For real orthonormal column vectors, $u^{T}u=v^{T}v=1$ and $u^{T}v=v^{T}u=0$. Expand using associativity of matrix multiplication:HenceMore precisely, it is the orthogonal projection onto the span of $u,v$: it fixes both vectors and annihilates every vector orthogonal to both.

The transpose convention presumes real vectors, or the explicitly stated bilinear identities above. For complex vectors orthonormal in the Hermitian inner product, the correct orthogonal projection is $u{u}^{†}+v{v}^{†}$. The printed transpose formula need not work in that setting: $u=(i,0{)}^T$ and $v=(0,1{)}^T$ are Hermitian-orthonormal, but $u{u}^T+v{v}^T=\mathrm{diag}(-1,1)$ has square $I_2$, which is different from itself.

Take determinants of the idempotence equation and use multiplicativity of the determinant:Thus $d=\det M$ satisfies $d(d-1)=0$, provingIf the determinant is one, $M$ is invertible and solution gives $M=I$. Every other idempotent matrix has zero determinant.

Using $M^2=M$,Thus $I-M$ is another idempotent matrix. In fact $M(I-M)=(I-M)M=0$: as linear projections, they select the complementary summands in the decomposition $x=Mx+(I-M)x$.

If the idempotent matrix $M$ is invertible, multiply $M^2=M$ on the left by its matrix inverse:Thus a nonidentity idempotent matrix must be singular; the identity itself is both invertible and idempotent.

If $M$ is an $r\times s$ matrix, the product $M^2=MM$ is defined in ordinary matrix multiplication only when $s=r$. Hence the defining equation of an idempotent matrix already requires equal row and column counts:Equivalently, an idempotent linear map is an endomorphism, so the same vector space is both its domain and codomain.

Expand the two linear combinations and use the Pauli matrix multiplication identity:Interchanging the vectors reverses the cross product, so their commutator is $2i\sigma (\mathbf{a}\times \mathbf{b})$. The three Pauli matrices are linearly independent; alternatively, trace against ${\sigma }_k$ to recover each coefficient. ThusFor nonzero real vectors, this means they are parallel or antiparallel, equivalently linearly dependent. The algebra also extends to complex vectors using bilinear dot and cross products, with the same linear-dependence criterion; the dot product there has no complex conjugation.

Besides their squares, direct multiplication yields ${\sigma }_x{\sigma }_y=i{\sigma }_z$, ${\sigma }_y{\sigma }_z=i{\sigma }_x$ and ${\sigma }_z{\sigma }_x=i{\sigma }_y$. Reversing each pair changes the sign. These identities combine into the Pauli matrix multiplication identityAdding the expression with $i,j$ exchanged cancels the antisymmetric term, proving $\{{\sigma }_i,{\sigma }_j\}=2{\delta }_{ij}I$. Taking the trace instead uses $\mathrm{tr}I=2$ and $\mathrm{tr}{\sigma }_k=0$, proving $\mathrm{tr}({\sigma }_i{\sigma }_j)=2{\delta }_{ij}$.

Each displayed Pauli matrix has zero trace and determinant $-1$. Direct multiplication gives ${\sigma }_i^2=I$, henceTheir two eigenvalues are therefore $+1$ and $-1$.

One has $(C^{†}C{)}^{†}=C^{†}C$, with no requirement that $C$ itself be Hermitian. Moreover,Thus $C^{†}C$ is Hermitian and positive semidefinite. The positive semidefinite property is stronger than the Hermitian conclusion alone.

Taking the Hermitian adjoint reverses the products and leaves their sum unchanged:Thus the symmetrized product is Hermitian, whether or not $A$ and $B$ commute. This is the Hermitian anticommutator divided by two.

The commutator is anti-Hermitian: $[A,B{]}^{†}=-[A,B]$. It is also Hermitian precisely when it is zero, so it is not Hermitian in general; it is Hermitian if and only if $A$ and $B$ commute. For example, $[{\sigma }_x,{\sigma }_y]=2i{\sigma }_z\ne 0$ for the Pauli matrices.

For finite-dimensional operators, use $(AB{)}^{†}=B^{†}{A}^{†}$ and the Hermitian assumptions. The commutator satisfies $[A,B{]}^{†}=-[A,B]$, soThus $i[A,B]$ is Hermitian. These adjoint identities are the intended matrix-level setting; for unbounded operators, domains and self-adjointness require additional analysis.

Substituting $\omega =U{\omega }^{\prime }$ and using the diagonalization givesFor $I_0>0$ and energy level one in the chosen units, its ellipse equation isThe major semiaxis has length $a=\sqrt{2/I_0}$ along $u_1=(1,1)/\sqrt{2}$; the minor semiaxis has length $b=\sqrt{2/(3I_0)}$ along $u_2=(-1,1)/\sqrt{2}$. The full axis lengths are $2a$ and $2b$. The original-coordinate curve is the same ellipse tilted by $45^{\circ }$, elongated along ${\omega }_y={\omega }_x$; an orthogonal coordinate change preserves lengths and does not change the ellipse itself.

The matrix has the standard rotation matrix form $\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$, with $\cos \theta =\sin \theta =1/\sqrt{2}$. ThusWith $\omega =U{\omega }^{\prime }$, the new first axis points along the original $(1,1)$ direction and is counterclockwise from the original first axis. Specifying this coordinate convention avoids confusing rotation of the axes with the inverse transformation of component columns.

The normalized vectors $u_1=(1,1{)}^T/\sqrt{2}$ and $u_2=(-1,1{)}^T/\sqrt{2}$ give eigenvalues $I_0$ and $3I_0$. ThereforeThe columns are orthonormal, so $U^{-1}=U^T=U^{\ast }$ and direct multiplication verifies $U^{-1}MU=D$. Signs of individual eigenvectors are free; the displayed signs also make $\det U=+1$.

In the quadratic form, the two equal off-diagonal contributions together supply the mixed term. Comparing coefficients givesFor a nontrivial physical moment of inertia scale take $I_0>0$; then this inertia tensor is positive definite.

Let $s=(x_0,y_0,z_0)$ be the sphere centre. The circle centre is its perpendicular projection onto the plane:For the specific plane this isFor a point on the section, its displacement from $h$ lies in the plane and is perpendicular to $s-h$. The Pythagorean theorem therefore givesA genuine circle requires $d<R$. At $d=R$ the section is a single tangent point, and at $d>R$ the plane misses the sphere. A central cut $d=0$ gives a great circle of radius $R$.

Provided $p,q$ are linearly independent, their linear combinations fill a two-dimensional direction space. Adding $r_0$ translates it to the plane through $r_0$. The unit normal and constant areThen $n\cdot (r_0+t_{1}p+t_{2}q)=c_0$ for every pair of parameters. Conversely a point satisfying $n\cdot r=c_0$ differs from $r_0$ by a vector in $n^{\perp }=\mathrm{span}\{p,q\}$, proving that the two descriptions give exactly the same plane.

For the three points, choose $r_0=(3,1,-2)$, $p=(-4,1,6)$ and $q=(-1,-2,3)$. Their cross product is $(15,6,9)=3(5,2,3)$, which is nonzero. The plane therefore has equationEach supplied point satisfies it. Its unit normal is $(5,2,3)/\sqrt{38}$, up to an overall sign, and $c_0=11/\sqrt{38}$.

For the quadratic form of a squared cross product, express the two scalar terms as quadratic forms in $\mathbf{a}$:Then ${\mathbf{a}}^{T}M\mathbf{a}=|\mathbf{b}{|}^2|\mathbf{a}{|}^2-(\mathbf{a}\cdot \mathbf{b}{)}^2$. In particular, for $\mathbf{b}=(b_1,b_2,0)$,The cancellation is structural: $M\mathbf{b}=0$, while its action on every vector perpendicular to a nonzero $\mathbf{b}$ is multiplication by $|\mathbf{b}{|}^2$. Its eigenvalues are therefore $0,|\mathbf{b}{|}^2,|\mathbf{b}{|}^2$; for $\mathbf{b}=0$ the whole matrix is zero.

For nonzero vectors with angle $\theta$, $|\mathbf{a}\times \mathbf{b}|=|\mathbf{a}||\mathbf{b}|\sin \theta$ and $\mathbf{a}\cdot \mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos \theta$. Hence ${\sin }^2\theta =1-{\cos }^2\theta$ gives Lagrange's identityIf either vector is zero both sides vanish, so this covers every case. Alternatively, expansion of the three cross product components gives the same identity without introducing an angle.