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A positive diagonal matrix $D=\mathrm{diag}(d_1,\dots ,d_n)$, with $d_i>0$, is an invertible matrix and changes the coordinate scales through the similarity transformation $B=D^{-1}AD$. Its diagonal entries are still $a_{ii}$, while $b_{ij}=a_{ij}{d}_j/d_i$. Thus the Gershgorin discs have the same centres but different radii. Matrix similarity preserves the characteristic polynomial and eigenvalues, so the original and scaled disk unions are both valid bounds and may be intersected. Suitable scaling can compensate for badly unbalanced off-diagonal entries; it need not shrink every radius.^[ref]

For example,The original Gershgorin discs are $|z-2|\le 100$ and $|z-3|\le 0.01$. Their union is simply the huge first disk; the small second disk contains no eigenvalue. After scaling, both radii are $1$, giving $|z-2|\le 1$ or $|z-3|\le 1$. The two Gershgorin discs overlap, so their individual eigenvalue counts are not certified. The scaled matrix is a real symmetric matrix, however, so matrix similarity now proves that the original matrix has real eigenvalues in $[1,4]$ and is an invertible matrix. Directly, both matrices have characteristic polynomial ${\lambda }^2-5\lambda +5$, henceThe scaling changes the available bounds, while these eigenvalues stay fixed.

For the Dirac representation of the gamma matrices, the unitary matrixsends ${\tilde{\gamma }}^0$ to $\left(\begin{array}{cc}0& I_2\\ I_2& 0\end{array}\right)$ while commuting with every spatial gamma matrix. This is the standard Weyl representation of the gamma matrices with spatial blocks $(0,{\sigma }^i;-{\sigma }^i,0)$. The similarity transformation preserves the Clifford algebra relations. Changing block-order or chirality conventions can change the displayed transformation, so its direction must be specified.

If a $k$-dimensional invariant subspace is mapped to the span of the first $k$ coordinate vectors by a similarity transformation, the transformed matrix has block formIts characteristic polynomial factors as $\det (\lambda I-B)\det (\lambda I-D)$, so its eigenvalues are the eigenvalues of the two diagonal blocks, counted with algebraic multiplicity.

In the standard Dirac representation of the gamma matrices, $({\gamma }^0{)}^{†}={\gamma }^0$ and $({\gamma }^i{)}^{†}=-{\gamma }^i$, so ${\gamma }^{a†}={\gamma }^0{\gamma }^a{\gamma }^0$. The transpose signs in that representation are positive for $a=0,2$ and negative for $a=1,3$. The charge-conjugation matrix expresses the basis-independent intertwining relation $C^{-1}{\gamma }^{a}C=-({\gamma }^a{)}^T$. Under an arbitrary nonunitary similarity transformation, the Hermitizing matrix changes by inverse Hermitian congruence; it need not equal the transformed temporal gamma matrix.

The virtual similarity transformation $A^i\mapsto X{A}^i{X}^{-1}$ leaves every periodic MPS amplitude invariant by cyclicity of the matrix trace. The fundamental theorem of matrix product states says that, up to an overall phase, this is the only ambiguity for injective uniform tensors.

A claimed inverse matrix C for a finite square matrix A must satisfy $AC=CA=I$. A single product entry differing from the identity matrix disproves the claim. Conversely, in finite dimension one exact identity such as $AC=I$ already implies that A has full rank, so A is invertible and $C=A^{-1}$; the other product then equals I as well. Checking the product is useful before using a supplied inverse in a similarity transformation.

Multiplication by each column of $B$ givesSince $B$ is invertible, its columns are an eigenbasis and the similarity transformation givesThere is an important source variant: the alternative scanned copy supplies an erroneous inverse with an extra factor $1/6$ and a different top-right entry. Calling that supplied matrix $C$, direct multiplication gives $(CB{)}_{22}=1/6$, so it cannot be $B^{-1}$. Inverting $B$ instead givesThis is the inverse in the cleaner copy used here. Using the scanned erroneous matrix would make $CAB$ nondiagonal, so correcting the inverse is necessary before applying the stated diagonalization. The inverse-matrix verification test catches the error without assuming the claimed outcome.

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.

The determinant is multiplicative and unchanged by a similarity transformation. HenceFor $n=1$, expansion of the original determinant givesin agreement with the product of its eigenvalues.

Use completing the square in the Hermitian form $x^{\ast }Ax$, rather than unitary diagonalization of a normal matrix. The first pivot is one, soA second square completion givesTherefore one answer isThe diagonal entries of $D$ are rational, $\det T=1$, and the displayed identity for every $x$ proves $T^{\ast }DT=A$.

If $T$ were a unitary matrix, this matrix congruence would also be a similarity transformation, so the diagonal entries of $D$ would be the eigenvalues of $A$. Its characteristic polynomial isThus its eigenvalues are $3,3+\sqrt{10},3-\sqrt{10}$, two of which are irrational. A unitary factor cannot give a rational diagonal here. Alternatively, the permitted rational root theorem implies that three rational eigenvalues would all be integers, whereas the quadratic factor has no integer roots.

Use the Leibniz definition of the determinant:For the permuted-column matrix, $a_{ij}^{\sigma }=a_{i,\sigma (j)}$. Reindex the sum by $\tau =\sigma \circ \pi$. The sign of a permutation is multiplicative, so $\mathrm{sgn}(\pi )=\mathrm{sgn}(\sigma )\mathrm{sgn}(\tau )$. It follows directly thatFor the matrix transpose, the products in its determinant are ${\prod }_i{a}_{\pi (i),i}$. Set $j=\pi (i)$ to obtain ${\prod }_j{a}_{j,{\pi }^{-1}(j)}$, and use $\mathrm{sgn}({\pi }^{-1})=\mathrm{sgn}(\pi )$. This proves $\det (A^t)=\det A$. These facts also show that equal columns, or equal rows by transposition, force the determinant to be zero: exchanging the equal pair leaves the matrix unchanged but changes the determinant's sign.

Let $M_{ij}$ be the matrix obtained by deleting row $i$ and column $j$, with the empty determinant taken as one for $n=1$. Define the cofactor $C_{ij}=(-1{)}^{i+j}\det M_{ij}$ and the adjugate matrix by $\mathrm{adj}(A{)}_{ij}=C_{ji}$. Grouping the Leibniz sum according to the column chosen in row $i$ gives the cofactor expansion $\det A={\sum }_k{a}_{ik}{C}_{ik}$: deleting that row and its chosen column leaves a permutation of $n-1$ entries, with sign factor $(-1{)}^{i+k}$. ThereforeWhen $i=j$ this is $\det A$. When $i\ne j$ it is the cofactor expansion along row $j$ of the matrix formed by replacing that row with row $i$; its remaining minors are unchanged, and its two equal rows give determinant zero. The column version proves the other product in the same way. Thus the adjugate identity is

For the real matrices $C,D$, let $p(t)=\det (C+tD)$. The determinant definition makes this a real polynomial of degree of a polynomial at most $n$. Since $C+iD$ is invertible, the allowed converse implies $p(i)\ne 0$, so $p$ is not the zero polynomial. It has at most $n$ real roots of a polynomial. Choose a real $\lambda$ outside those roots; the adjugate argument proves that $C+\lambda D$ is invertible.

Finally write the given complex similarity transformation as $P=C+iD$. The relation $AP=PB$, with $A,B$ real, yields $AC=CB$ and $AD=DB$ by comparing real and imaginary parts. Choose the preceding $\lambda$ and put $Q=C+\lambda D$. Then $AQ=QB$, soThis proves real similarity from complex similarity by constructing a real intertwiner, rather than requiring $C$ or $D$ separately to be invertible.

A trace-zero Hermitian matrix has the unique formSet $x_3=h$, $x_1=\mathrm{Re}w$ and $x_2=-\mathrm{Im}w$. Then it is exactly $R(x)$; these assignments also prove uniqueness. This is the real Pauli matrix representation $R(x)=x_1{\sigma }_1+x_2{\sigma }_2+x_3{\sigma }_3$.

For a unitary matrix, $U^{-1}=U^{†}$. Thusand invariance of the matrix trace under similarity transformations gives $\mathrm{tr}(U^{-1}AU)=0$. Therefore conjugation maps the space to itself.

Finally,The determinant is invariant under similarity transformations, so $\det R(y)=\det R(x)$ implies

A Jordan normal form of a complex $n\times n$ matrix $A$ is a similarity transformation $P^{-1}AP=J$ in which $J$ is a direct sum of Jordan blocksFor each eigenvalue, the multiset of block sizes is uniquely determined by $A$; only the order of the blocks is arbitrary. The change-of-basis matrix $P$ is generally not unique.

For the given matrix, expansion of the characteristic polynomial givesThe upper-left $2\times 2$ minor of $A-I$ is $-5$, so its rank is $2$ and the eigenspace for $1$ has dimension $1$. Thus the two occurrences of the eigenvalue $1$ belong to a single size-$2$ Jordan block. More explicitly, the Jordan chain and the remaining eigenvector can be chosen asConsequently, with $P=(v_1{v}_2{v}_3)$, for which $\det P=1$, the answer is

For this question use the mostly-plus Dirac convention: the Minkowski metric is ${\eta }_{ab}=\mathrm{diag}(-1,1,1,1)$ andThis convention matches the printed plane-wave phase and final identity. It is related to the usual mostly-minus Dirac equation by reversing the metric and taking the negatives of the usual gamma matrices. In particular, the resulting Dirac action and its Dirac adjoint describe the same physical massive field. The Dirac gamma matrices are four complex $4\times 4$ matrices representing the spacetime Clifford algebra with quadratic form $-\eta$. The irreducible complex representation has dimension four. A convenient explicit choice is the negative of the standard Dirac representation of the gamma matrices:where ${\sigma }^i$ are the Pauli matrices. Their multiplication law verifies the displayed anticommutator.

In this representation the gamma matrix adjoint and transpose identities areandThe invariant way to express the latter pattern uses the charge-conjugation matrix:Individual transpose signs depend on the basis. More generally, a similarity transformation ${\gamma }^{\prime a}=M{\gamma }^a{M}^{-1}$ changes the Hermitizing matrix to $H^{\prime }=M^{-†}{\gamma }^0{M}^{-1}$ and the charge-conjugation matrix to $C^{\prime }=MC{M}^T$. Then ${\gamma }^{\prime a†}=H^{\prime }{\gamma }^{\prime a}{H}^{\prime -1}$ and $C^{\prime -1}{\gamma }^{\prime a}{C}^{\prime }=-{\gamma }^{\prime aT}$. Thus the simple formula with ${\gamma }^{\prime 0}$ itself presumes a compatible Hermitian basis, rather than an arbitrary nonunitary similarity transformation.

Applying $(i{\gamma }^a{\partial }_a-m)$ to the Dirac equation gives $(\square -m^2)\Psi =0$. Its mass shell is $p^2=-m^2$, so the frequencies are $p^0=\pm \sqrt{{\mathbf{p}}^2+m^2}$. The Dirac spinor transforms in the four-component Spinor representation of the Lorentz group. Under spatial rotations, the two upper and the two lower components each transform as a two-component spin-$1/2$ representation: the spin angular momentum matrices are $\mathrm{diag}({\sigma }^i,{\sigma }^i)/2$. At rest the positive-energy equation selects the upper two components, giving two independent spin polarizations, and the negative-frequency equation selects the lower two.

In the quantum theory a mode expansion isWith the mostly-plus Dirac convention, $e^{ip\cdot x}=e^{-iEt+i\mathbf{p}\cdot \mathbf{x}}$ is positive frequency. The negative-frequency coefficient obeys $(/p+m)v_s(p)=0$. The fermionic annihilation operators $b_s$ and $d_s$ satisfy the canonical anticommutation relations, with their respective fermionic creation operators. The $b_s^{†}$ excitations are particles; the $d_s^{†}$ excitations are antiparticles with the same positive energy, mass and spin-$1/2$ but opposite charge. After normal ordering, the Hamiltonian operator contains positive multiples of $b_s^{†}{b}_s+d_s^{†}{d}_s$. Reinterpreting the negative-frequency part as antiparticle creation supplies a spectrum bounded below rather than a physical tower of negative-energy particles.

For the printed $e^{ip\cdot x}$ wave, ${\partial }_a\Psi =i{p}_a\Psi$. Substitution givesThe spin label $s$ indexes the two states of a spin-$1/2$ particle, rather than varying the particle's total spin. For real on-shell $p$, Hermitian conjugation and ${\gamma }^{a†}={\gamma }^0{\gamma }^a{\gamma }^0$ giveThese are right and left null-vector equations for the same on-shell matrix.

To obtain the Gordon identity, take both external Dirac spinors to have the same real mass $m$. Their two equations implyDefine ${\gamma }^{ab}=[{\gamma }^a,{\gamma }^b]/2$. The Clifford algebra relation yieldsThereforeMultiplying the previous null-vector equation by $-1$ proves the required formula exactly:For $m\ne 0$ it can be solved for the vector-current matrix element, separating a momentum term from the antisymmetric Dirac spinor term. The identity before division also holds at $m=0$. The signs depend jointly on the metric, Clifford relation, Dirac mass term and plane-wave phase. In the mostly-minus convention of Questions 1 and 3, the printed phase instead gives $(/p+m)u=0$; the corresponding identity uses $(p_b-p_b^{\prime })$ in the antisymmetric term. Mixing that convention with the formula proved here would produce an apparent sign error.

A Dirac-to-Weyl gamma-matrix similarity transform with the required signs isIt is a unitary matrix, with $U^{-1}=U^{†}=2^{-1/2}\left(\begin{array}{cc}{I}_2& I_2\\ -I_2& I_2\end{array}\right)$. Multiplication givesThe matrix $U$ commutes with every ${\tilde{\gamma }}^i$: their common block structure uses the same antisymmetric $2\times 2$ block matrix, and the scalar blocks of $U$ commute with each Pauli matrix. Hence $U{\tilde{\gamma }}^i{U}^{-1}={\tilde{\gamma }}^i$, exactly the spatial matrices specified in the Weyl representation of the gamma matrices. A similarity transformation preserves every Clifford algebra relation, so no new consistency check is needed for the transformed representation.

Since $S\mathbf{w}=c{\mathbf{e}}^{(1)}$ and $A\mathbf{w}={\lambda }_1\mathbf{w}$,Thus the first column of $\hat{A}$ is $({\lambda }_1,0,\dots ,0{)}^T$, and $\hat{A}$ is a block upper triangular matrix:where $B$ is its bottom-right $(n-1)\times (n-1)$ submatrix. ThereforeA similarity transformation preserves the characteristic polynomial, sowith algebraic multiplicities included.

To construct $S$, use a Householder transformation. With $\alpha =\Vert \mathbf{w}{\Vert }_2$, choose the sign of $\alpha$ to avoid cancellation and setThen $S$ is an orthogonal matrix and maps $\mathbf{w}$ to $\alpha {\mathbf{e}}^{(1)}$ up to the chosen sign. If $\mathbf{w}$ already lies on the first coordinate axis, take a suitable diagonal sign matrix. This is the one-vector case of orthogonal coordinate reduction of a subspace.

For a nonzero vector $v\in {\mathbb{R}}^n$, a Householder reflection isIt is immediately symmetric. If $P=v{v}^T/(v^{T}v)$, then $P^2=P$, soHence $H$ is also an orthogonal matrix, with $H^{-1}=H$.

The similarity transformation can be expanded asComputing $v^{T}A$, $Av$, and the scalar $v^{T}Av$ costs $O(n^2)$ arithmetic operations, after which the remaining updates are outer products and scalar multiples, also costing $O(n^2)$. Thus $HA{H}^{-1}$ can be formed in $O(n^2)$ operations rather than by two general $O(n^3)$ matrix multiplications.

The similarity transformation gives $A=PD{P}^{-1}$. In the matrix power, each adjacent pair $P^{-1}P$ cancels to the identity matrix:This also follows by mathematical induction: the formula holds for $n=1$, and multiplication by $A=PD{P}^{-1}$ turns its $n$th instance into the $(n+1)$st.

If $h=ag{a}^{-1}$, the representation law gives $D(h)=D(a)D(g)D(a{)}^{-1}$. Trace invariance under this similarity transformation givesThus the character of a representation is constant on every conjugacy class.

The trace is invariant under a similarity transformation, by cyclicity of the trace:Hence equivalent representations have the same character of a representation.