Question bank — Orthogonal matrices — properties, det = ±1

False — perpendicular is only half of it; the columns must also each have length 1. Columns $\left(\genfrac{}{}{0ex}{}{2}{0}\right),\left(\genfrac{}{}{0ex}{}{0}{3}\right)$ are perpendicular but $Q^{\mathsf{T}}Q=\text{diag}(4,9)\ne I$.

True — taking $\det$ of $Q^{\mathsf{T}}Q=I$ gives $(\det Q{)}^2=1$, so $\det Q=\pm 1$. This is a genuine property of all orthogonal matrices.

False — $\det =\pm 1$ is necessary, not sufficient. The shear $\left(\begin{array}{cc}1& 5\\ 0& 1\end{array}\right)$ has $\det =1$ yet skews space; its columns aren't orthonormal.

False — "orthogonal matrix" is defined only for square $Q\in {\mathbb{R}}^{n\times n}$; a $3\times 2$ matrix with orthonormal columns is called having orthonormal columns, but it has no inverse or determinant, so it isn't an orthogonal matrix.

True — that is the definition rearranged: $Q^{\mathsf{T}}Q=I$ says $Q^{\mathsf{T}}$ undoes $Q$. This is why orthogonal maps are so cheap to invert.

False — it holds exactly when $Q$ is orthogonal. Always test $Q^{\mathsf{T}}Q\stackrel{?}{=}I$ before trusting it; see Orthogonal matrices — properties, det = ±1.

True — since $(Qx{)}^{\mathsf{T}}(Qy)=x^{\mathsf{T}}y$ and lengths are unchanged, the ratio $\cos \theta =\frac{x^{\mathsf{T}}y}{\Vert x\Vert \Vert y\Vert }$ stays the same, so the angle survives — exactly the rigidity shown in the figure above.

False — it preserves the length of every vector individually ($\Vert Qx\Vert =\Vert x\Vert$), so no direction is stretched at all. Volume preservation alone would allow stretch-and-shrink; length preservation forbids it.

True — $(Q_1{Q}_2{)}^{\mathsf{T}}(Q_1{Q}_2)=Q_2^{\mathsf{T}}{Q}_1^{\mathsf{T}}{Q}_1{Q}_2=Q_2^{\mathsf{T}}I{Q}_2=I$. This closure is why $O(n)$ forms a group.

False — signs multiply: two reflections ($-1$ each) give $(-1)(-1)=+1$, but a rotation times a reflection gives $-1$. Only "same sign" pairs land in $SO(n)$.

False — a genuine rotation like $\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$ has complex eigenvalues $e^{\pm i\theta }$ (no real arrow keeps its direction), see Eigenvalues and eigenvectors.

The determinant can also be $-1$. Writing "$=1$" silently assumes a rotation; reflections are orthogonal too and have $\det =-1$.

For a square matrix the two are equivalent — a one-sided inverse of a square matrix is automatically two-sided. Both automatically hold together.

For a square orthogonal matrix rows are orthonormal too, because $Q{Q}^{\mathsf{T}}=I$ says exactly that about the rows. Columns orthonormal $\iff$ rows orthonormal.

The magnitude $|\det Q|=1$ is the area factor, so area is preserved. The minus sign is about orientation flip (handedness), not size — the flip shown in the second panel of the figure.

Preserving length does not mean fixing the vector — a rotation moves every nonzero vector while keeping its length. Equal length $\ne$ equal vector.

The determinant says nothing about column length or perpendicularity here: column $\left(\genfrac{}{}{0ex}{}{5}{1}\right)$ has length $\sqrt{26}\ne 1$ and isn't perpendicular to $\left(\genfrac{}{}{0ex}{}{1}{0}\right)$.

Preserving dot products makes $Q$ an isometry, not the identity. Every rotation and reflection preserves all dot products yet moves vectors.

The $(i,j)$ entry of $Q^{\mathsf{T}}Q$ is the dot product $q_i^{\mathsf{T}}{q}_j$ of columns $i$ and $j$; setting it equal to ${\delta }_{ij}$ (1 on the diagonal, 0 off it) means each column has length 1 ($i=j$) and distinct columns are perpendicular ($i\ne j$).

Because $(\det Q{)}^2=1$ has only the two real solutions $\pm 1$; geometrically length preservation kills all scaling, so $|\det Q|$ must be exactly $1$.

An eigenvalue $\lambda$ satisfies $Qx=\lambda x$ for some nonzero arrow $x$; then $\Vert x\Vert =\Vert Qx\Vert =|\lambda |\Vert x\Vert$, forcing $|\lambda |=1$ — the map can't change the length of an eigenvector, so its stretch factor lies on the unit circle.

$Q$ is real, so its characteristic polynomial has real coefficients; complex roots of a real polynomial always pair with their conjugates, and $|\lambda |=1$ pins them to $e^{\pm i\theta }$.

$\det$ is the signed volume factor: $+1$ keeps orientation (spin, right panel of the figure stays a spun frame), $-1$ reverses it (mirror flip), linking to [[Rotations and reflections in ${\mathbb{R}}^2$ and ${\mathbb{R}}^3$]].

Because orthonormal columns make $Q^{\mathsf{T}}Q=I$ directly, so $Q^{\mathsf{T}}$ is the inverse — no elimination needed, which is exactly what makes QR decomposition so efficient.

It outputs mutually perpendicular unit vectors — precisely the cyan arrows in the figure — and stacking them as columns is exactly the condition $Q^{\mathsf{T}}Q=I$; see Orthonormal bases & Gram–Schmidt.

Yes — its columns are the standard orthonormal basis (the grey arrows in the figure), so $I^{\mathsf{T}}I=I$, and $\det I=+1$ (the trivial rotation that moves nothing).

Yes, orthogonal; $\det (-I)=(-1{)}^2=+1$ in $2\times 2$, so it's a rotation by $180^{\circ }$ — not a reflection, despite the minus sign.

$\det (-I)=(-1{)}^3=-1$, so in odd dimensions $-I$ reverses orientation and counts as a reflection-type map — the parity of the dimension decides.

Yes: $Q=[q]$ with $q^2=1$, so $q=\pm 1$. The only $1\times 1$ orthogonal matrices are $[1]$ (identity) and $[-1]$ (reflection through the origin on the line).

No — $0^{\mathsf{T}}0=0\ne I$, and its columns have length $0$, not $1$. An orthogonal matrix must be invertible, but the zero matrix is not.

Never — $\det Q=\pm 1\ne 0$, so every orthogonal matrix is invertible (with inverse $Q^{\mathsf{T}}$). Singularity is impossible.

Yes — its columns are the standard basis vectors in some order, hence orthonormal; the determinant is $\pm 1$ depending on whether the permutation is an even or odd number of swaps.

No — it lacks the identity ($\det I=+1$) and isn't closed (two of them multiply to $\det =+1$). Only $SO(n)$ with $\det =+1$ is a subgroup.