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A matrix entry whose row and column indices differ. These entries lie outside the diagonal matrix entries. A diagonal matrix has all its off-diagonal entries equal to zero. Cancellation of these entries when averaging density matrices does not require the individual preparations to have zero entries there.

A real skew-symmetric matrix is an orthogonal matrix exactly when its square is minus the identity matrix. Indeed, $B^{\mathsf{T}}B=-B^2$. Such a matrix can exist only in even dimension: taking determinants of $B^2=-I$ would otherwise give $(\det B{)}^2=-1$. In two dimensions every example isThese are rotation matrices through opposite right angles. Their orthogonal similarity transforms preserve eigenvalues, but do not in general produce a diagonal matrix.

For steady uniaxial extension of the Giesekus model, the component-first velocity gradient is diagonal, $G=\mathrm{diag}(\gamma ,-\gamma /2,-\gamma /2)$. For a steady spatially uniform diagonal stress, the material derivative vanishes and the convected terms are $-2\gamma {\sigma }_{11}$ and $+\gamma {\sigma }_{22}$. The off-diagonal equations vanish identically, and symmetry permits equal transverse components. The remaining equations areSet $W=\lambda \gamma \ge 0$, $b=\alpha p_T$ and $s_i={\sigma }_{ii}/p_T$, with $\mu =\lambda p_T$. For $b>0$, the branches continuously connected to zero stress at zero rate areA sufficient nondegenerate range for physical solutions at every nonnegative rate is $0<b<1$. Indeed $M=(1-b)I+bA$ with $A=HC/(k_{B}T)$, so positive conformation implies positive mobility. The longitudinal quadratic equation is nonpositive at $s_1=0$ and has a nonnegative root of a polynomial. The transverse quadratic equation has values $b-1<0$ at $s_2=-1$ and $W\ge 0$ at $s_2=0$, so its larger root of a polynomial lies in $(-1,0]$. Thus the displayed branches satisfy $s_i>-1$, giving $A_i=1+s_i>0$. The other algebraic roots do not provide a physical branch connected to the equilibrium distribution. For constant positive diagonal $A$, the associated stationary configuration distribution is a Gaussian with covariance $C=(k_{B}T/H)A$.

For positive $\alpha$ on the physical branch, the stretching stress remains finite through $W=1/2$. At that rate,At large extension rate,Thus the stress continues to grow, but the uniaxial extensional viscosity ${\eta }_E=({\sigma }_{11}-{\sigma }_{22})/\gamma$ approaches the finite plateau $2\lambda /\alpha$ rather than developing a finite-rate pole. The transverse stress approaches $-p_T$ as its conformation tensor variance becomes small. At small rate the two stresses are $2\mu \gamma$ and $-\mu \gamma$, so ${\eta }_E\to 3\mu$.

When $\alpha =0$, the physical steady branch instead isThe stretching stress diverges as $W\to 1/2$ from below. Above that threshold its algebraic continuation has $1+s_1=1/(1-2W)<0$ and cannot be a conformation covariance. There is no physical finite steady stretching state. The transient equationgives linear growth at $W=1/2$ and exponential growth at larger rates, the Hookean coil-stretch failure of the upper-convected Maxwell limit.

Parameter qualification. Positivity of $\alpha$ alone, without a restriction on $b=\alpha n{k}_{B}T$, does not ensure real physical steady solutions for the entire tensor. For example $b=2$, $W=1$ makes the transverse equation $2s_2^2+2s_2+1=0$, which has no real root. Its diffusion/resistance interpretation is already inadmissible for some conformations. The physically positive-mobility regime stated above avoids that counterexample; the source's positive-$\alpha$ case must not be taken as a guarantee for arbitrary parameters.

Write $K=\beta J$, where $\beta =1/(k_{B}T)$ is the inverse temperature, $T$ is temperature and $k_B$ is the Boltzmann constant. In the canonical ensemble, each spin configuration has probability $e^{-\beta H}/Z_N$. The nearest-neighbor Boltzmann factors multiply, so the spin-chain transfer matrix, in the order $+1,-1$, isMultiplying its entries around the ring and summing over every Ising spin yields the partition function:The trace implements the periodic boundary conditions; replacing it by an endpoint sum would describe a different partition function. The normalized eigenvectors and corresponding eigenvalues areThe negative eigenvalue is essential for the antiferromagnetic Ising model. Introduce their ratioThe exact finite-$N$ partition function and Helmholtz free energy areFor finite $K>0$, $|t|<1$, so $Z_N>0$ for both even and odd $N$. The last term retains the finite-ring effect of geometric frustration.

To insert an Ising spin into this trace, use the diagonal matrix $\mathsf{S}=\mathrm{diag}(1,-1)$. In the eigenbasis of $\mathsf{T}$, it exchanges $|+⟩$ and $|-⟩$, and hence has vanishing diagonal entries. The one-point correlation function is thereforeThe same conclusion follows directly from spin inversion symmetry: reversing every Ising spin preserves the energy and the Boltzmann factor while reversing $S_1$.

For the two-point correlation function, the two insertions divide the ring into paths of $n$ and $N-n$ bonds. ThusIn particular, the exact finite-ring antiferromagnetic Ising correlation isBecause the one-point correlation functions vanish, this is also the connected correlation function. It gives $C_0=1$ and $C_n=C_{N-n}$, as required by $S_1^2=1$ and exchanging the two paths around the ring. At $K=0$, distinct Ising spins are independent and $C_n=0$ for $1\le n\le N-1$; $C_0=1$ is understood separately if a power $0^0$ occurs in the displayed expression.

The quadratic polynomial $x^2+4x+10=(x+2{)}^2+6$ has no real roots of a polynomial. Thus the characteristic polynomial does not split over the real numbers. A real diagonal matrix would have only real eigenvalues, and similar matrices have the same characteristic polynomial. Therefore $A$ is not diagonalizable over $\mathbb{R}$.

Suppose a single basis gives the diagonal matrices $D_S,D_T$. Entrywise multiplication gives $D_S{D}_T=D_T{D}_S$. If $P$ is the change-of-basis matrix, thenThus simultaneous diagonalization implies commutation.

Yes, provided “commute” means pairwise commute. Necessity follows because three diagonal matrices commute pairwise.

For sufficiency, first choose the common eigenbasis for the first two linear maps, say $S,T$, obtained in part (b). Their joint eigenspacesgive a direct sum decomposition of $V$. The third map $U$ preserves each $W_{\lambda ,\mu }$, since $SUv=USv=\lambda Uv$ and $TUv=UTv=\mu Uv$. By the diagonalizable restriction lemma, $U{|}_{W_{\lambda ,\mu }}$ has a basis of eigenvectors there. Every such vector already has fixed eigenvalues for $S$ and $T$.

Joining these bases therefore gives a common eigenbasis for all three maps. The same induction proves simultaneous diagonalization for any finite pairwise commuting family of diagonalizable linear endomorphisms.

Use the monomial basis $e_k=x^{n-k}{y}^k$, $0\le k\le n$. Direct differentiation giveswith the endpoint terms zero. Repeated application therefore gives $P^{n+1}=Q^{n+1}=0$, while $P^n{e}_n=n!e_0\ne 0$ and $Q^n{e}_0=n!e_n\ne 0$. Each nilpotency index is exactly $n+1$, soNextand hence $(PQ-QP)e_k=(n-2k)e_k$. These $n+1$ real eigenvalues are distinct. A polynomial annihilates this operator exactly when it vanishes at all of them, givingThe commutator is diagonal in the displayed monomial basis. Since $n\ge 1$, $P,Q$ are nonzero nilpotent linear map; a diagonalizable nilpotent map would have a diagonal matrix with all eigenvalues zero and thus be the zero map. Therefore only $PQ-QP$ is diagonalizable. The calculation realizes raising and lowering on binary homogeneous polynomials.

Choose representative vectors $x_1,\dots ,x_{n+1}$ of the first $n+1$ points; they form a basis. Write a representative of the last point as ${\sum }_i{a}_i{x}_i$. General position forces each $a_i\ne 0$, since otherwise $n+1$ of the points would lie in a hyperplane. Rescale the basis vectors to $a_i{x}_i$. The tuple is then projectively equivalent to the standard frameApply the same normalization to the second tuple and compose the two transformations to obtain existence. An automorphism fixing the coordinate points is represented by a diagonal matrix. Fixing their sum forces all its diagonal entries to agree, so its projective action is the identity. This proves existence and uniqueness. At $n=1$ it says the Möbius group acts sharply triply transitively on the projective line: any ordered triple of distinct points can be sent to any other in exactly one way.

By irreducible traceless matrix Lie algebra is semisimple, let $R$ be the solvable radical. Lie's theorem triangularizes its action, with finitely many diagonal characters ${\lambda }_i:R\to \mathbb{C}$, and supplies a nonzero common eigenspace for one of them. For $x\in L$, conjugation by $e^{tx}$ preserves $R$ and gives an equivalent representation of it. Thus the multiset of characters ${\lambda }_i\circ e^{t\mathrm{ad}x}$ is constant. Each character varies continuously in $t$ and has values in that finite set of characters, so it is individually constant. Differentiating at zero gives ${\lambda }_i([x,r])=0$ for all $r\in R$.

Choose a common eigenspace $V_{\lambda }\ne 0$. For $v\in V_{\lambda }$,Thus $V_{\lambda }$ is $L$-invariant, so irreducibility makes it all of ${\mathbb{C}}^n$. Every element of $R$ acts as a scalar. Its trace is zero because $L\subseteq {\mathfrak{sl}}_n$, so that scalar is zero. Faithfulness of the matrix inclusion gives $R=0$. Hence $\boxed{L\text{ is semisimple}}$.

One sufficient diagonal-matrix condition for simplicity is that $L$ contain all traceless diagonal matrices. Indeed invariance under their adjoint actions then decomposes $L$ into that diagonal algebra and some matrix-unit spaces $\mathbb{C}{e}_{ij}$. Put a directed edge $j\to i$ when $e_{ij}\in L$. Irreducibility makes the directed graph strongly connected; otherwise a terminal component spans a proper invariant coordinate subspace. Bracketing matrix units along a simple directed path supplies the matrix unit between its endpoints. Strong connectivity therefore supplies every off-diagonal matrix unit, so $L={\mathfrak{sl}}_n$, which is simple. The mere existence of a diagonal matrix with distinct diagonal entries would not be a sufficient condition.

For the specified construction, $H$ has dimension two. Its diagonal adjoint action and the twelve distinct nonzero weights give a direct sum $L=H\oplus {⨁}_{i=1}^{12}\mathbb{C}{g}_i$, of dimension fourteen. The stated bracket condition and the weight-vector relations make it a Lie algebra. A subspace invariant under it is invariant under all the $g_i$ and hence their associative algebra, so it acts irreducibly on ${\mathbb{C}}^7$. The preceding result makes it semisimple.

Use prime-dimensional faithful irreducible module forces simplicity. To see this, decompose the algebra into simple ideals. Complete reducibility for one ideal and commutation with the others make its isotypic component invariant under the whole algebra, hence all of the representation. Writing that component as $W\otimes \mathrm{Hom}(W,V)$ and repeating produces a tensor product of irreducibles for the simple ideals. Faithfulness makes every factor nontrivial, of dimension at least two. Their dimensions multiply to seven, so only one factor is possible. Thus the constructed $L$ is simple.

The diagonal algebra $H$ is toral, and its centralizer inside $L$ is just $H$, since no $g_i$ has zero weight. A maximal toral subalgebra containing $H$ lies in that centralizer, so equals $H$; it is a Cartan subalgebra. Thus the rank is two and the twelve weights are the roots. Pair them as ${\alpha }_{i+6}=-{\alpha }_i$. The opposite-root brackets have weight zero and belong to $H$. For distinct $i,j\le 6$, the sum cannot be zero, and the supplied root-bracket result says their bracket is nonzero exactly when ${\alpha }_i+{\alpha }_j$ is a root. This is equivalent to $-{\alpha }_i-{\alpha }_j$ being a root, proving the requested equivalence for the paired brackets.

Among rank-two simple root systems, $A_2,B_2,G_2$ have respectively six, eight and twelve roots. ThereforeChoose simple roots $\alpha$ short and $\beta$ long. In $G_2$, $\beta +\alpha$ and $\beta +2\alpha$ are roots. Renumber with $g_1\in L_{\alpha }$, $g_3\in L_{\beta }$, $g_4\in L_{\beta +\alpha }$ and $g_5\in L_{\beta +2\alpha }$, retaining opposite partners in the other six positions. Root spaces are one dimensional, and the allowed bracket equality gives

Use the standard structural fact that $\mathfrak{sp}(V)$ is a semisimple Lie algebra over an algebraically closed field of characteristic zero. If $\mathfrak{h}$ is a Cartan subalgebra, part (a)(ii) gives $C_{\mathfrak{sp}(V)}(\mathfrak{h})=\mathfrak{h}$, and $\mathfrak{h}$ is an abelian Lie algebra. Part (c) therefore shows that its weight spaces occur as $n$ pairs of one-dimensional spaces, with no zero weight space.

Choose representatives ${\lambda }_1,\dots ,{\lambda }_n$ of its nonzero weights of a representation. Pick $p_i\ne 0$ in $V_{{\lambda }_i}$ and $q_i\ne 0$ in $V_{-{\lambda }_i}$, scaling $q_i$ so that $\omega (p_i,q_i)=1$. Part (b) shows that all other pairings among different pairs are zero. Each individual weight space is one-dimensional and the bilinear form is alternating, so the ordered basisis a symplectic basis. Every $h\in \mathfrak{h}$ is a diagonal matrix in this basis, with entries ${\lambda }_1(h),\dots ,{\lambda }_n(h),-{\lambda }_1(h),\dots ,-{\lambda }_n(h)$.

Let $D_{\mathcal{B}}$ be the Lie subalgebra consisting of all diagonal matricesEach such matrix lies in the symplectic Lie algebra and commutes with $\mathfrak{h}$. Henceand $\mathfrak{h}=D_{\mathcal{B}}$. In particular, every Cartan subalgebra has dimension $n$; the ${\lambda }_i$ are a basis of ${\mathfrak{h}}^{\ast }$.

Carry out the same construction for ${\mathfrak{h}}^{\prime }$, obtaining a symplectic basis ${\mathcal{B}}^{\prime }=(p_1^{\prime },\dots ,p_n^{\prime },q_1^{\prime },\dots ,q_n^{\prime })$ with ${\mathfrak{h}}^{\prime }=D_{{\mathcal{B}}^{\prime }}$. The unique linear map $g$ sending $p_i$ to $p_i^{\prime }$ and $q_i$ to $q_i^{\prime }$ is invertible and preserves all pairings of basis vectors. By bilinearity, it preserves $\omega$ on all of $V$, so $g$ belongs to the symplectic group $\mathrm{Sp}(V)$. The change of basis givesThus conjugacy of Cartan subalgebras of the symplectic Lie algebra follows by explicitly matching their symplectic bases. If $n=0$, the Lie algebra is zero and the same conclusion holds with $g$ the identity on the zero vector space.

In the monomial basis $\mathcal{B}=(1,x,x^2,x^3)$ of the complex vector space $V=\mathbb{C}[x{]}_{\le 3}$, the binomial theorem givesThe characteristic polynomial of this upper triangular matrix is the product of $t$ minus its diagonal entries, so ${\chi }_T(t)=(t-1{)}^4$.

For the minimal polynomial, set $\Delta =T-I$, the forward difference with step one. For a nonconstant polynomial of polynomial degree $d$, the binomial theorem shows that $f(x+1)-f(x)$ has polynomial degree $d-1$. Constant polynomials have zero forward difference. Consequently ${\Delta }^4=0$ on $V$. To prove that the nilpotency index is exactly four, computeIt follows thatIf $T$ were diagonalizable, its only eigenvalue would be $1$, and its diagonal matrix in an eigenbasis would be $I$. This would make $T=I$, whereas $T(x)=x+1\ne x$. Thus $T$ is not diagonalizable. Equivalently, its minimal polynomial has a repeated root.

Choose the diagonal matrix $W=\mathrm{diag}(w_1,\dots ,w_m)$. Multiplying a vector by $W$ scales its $i$th component by $w_i$. ThereforeThe Euclidean norm objective for ordinary linear least squares on $(WA,Wb)$ is exactly the weighted least squares objective, so the sets of minimizing vectors coincide. Zero weights simply discard the corresponding residuals.

A power of a diagonal matrix acts entrywise. ThusFrom $A=BD{B}^{-1}$, adjacent $B^{-1}B$ factors cancel in repeated multiplication, givingThis is the matrix power rule for a similarity transformation, proved directly rather than by taking powers of separate noncommuting factors.

For a diagonal matrix, each diagonal entry separately obeys $d^2+2d-15=(d-3)(d+5)=0$. Hence all four possibilities areFor the second form, $X^2=I$ givesBecause $I,X$ are linearly independent and $b\ne 0$, the $X$ coefficient forces $a=-1$; the $I$ coefficient then gives $b^2=16$. ThusBoth constants are nonzero, as required.

For a finite matrix, the series defining the matrix exponential converges absolutely in any matrix norm, because $\Vert A^m/m!\Vert \le \Vert A{\Vert }^m/m!$. Thus the Hermitian adjoint may be taken term by term. Since $A^{†}=A$,Therefore the matrix exponential of a Hermitian matrix is Hermitian.

If $A=\mathrm{diag}(a_1,\dots ,a_n)$ is a diagonal matrix, then $A^m=\mathrm{diag}(a_1^m,\dots ,a_n^m)$. Summing each entry separately givesFor the Hermitian matrix case each $a_i$ is real. The diagonal entry formula itself remains valid for complex diagonal entries.

The characteristic polynomial isThus the eigenvalues are $i$ and $-i$. There is no real diagonalization, because a real diagonal matrix would have real eigenvalues. Over the complex numbers, the two distinct eigenvalues give linearly independent eigenvectors, so the matrix is diagonalizable.

For $\lambda =i$, the first row gives $(1-i)x+2y=0$. Choose $x=2$, giving $y=-1+i$. Its standard complex norm is $\sqrt{|2{|}^2+|-1+i{|}^2}=\sqrt{6}$; the conjugate eigenvector belongs to $-i$. Hence normalized eigenvectors and a suitable change of basis areDirectly, $AP=P\mathrm{diag}(i,-i)$ and $\det P=-2i/3\ne 0$, soNormalization does not assert that these eigenvectors are orthogonal.

For distinct coordinates $i,j$, let $E_{ij}(t)=I+t{e}_{ij}$, where $e_{ij}$ is the matrix unit. For $t\ne 0$ this is a transvection: it fixes the hyperplane with $j$th coordinate zero pointwise, has determinant one and inverse $E_{ij}(-t)$. We show these elementary transvection matrices generate $SL(n,F)$ over any field.

Let $M\in SL(n,F)$. Row addition is left multiplication by an $E_{ij}(t)$. If the current diagonal pivot is zero, a nonzero entry occurs below it in the same remaining column, since the remaining square block is invertible; adding that row makes the pivot nonzero. Row additions then clear the entries below the pivot. Repeating produces an upper-triangular matrix with nonzero diagonal. Clearing the entries above the diagonal, starting from the last column, gives a diagonal matrix $D=\mathrm{diag}(a_1,\dots ,a_n)$ with ${\prod }_i{a}_i=1$. Thus it remains to realize determinant-one diagonal matrices using elementary transvections.

In a two-coordinate block putDirect multiplication givesThe diagonal matrix $D$ is the product, for $i=1,\dots ,n-1$, of these blocks on coordinates $i,n$, with entries $a_i,a_i^{-1}$. Their final $n$th entry is ${\prod }_{i<n}{a}_i^{-1}=a_n$. Therefore $D$, and hence $M$, is a product of elementary transvections. All row operations used are invertible transvections, and no division except by a nonzero pivot or $a_i$ has been made. This provesThe identity element $E_{ij}(0)$ can be omitted from the generating set. This is transvections generate the special linear group; ordinary row scaling alone would not prove it, because such a scaling can leave $SL(n,F)$.

Choose an invertible change-of-basis matrix $P$ so that $J=PT{P}^{-1}$ is the Jordan normal form of $T$. Write each block aswhere $N_{m_j}$ has ones on its first superdiagonal and all other entries zero. DefineEvery ${\lambda }_j\ne 0$ because $T$ is invertible, so $D_0$ is an invertible diagonal matrix. The matrix $N$ is strictly upper triangular and hence nilpotent. On each block $D_0$ is a scalar multiple of the identity, so it commutes with $N$ on that block and therefore on the whole space. This gives

False; in fact no matrix with this spectrum is diagonalizable over $\mathbb{R}$. If a real invertible $P$ gave $P^{-1}AP=D$ with $D$ a real diagonal matrix, all eigenvalues of $A$ would be the real diagonal entries of $D$. But two of the required eigenvalues have nonzero imaginary parts. This contradiction provesThe real block form in the initial example is consistent with this: a two-dimensional rotation block is not a real diagonalization.