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Fordefine its action on the projective line $F\cup \{\infty \}$ by the Möbius transformationwith the usual conventions when the denominator vanishes or $x=\infty$. Matrix multiplication agrees with composition, so this is a group action. If $A$ fixes every projective point, then it fixes $0$ and $\infty$, forcing $b=c=0$, and fixing $1$ then gives $a=d$. Thus the kernel consists exactly of the nonzero scalar matrices $Z$. The induced group homomorphismis therefore injective. This is the projective general linear group action on the projective line.

For $F={\mathbb{F}}_4$,The scalar subgroup has order $|Z|=3$, soThe four classes represented byform a subgroup $P$ of order four, hence a Sylow subgroup for the prime two. Its action is translation $x\mapsto x+b$ on ${\mathbb{F}}_4$ and fixes $\infty$. For $b\ne 0$, characteristic two makes this permutation a product of two disjoint transpositions on the four finite points. It is therefore even, andThis realizes the Sylow 2-subgroup of PGL2 over F4.

A function $f:D\to \mathbb{R}$ on a convex set $D$ is convex whenfor all $x,y\in D$ and $0\le t\le 1$.

If $f$ is once differentiable, this is equivalent to monotonicity of its gradient:Equivalently, every tangent hyperplane supports the graph:If $f$ is twice differentiable, convexity is equivalent to the Hessian matrix being positive semidefinite throughout $D$.

Forthe Hessian isA real symmetric two-by-two matrix is positive semidefinite exactly when its two diagonal entries and its determinant are nonnegative. ThusThe largest convexity domain is thereforeFor $A\ne 0$, its boundary is the hyperbola $y=A^2/(36x)$ in the first quadrant and the domain lies above it. For $A=0$, it is the closed first quadrant. This is the convexity domain of x cubed plus y cubed plus Axy.

A linear operator $\alpha$ on a real inner-product space is self-adjoint whenfor all $v,w$.

To prove the finite-dimensional spectral theorem, first note that the continuous quadratic function $⟨\alpha v,v⟩$ has a maximum on the unit sphere. The Lagrange multiplier equation at a maximizing vector $v$ gives $\alpha v=\lambda v$, so $\alpha$ has a real unit eigenvector. Its orthogonal complement is invariant becauseInduction on the dimension supplies an orthonormal eigenbasis of that complement and hence of $V$.

For $f\ne 0$,while symmetry and bilinearity are immediate, so the displayed formula defines an inner product on $P_n$. Sinceintegration by parts givesThe boundary term vanishes for polynomials, proving that $\alpha$ is self-adjoint.

On the monomial $x^k$,Thus the matrix of $\alpha$ in the monomial basis is triangular with diagonalThese are therefore its eigenvalues. For $n=2$, corresponding eigenvectors arefor eigenvalues $0,-1,-2$, respectively. They are the first Laguerre polynomials up to normalization, as described by the Laguerre differential operator on polynomials.

For a nonzero real vector $x$, the Rayleigh quotient of the matrix $A$ isThe Rayleigh quotient iteration starts from a unit vector $v_0$. Given $v_k$, compute the shiftsolve the linear systemand normalize in the Euclidean norm:The next eigenvalue estimate is ${\mu }_{k+1}=R_A(v_{k+1})$. The process stops when the eigenpair residual is sufficiently small.

Since $P=P^{\prime }=0$, it suffices to map the two edge vectors. Let the columns of the required matrix be $u_1,u_2,u_3$. The condition $MQ=Q^{\prime }$ givesThe condition $MR=R^{\prime }$ then givesand henceThese are orthonormal. Taking $u_3=u_1\times u_2=(0,-1,0{)}^T$ produces the orthogonal matrixIts columns are orthonormal, and direct substitution verifies $MP=P^{\prime }$, $MQ=Q^{\prime }$, and $MR=R^{\prime }$.

The matrix trace isUsing matrix multiplication and interchanging dummy indices,

The proposed noncyclic interchange is false. With matrix units,one hasTrace is invariant under cyclic permutations, but an arbitrary swap need not be cyclic.

Write the complex similarity matrix aswith $X,Y$ real. From $C=SD{S}^{-1}$ we obtain $CS=SD$, and equating real and imaginary parts givesThereforefor every real $t$.

Now $p(t)=\det (X+tY)$ is a real polynomial andThus some real $t$ has $p(t)\ne 0$. Taking $T=X+tY$, we have a real invertible matrix satisfying $CT=TD$, orThis proves real similarity from complex similarity.

Eigenvectors belonging to distinct eigenvalues are linearly independent. Indeed, supposeis a shortest nontrivial relation, where $A{v}_j={\lambda }_j{v}_j$ and the ${\lambda }_j$ are distinct. Applying $A-{\lambda }_{m}I$ givesa shorter nontrivial relation, a contradiction.

Thus $n$ distinct eigenvalues provide a basis of $n$ eigenvectors, proving that $A$ is diagonalizable over $\mathbb{C}$. If $A$ is real and all its eigenvalues are real, each real singular matrix $A-\lambda I$ has a nonzero real null vector. Choosing one for each eigenvalue gives a real eigenbasis. Hence the real matrix is diagonalizable over $\mathbb{R}$. This is the distinct eigenvalues imply diagonalizability criterion.

Suppose $A$ is diagonalized by a unitary matrix:where $D$ is diagonal. ThenSince the diagonal matrices $D$ and $D^{†}$ commute,Thus $A$ is a normal matrix.

A complex matrix $U$ is unitary matrix whenIf $Uv=\lambda v$ with $v\ne 0$, unitarity preserves norms, soTherefore every eigenvalue satisfies

LetThis permutation matrix is unitary. The actionsgivewhich is upper triangular.

Both sides are alternating in $i,j,k$ and in $l,m,n$. If either triple repeats an index, both sides vanish. Otherwise each triple is a permutation of $(1,2,3)$; the determinant of the corresponding permutation matrix is exactly the product of the two permutation signs. HenceThere are $3!=6$ nonzero components of ${ϵ}_{ijk}$, each equal to $\pm 1$, so

The special orthogonal group isIfits first column is a unit vector, say $(\cos \theta ,\sin \theta {)}^T$. The second column is the unique unit vector perpendicular to it that gives positive determinant, namely $(-\sin \theta ,\cos \theta {)}^T$. Thusso $A$ is rotation through $\theta$ about the origin.

For $A\in SO(3)$, its real characteristic polynomial of odd degree has a real eigenvalue. Every eigenvalue of an orthogonal matrix has modulus one, so every real eigenvalue is $\pm 1$. Nonreal eigenvalues occur in conjugate pairs whose product is one, while $\det A=1$; if all eigenvalues are real, their product likewise forces at least one to be $+1$. Hence $A$ fixes a nonzero vector $u$. The plane $u^{\perp }$ is $A$-invariant, and the restriction to it is an orientation-preserving orthogonal map. By the $SO(2)$ result it is a planar rotation. Therefore $A$ is a rotation about the axis $\mathbb{R}u$.

For $\sigma \in S_n$, let $P_{\sigma }$ be its permutation matrix and define the sign of a permutation bySince $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$,so this is a homomorphism. A transposition interchanges two columns of the identity matrix and therefore has determinant $-1$. If $\sigma$ is expressed as a product of $r$ transpositions, then $\mathrm{sgn}(\sigma )=(-1{)}^r$; because the determinant depends only on $\sigma$, the parity of $r$ is independent of the chosen expression.

There are four main diagonals and the action is transitive, so the stabilizer of one has orderFor the diagonal spanned by $(1,1,1)$, the stabilizing signed permutation matrices are $P$ and $-P$ with $P$ a permutation matrix. HenceThe action is not faithful: both $I$ and the central inversion $-I$ fix every main diagonal setwise. In fact its kernel is exactly $\{I,-I\}$.

A Givens rotation ${\Omega }^{[p,q]}$ is the identity except in rows and columns $p,q$, where it has the blockFor $a=A_{pj}$ and $b=A_{qj}$, choosewhen $(a,b)\ne (0,0)$. ThenIf both entries vanish, any angle works.

For the given matrix, first useIt givesThen useThus $R={\Omega }^{[2,3]}{\Omega }^{[1,3]}A$ isAll leading row entries are positive. The resulting QR decomposition by Givens rotations iswhereand $R$ is the matrix above. Since it is a product of transposed rotations, $Q$ is orthogonal.

For a linear map $T:V\to W$, its rank is $\dim \mathrm{im}T$ and its nullity is $\dim \ker T$. The rank-nullity theorem states that, when $V$ is finite-dimensional,To prove it, take a basis $v_1,\dots ,v_k$ of $\ker T$ and extend it to a basisof $V$. The vectors $T{v}_{k+1},\dots ,T{v}_n$ span $\mathrm{im}T$. They are also linearly independent: if ${\sum }_{j=k+1}^n{a}_{j}T{v}_j=0$, then ${\sum }_{j=k+1}^n{a}_j{v}_j\in \ker T$, and independence of the chosen basis forces every $a_j$ to vanish. Thus the rank is $n-k$ and the nullity is $k$.

For the given subspace, row reduction of the coefficient matrix givesTaking $x_3=s$ and $x_5=t$ therefore givesHenceis a basis of $W$.

The characteristic polynomial isTo prove triangularizability, use induction on $n$. The result is immediate for $n=1$. Over $\mathbb{C}$, ${\chi }_A$ has a root $\lambda$, so $A$ has an eigenvector $v_1$. Extend it to a basis. In this basis,By induction, a change among the remaining $n-1$ basis vectors makes $B$ upper triangular. Thus the triangularization over an algebraically closed field gives

The minimal polynomial $m_A$ is the monic polynomial of least degree satisfying $m_A(A)=0$. To establish existence without quoting the Cayley-Hamilton theorem, let $T$ be an upper-triangular matrix similar to $A$, with diagonal entries ${\lambda }_1,\dots ,{\lambda }_n$, and putThenThe factors $T-{\lambda }_{k}I$ commute, so applying all of them successively lowers the invariant flag to zero:Similarity gives the same polynomial identity for $A$. Hence a nonzero monic annihilating polynomial of degree $n$ exists, and a least-degree one exists.

If $m$ and $\tilde{m}$ were two monic annihilating polynomials of the same least degree, then $m-\tilde{m}$ would be an annihilating polynomial of smaller degree unless it were zero. Thus the minimal polynomial is unique, and the minimal polynomial bound from a triangular invariant flag gives

Integer row and column operations giveThus the Smith normal form isThe two relation vectors are the columns of the displayed matrix, so its cokernel is the presented module. Consequently

The Smith normal form of an integer $m\times n$ matrix $A$ is a diagonal matrixfor whichwith $U\in G{L}_m(\mathbb{Z})$ and $V\in G{L}_n(\mathbb{Z})$. It exists and is unique.

The structure theorem for finitely generated modules over a principal ideal domain, specialized to $\mathbb{Z}$, says that every finitely generated abelian group has a unique invariant-factor decompositionEquivalently, its finite part is a direct sum of cyclic groups of prime-power order.