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An invariant subspace $W$ satisfies $D(g)W\subseteq W$ for every $g\in G$; applying the inverse gives equality. The representation is irreducible when the only invariant subspaces are $0$ and the full space.

The inverse and conjugate transpose each reverse multiplication order, so their combination preserves it:Its values are invertible, so this is the conjugate contragredient representation.

If $W$ is invariant under $D_2$, take $w\in W$ and $v\in W^{\perp }$. For the standard Hermitian inner product,since $D_2(g^{-1})w\in W$. Hence $D_1(g)v\in W^{\perp }$, proving its invariance. If $D_1$ is irreducible, $W^{\perp }$ is either zero or the whole space; finite-dimensional orthogonal complementation makes $W$ the whole space or zero, respectively. Thus $D_2$ is irreducible whenever $D_1$ is irreducible.

A square matrix is invertible precisely when its determinant is nonzero. Equivalent conditions are full rank, a trivial kernel, and zero not being an eigenvalue.

For a lower triangular matrix, the first row has only the possibly nonzero entry $A_{11}$. A Laplace expansion along that row gives $\det A=A_{11}\det A^{(1)}$, where the remaining principal submatrix is again lower triangular. Induction from the one-by-one case yieldsThis proof includes singular triangular matrices; it makes no division by diagonal entries.

For the diagonal example, put $N=n(n+1)/2$. Its determinant is $b^{1+2+\cdots +n}=b^N$, and invertibility requires $b\ne 0$. Write the complex number as $b=r{e}^{i\theta }$ with $r>0$. A nonzero purely imaginary determinant requires $\cos (N\theta )=0$. ThusEquivalently $b^N\in i\mathbb{R}\setminus \{0\}$; the angle is understood modulo $2\pi$.

For skew-Hermitian matrices, reversing products under the conjugate transpose gives $(CD{)}^{†}=D^{†}{C}^{†}=DC$. ConsequentlyThe latter condition is vanishing of the anticommutator. A universal choice for the last request isIndeed $(CD-DC{)}^{†}=DC-CD=-(CD-DC)$, so the commutator of two skew-Hermitian matrices is skew-Hermitian.

Taking the conjugate transpose of $M{M}^{-1}=I$ and $M^{-1}M=I$ reverses the product order and givesThus $(M^{-1}{)}^{†}$ is the two-sided matrix inverse of $M^{†}$, proving

Multiplying the defining equation on the right by $I+A$ gives $(I+U)A=I-U$. Consequently the only possible answer isThe factors commute because they are rational functions of $U$. Its conjugate transpose, using the unitarity $U^{†}=U^{-1}$, isThus $A$ is Skew-Hermitian. Moreover $I+A=2(I+U{)}^{-1}$ and $I-A=2U(I+U{)}^{-1}$, so substitution recovers $U$ and establishes existence as well as uniqueness. The unitary matrix $U=-I$ is a counterexample to invertibility of $I+U$; more generally the obstruction is an eigenvalue $-1$.

A Hermitian matrix satisfies $H^{†}=H$, where $†$ denotes the conjugate transpose. For a nonzero eigenvector $v$ with $Hv=\lambda v$, the scalar $v^{†}Hv$ equals its own complex conjugate, andThus every Hermitian eigenvalue is real.

A unitary matrix satisfies $U^{†}U=U{U}^{†}=I$. If $Uv=\mu v$, preservation of the squared norm givesSince $v\ne 0$, $\boxed{|\mu |=1}$. These statements concern complex eigenvectors, even when the matrix entries happen to be real.

An $n\times n$ matrix is diagonalizable over a chosen field exactly when it has a basis of $n$ linearly independent eigenvectors. Equivalently, $P^{-1}AP$ is diagonal for some invertible $P$. Distinct eigenvalues suffice but are not necessary. For example,is not diagonalizable: its only eigenvalue is $1$ and its eigenspace is the one-dimensional span of $(1,0{)}^T$. It is a nontrivial Jordan block.

A Hermitian matrix satisfies $A^{†}=A$, where the dagger denotes conjugate transpose. If $Av=\lambda v$ with $v\ne 0$, thenThe denominator is positive real, soFor $Av=\lambda v$ and $Aw=\mu w$, Hermiticity also givesThusThis is the necessary qualification to the PDF's orthogonality wording: arbitrary eigenvectors belonging to a repeated eigenvalue need not be orthogonal. Within each repeated eigenspace, Gram-Schmidt orthogonalization produces an orthonormal basis; together the eigenspaces give an orthonormal eigenbasis by the finite-dimensional spectral theorem. For example, $(1,0{)}^T$ and $(1,1{)}^T$ are nonorthogonal eigenvectors of the identity matrix.

Using $(AB{)}^{†}=B^{†}{A}^{†}$, the conjugate transpose of the proposed matrix isThusThe commutator of two Hermitian matrices is skew-Hermitian, and multiplication by $i$ changes it into a Hermitian matrix. No simplicity assumption on the spectra is needed for this part.

The finite-dimensional spectral theorem gives a spectral decomposition of the Hermitian matrix:The matrix exponential respects similarity, soThe diagonal entries are real, so the conjugate transpose of this expression equals itself. Its eigenvalues are $e^{{\lambda }_j}>0$, provingIn particular, for any nonzero $v$, $v^{†}{e}^{A}v={\sum }_j{e}^{{\lambda }_j}|(Q^{†}v{)}_j{|}^2>0$. Thus the exponential of a Hermitian matrix is a Hermitian positive-definite matrix, even when $A$ itself has negative eigenvalues.

Every $D_2(g)$ is invertible because both inversion and the conjugate transpose preserve invertibility. For $g,h\in G$, the group homomorphism property of $D_1$ and reversal of multiplication under the adjoint giveAlso $D_2(e)=I$. Hence $D_2$ is a group representation. It is the inverse-adjoint representation of $D_1$; the inverse and adjoint each reverse product order, so together they preserve it.

A skew-Hermitian matrix satisfies $A^{†}=-A$, where $†$ denotes the conjugate transpose. If $Av=\lambda v$ with $v\ne 0$, thenSince $v^{†}v>0$, this gives $\boxed{\overline{\lambda }=-\lambda ,\text{ so }\lambda \in i\mathbb{R}}$.

A unitary matrix satisfies $U^{†}U=U{U}^{†}=I$. It preserves the Hermitian inner product and norm. Thus if $Uv=\mu v$,and therefore $\boxed{|\mu |=1}$. Zero is allowed among the purely imaginary eigenvalues in the first statement, but not among the unitary eigenvalues.

Taking the conjugate transpose term by term in the convergent matrix exponential givesThe matrices $A$ and $-A$ commute, so the matrix exponential product law gives $U^{†}U=e^{-A}{e}^A=I$ and likewise $U{U}^{†}=I$. Hence $\boxed{U\text{ is unitary}}$.

For $V$, use the inverse and product rules for the conjugate transpose:The factors $I\pm A$ commute, as do their inverses, since they are expressions in the same matrix. Consequentlyand similarly $V{V}^{†}=I$. Therefore $\boxed{V\text{ is unitary}}$. This rational construction is the skew-Hermitian form of the Cayley transform of a Hermitian matrix, obtained by writing $A=iH$ with $H$ a Hermitian matrix.

Using the conjugate transpose and the Hermitian matrix assumptions,Thus $ABA$ is a Hermitian matrix, whether or not $A$ and $B$ commute.

Take $\boxed{c=1}$. Indeed the conjugate transpose isso the anticommutator is a Hermitian matrix. The original PDF, p. 6, supplies the factor $BA$ that is corrupted in the local OCR.

The conjugate transpose reverses the order of matrix multiplication. Since $A$ and $B$ are Hermitian matrices,Consequently $\boxed{AB\text{ is Hermitian}⟺AB=BA}$: the condition is exactly that the two matrices commute.

Put $D=A-iB$, so $D^{†}=A+iB$. Because the Hermitian matrices $A,B$ commute,For every nonzero vector $v$,since $A$ is an invertible matrix. Thus $D$ and $D^{†}$ are invertible matrices, justifying the inverse in the definition of $U$. The conjugate transpose of $U=D^{†}{D}^{-1}$ is $(D^{†}{)}^{-1}D$. Since $D$ and $D^{†}$ commute, their inverses do too, andHence $U$ is a unitary matrix.

For every vector $v$, the Hermitian inner product givesThe norm vanishes only on the zero vector, so $Cv=0$ for every $v$ and $\boxed{C=0}$. In contrast,is a nonzero nilpotent matrix. Multiplying a matrix by its conjugate transpose detects column norms; multiplying it by itself has no corresponding positivity property.

A QR factorization of a real $n\times n$ matrix isThus $Q$ is an orthogonal matrix and $R$ is an upper triangular matrix. It exists even if $A$ is singular; requiring positive diagonal entries of $R$ is a convenient uniqueness convention for nonsingular $A$, rather than part of the general existence statement. Over the complex numbers, replace orthogonal by unitary and transpose by conjugate transpose.

The Moore--Penrose inverse is characterized by

Here $A^{†}=A^{+}$ denotes the Moore-Penrose inverse, while the star denotes the conjugate transpose. The first two identities express consistency when applying and reversing the linear map; the last two make the resulting orthogonal projection matrices Hermitian. They must hold together. For example, $A=(12)$ and $B=(10{)}^T$ satisfy $ABA=A$ and $BAB=B$, butis not Hermitian. This $B$ is a right inverse, yet $Bb=(3,0{)}^T$ is not the minimum-norm least-squares solution for $b=3$. The last two Penrose equations impose the orthogonality that selects the pseudoinverse.

The strict positivity of a quadratic form on every nonzero vector. A real symmetric matrix has positive definiteness exactly when its associated quadratic form does, making it a positive-definite matrix. A Hermitian form is positive definite when its value on $(v,v)$ is real and strictly positive for every nonzero vector.

Think of a positive-definite quadratic form as an energy cost for a displacement: every direction costs something, and zero cost identifies the origin. For example, the real symmetric matrixis a positive-definite matrix because the two squares can vanish together only at $(0,0)$. The mixed term couples the coordinates, but cannot overwhelm their positive contributions: $q(1,1)=8$ and $q(1,-1)=4$. The graph is a bowl, and its positive level sets are ellipses with axes along the eigenvectors, rather than along the original coordinates.

The real and complex settings require different symmetry assumptions. For a real symmetric matrix use $q(x)=x^{T}Ax$ with $x\in {\mathbb{R}}^n$; for a complex Hermitian matrix use $q(z)=z^{\ast }Az$ with $z\in {\mathbb{C}}^n$, where $z^{\ast }$ is the conjugate transpose. The latter expression is real, so strict positivity is meaningful. A complex quadratic form such as $z_1^2+z_2^2$ has no such positivity: it vanishes at $(1,i)$ and takes negative real values at $(i,0)$. A Hermitian form uses complex conjugation, which changes this example to $|z_1{|}^2+|z_2{|}^2$.

For a real matrix $M$ without a symmetry hypothesis, its quadratic form depends only on its symmetric part of a matrix:Positive eigenvalues of $M$ alone do not establish positive definiteness of this quadratic form. For example, $M=\left(\begin{array}{cc}1& 4\\ 0& 1\end{array}\right)$ has both eigenvalues equal to $1$, yet $q(1,-1)=1-4+1=-2$. The relevant test is positivity of the symmetric part of a matrix, not the spectrum of an arbitrary matrix.^[ref]

For real vectors $x_i$ in a Euclidean inner product space, their Gram matrix $G_{ij}=x_i^T{x}_j$ satisfies the displayed identity by expanding the squared norm. Thus every Gram matrix is a positive semidefinite matrix, and it is a positive-definite matrix exactly when the vector family is linearly independent. The complex version replaces transposes by conjugate transposes and uses $c^{\ast }Gc=\Vert {\sum }_i{c}_i{x}_i{\Vert }^2$.