top }\mathbf{x}=\Vert \mathbf{x}{\Vert }^2$; the orthonormality collapses the middle to identity, so $Q$ acts like a rotation/reflection.

Gram–Schmidt is the step-by-step procedure; writing each ${\mathbf{a}}_k={\sum }_i{r}_{ik}{\mathbf{q}}_i$ and stacking those relations column-by-column is literally the matrix equation $A=QR$.

$Q=[{\mathbf{a}}_1/\Vert {\mathbf{a}}_1\Vert ]$ and $R=[\Vert {\mathbf{a}}_1\Vert ]$ — a $1\times 1$ triangular matrix; there are no later columns to orthogonalize.

The second one has zero leftover after projection (${\mathbf{u}}_2=0$), so $r_{22}=0$ and ${\mathbf{q}}_2$ is undefined — QR with positive diagonal doesn't exist because the columns are dependent.

$Q=I$ and $R=I$ — the columns are already orthonormal, so Gram–Schmidt does nothing, matching the "orthogonal in ⇒ do nothing" rule.

That column is dependent (it's a trivial combination of the others), ${\mathbf{u}}_k=0$, and no positive-diagonal QR exists; the matrix lacks independent columns.

Yes — when $m=n$ there are no extra rows to pad, so $Q$ is already $n\times n$ orthogonal and $Q{Q}^{\top }=I$ holds too.

Zero — the projection is exact, $Ax=b$ has a true solution, and solving $Rx=Q^{\top }b$ recovers it with no leftover error.

It is $\Vert {\mathbf{a}}_1\Vert$, the length of the first column, since ${\mathbf{u}}_1={\mathbf{a}}_1$ with nothing subtracted yet — the first diagonal of $R$ is just that column's length.