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By diagonalizability, $V$ has an eigenbasis $e_1,\dots ,e_m$ with $T{e}_j={\lambda }_j{e}_j$. Arrange the basis so that ${\lambda }_1=\cdots ={\lambda }_r=0$ and the remaining eigenvalues are nonzero. If there are no zero or no nonzero eigenvalues, the corresponding linear span below is simply the zero vector subspace.

For $v={\sum }_j{a}_j{e}_j$, the equation $Tv=0$ holds precisely when $a_j=0$ for every $j>r$. ThusEvery output is in $\mathrm{span}\{e_{r+1},\dots ,e_m\}$, and conversely $e_j=T({\lambda }_j^{-1}{e}_j)$ for each $j>r$. ThereforeThese two disjoint parts of a basis span $V$ and their linear spans intersect only at zero. This is exactly the direct sum of vector spaces assertion

Find a stationary distribution $\mathbf{\pi }=(u,v,w{)}^T$ from $P\mathbf{\pi }=\mathbf{\pi }$ and $u+v+w=1$. The first two independent linear equations areThey give $u=9w/8$ and $v=7w/4$. Normalization gives $w=8/31$, hence

To justify the limit using the assumptions, distinct eigenvalues give an eigenbasis, over the complex numbers if necessary. Choose $\mathbf{\pi }$ as the eigenvector for 1 and let ${\mathbf{v}}_2,{\mathbf{v}}_3$ correspond to the other eigenvalues ${\lambda }_2,{\lambda }_3$. Expand the initial probability distribution asSince $1^{T}P=1^T$, each ${\lambda }_j\ne 1$ satisfiesMultiplying the expansion by $1^T$ therefore gives $\alpha =1$. It follows thatbecause $|{\lambda }_2|,|{\lambda }_3|<1$. The components of the limiting probability distribution are approximately $0.2903,0.4516,0.2581$.

For the spectral error bound for a diagonalizable fixed-point iteration, the exact solution $x$ is a fixed point: $x=Gx+c$. Subtracting it from the iteration gives the error recurrence $e_n=G{e}_{n-1}$, with $e_n=x_n-x$. Multiplying by $S^{-1}$ and using $G=SD{S}^{-1}$ givesIterating proves $f_n=D^n{f}_0$, including $n=0$ because $D^0=I$. This argument does not assume $S$ is orthogonal; it simply expresses the error in an eigenbasis.

Write $\hat{A}=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$ and $v_i=P^{-1}{e}_i$, where $e_i$ are the standard basis vectors. Since $A{P}^{-1}=P^{-1}\hat{A}$,Thus the columns of $P^{-1}$ form an eigenbasis, and the requested ordered basis is $D=(P^{-1}{e}_1,\dots ,P^{-1}{e}_n)$. Its basis matrix is $P^{-1}$, giving $[t{]}_D=PA{P}^{-1}=\hat{A}$. More completely, for each distinct eigenvalue $\lambda$, its eigenspace isAll nonzero vectors in these eigenspaces are eigenvectors; when an eigenvalue repeats, they need not be individual columns of $P^{-1}$.

Let $PA{P}^{-1}=\mathrm{diag}({\alpha }_1,\dots ,{\alpha }_n)$ and $PB{P}^{-1}=\mathrm{diag}({\beta }_1,\dots ,{\beta }_n)$. ThenBy matrix similarity, every eigenvalue of $A+B$ is ${\alpha }_i+{\beta }_i$ for some $i$. These are corresponding eigenvalues on the common eigenbasis, rather than every possible pairwise sum. This is eigenvalue addition in a common eigenbasis.

Work over the field containing the stated $n$ distinct eigenvalues of $B$; in a real formulation these must be real to obtain real eigenvectors. Choose one eigenvector for each distinct eigenvalue. By independence of eigenvectors for distinct eigenvalues, these $n$ vectors are a basis. Consequently each eigenspace $E_{\lambda }(B)$ has dimension one: in an expansion in this eigenbasis, $Bw=\lambda w$ forces every coefficient with a different eigenvalue to vanish.

Now let $v\ne 0$ be any eigenvector of $B$. By part (d), $Av$ belongs to its one-dimensional eigenspace, so $Av=\mu v$ for some scalar $\mu$. This proves that $v$ is an eigenvector of $A$, including the case $Av=0$, where $\mu =0$. This is the crucial one-dimensional step in simple spectrum gives simultaneous diagonalization.

For $n=2$, takeThey are commuting operators because every matrix commutes with the identity matrix. However $A$ has only the eigenvalue zero, and $\ker A=\mathrm{span}\{(1,0{)}^T\}$ has dimension one. It has no eigenbasis, so it cannot appear in any simultaneous diagonalization. Equivalently, a diagonal matrix similar to this nonzero nilpotent matrix would have to be zero, an impossibility. For any prescribed $n\ge 2$, use $A=E_{12}$ and $B=I_n$; its zero eigenspace has dimension $n-1$. For $n=1$ no such example exists.

The initial vector decomposes especially simply in the eigenbasis:because $\phi +\psi =1$. Applying $A^n$ multiplies the two eigenvectors by ${\phi }^n$ and ${\psi }^n$. Part (b) then yieldsReading the second coordinate gives the Lucas closed formAt $n=0$ this is 2 and at $n=1$ it is 1. Each summand obeys the same recurrence relation, which independently verifies the formula and explains why the irrational parts cancel to give an integer.

The all-ones matrix is diagonalizable for every $n\ge 1$. The vector $u$ together with the $n-1$ vectors from part (ii) is an eigenbasis. Independence follows because a multiple of $u$ has coordinate sum $n$ times that multiple, whereas every other basis vector has coordinate sum zero. In this basis the diagonal matrix is $\mathrm{diag}(n,0,\dots ,0)$. At $n=1$ the one-vector basis suffices. Alternatively, the spectral theorem for real symmetric matrices applies. If dimension zero is allowed by the convention for $\mathbb{N}$, the empty $0\times 0$ matrix has no eigenvalues, an empty basis and a vacuous diagonalization.

Suppose $A\in {\mathbb{C}}^{n\times n}$ is a diagonalizable matrix, $n\ge 2$, with an eigenbasis $v_1,\dots ,v_n$ and spectral decompositionAssume the dominant eigenvalue ${\lambda }_1$ is a simple eigenvalue andThen the normalized power iteration approaches the one-dimensional eigenspace of ${\lambda }_1$. Its asymptotic direction-error bound has factor $\gamma =|{\lambda }_2/{\lambda }_1|<1$. This spectral gap is a gap in modulus: if $g=|{\lambda }_1|-|{\lambda }_2|$, then $\gamma =1-g/|{\lambda }_1|$. A small relative gap gives slow convergence.^[ref]

The short proof is the exact eigenbasis identityThe remaining sum has norm bounded by a constant times ${\gamma }^k$. This proves convergence of direction, allowing a changing sign or complex factor. A tiny $c_1$ or a poorly conditioned, nonorthogonal eigenbasis can delay visible convergence: the condition number of $V$ affects constants, while $\gamma$ controls the eventual rate. The bound need not be an equality; if the initial coefficient of the second-largest eigenvalue vanishes, the active spectrum of the power method gives a faster rate. When $\gamma =0$, all unwanted components disappear after one multiplication.^[ref]

The hypothesis is $c_1=(V^{-1}{x}_0{)}_1\ne 0$, not an ordinary nonzero inner product with the right eigenvector. Let $w_1^{\ast }$ be the first row of $V^{-1}$, a left eigenvector normalized so $w_1^{\ast }{v}_j$ is $1$ for $j=1$ and $0$ otherwise; $\ast$ denotes conjugate transpose. Then $c_1=w_1^{\ast }{x}_0$. For a Hermitian matrix, choose an orthonormal eigenbasis and the criterion becomes $v_1^{\ast }{x}_0\ne 0$: the starting vector must be nonorthogonal to the dominant eigenvector. For a general diagonalizable matrix, the nonorthogonal eigenbasis makes these two tests different, as illustrated under power iteration failure cases.

For a unit vector $u$ spanning the target eigenspace and a normalized iterate $x_k$, defineThis is the norm of the component in the orthogonal complement of $u$, equivalently the sine of the acute angle to its line. It ignores the arbitrary sign or complex unit factor of an eigenvector. Under power iteration convergence for a diagonalizable matrix, ${\epsilon }_k=O({\gamma }^k)$ in Big O notation. The orthogonal projection defines how we measure direction; it does not assume that the eigenbasis of $A$ is orthogonal. In computation, the projection form avoids subtracting nearly equal numbers inside the square root.

A tied largest modulus can prevent even direction convergence. With $D=\mathrm{diag}(2,-2)$ and $x_0=(1,1{)}^T/\sqrt{2}$,The two directions form a cycle; neither is an eigenvector. There is no dominant-modulus spectral gap. A tied modulus is not a universal failure: an initial eigenvector remains on its line, and $A=2I$ preserves every direction. Subspace iteration addresses a separated cluster by targeting a dominant invariant subspace instead of one eigenvector.

An orthogonal start can fail despite a strict spectral gap. For $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ and $x_0=(1,-1{)}^T/\sqrt{2}$, the dominant component is zero, so $A{x}_0=x_0$ and every iterate equals $x_0$. The Rayleigh quotient is exactly $1$ and the eigenpair residual is zero, although the dominant eigenvalue is $3$. Exact arithmetic never creates a missing component.^[ref]

The third panel contrasts these failures with $N=\mathrm{diag}(-3,1)$, $x_0=(1,1{)}^T/\sqrt{2}$, for whichHere the horizontal eigenvector direction converges with a strict spectral gap, while the normalized vectors alternate signs. This is the successful direction convergence described under sign and phase in power iteration.

For a nonorthogonal eigenbasis, even the word “orthogonal” must be used carefully. The diagonalizable matrix $C=\left(\begin{array}{cc}3& 2\\ 0& 1\end{array}\right)$ has right eigenvectors $v_1=(1,0{)}^T$, $v_2=(-1,1{)}^T$, and dominant left eigenvector $w_1=(1,1{)}^T$. For $x_0=(a,b{)}^T$, the dominant coefficient is $c_1=w_1^T{x}_0=a+b$. The starting vector $v_2$ is nonorthogonal to $v_1$ but stays at the smaller eigenvalue $1$; $(0,1{)}^T$ is orthogonal to $v_1$ but has $c_1=1$ and converges to the dominant direction. The left eigenvector test, not the right-eigenvector inner product, controls the outcome.

A random vector drawn from a continuous probability distribution with a probability density function on the full real or complex vector space has $c_1\ne 0$ with probability one. This avoids the exact missing-component hyperplane, but does not guarantee a large component or fast convergence. Floating-point arithmetic may seed a missing component, yet a structurally preserved zero can remain zero, so accidental perturbations are not a convergence guarantee.

Iterate $q^{(k+1)}=A{q}^{(k)}/\Vert A{q}^{(k)}\Vert$. For a self-adjoint operator with a unique dominant eigenvalue separated in modulus from the others, a starting vector that is nonorthogonal to its dominant eigenvector gives convergence of the eigendirection at the spectral-ratio rate. For a general diagonalizable matrix, the condition is a nonzero coefficient in the dominant component of its eigenbasis, detected by the corresponding left eigenvector; an ordinary dot product with the right eigenvector does not generally detect that coefficient.

The power method is power iteration; its convergence depends on the dominant spectral component of the starting vector.

For example, $B=\left(\begin{array}{cc}2& 1\\ 0& 1\end{array}\right)$ has dominant right eigenvector $(1,0{)}^T$ and a dominant left eigenvector $(1,1{)}^T$. The starting vector $q^{(0)}=(-1,1{)}^T/\sqrt{2}$ has nonzero dot product with the right eigenvector, but $B{q}^{(0)}=q^{(0)}$: it stays forever in the eigenspace for the smaller eigenvalue $1$. Its coefficient along the dominant direction is zero, as the left eigenvector detects. An orthonormal eigenbasis makes the two tests agree in the self-adjoint case.

For a fixed Hilbert space, a bounded self-adjoint Hamiltonian operator with a finite step $H_i\to H_f$ at $t=0$ does not deliver an instantaneous rotation of the quantum state. Integrating the Schrödinger equation across a short interval givesFor a normalized quantum state and $\Vert H(t)\Vert \le M$ in operator norm, the difference has norm at most $2ϵM/\hslash$, which tends to zero. The time derivative can jump even though the quantum state is continuous. Here $\hslash$ is the reduced Planck constant. An instantaneous replacement of the eigenbasis changes coordinates, not the vector being represented.

For a real switch occupying $0\le t\le \tau$, remove any real scalar contribution $a(t)I$ that produces only a global phase, and write $|\tilde{\psi }(t)⟩=e^{i{\int }_0^{t}a(s)ds/\hslash }|\psi (t)⟩$. The same integral argument gives the sufficient bounded-model estimateThus $\epsilon \ll 1$ controls the sudden approximation up to a global phase. For any normalized final energy eigenstate, its probability amplitude differs from the instantaneous-switch value by at most $\epsilon$, and its Born rule probability differs by at most $2\epsilon$. These are sufficient error bounds, not necessary criteria. For the fixed-gap two-level sudden quench, any switching path $H(t)=-\Delta \mathbf{n}(t)\cdot \mathbf{\sigma }/2$ with a unit vector $\mathbf{n}(t)$ has operator norm $\Delta /2$, giving $\epsilon \le \Delta \tau /(2\hslash )$.

This estimate is deliberately for bounded Hamiltonian operators. An unbounded energy spectrum does not supply a finite global $M$; relevant quantum states, operator domains and excited energy eigenvalues require separate control. The adiabatic theorem also requires hypotheses about the path and its spectral gaps; there is no universal switch-duration criterion covering all spectra. The cold-atom collapse and revival after a lattice quench illustrates how a single switch can be sudden for one degree of freedom and slow for another.

Choose the Pauli matrices with ${\sigma }_z=\mathrm{diag}(1,-1)$, and let $\omega >0$ be an angular frequency. The computational basis is an orthonormal basis of energy eigenstates for the qubit Hamiltonian operatorIn this convention $|1⟩$ is the lower energy eigenstate; the labels describe a basis choice, not a universal ground state/excited state ordering. Start in the Hadamard basis pure state $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$. ThenThe first factor in the second expression is a global phase. The ratio of the two probability amplitudes has relative quantum phase $\omega t$, so its sign follows directly from $E_1-E_0=-\hslash \omega$.

Multiplying this density matrix by each Pauli matrix and taking its trace gives the Bloch vector:The Bloch vector travels counterclockwise around the equator of the Bloch sphere, from $+x$ toward $+y$ when viewed from $+z$. A computational-basis measurement always has $P_Z(0)=P_Z(1)=1/2$, while the Hadamard basis and ${\sigma }_y$ eigenbasis measurements giveAt $\omega t=\pi /2$, the quantum state is $|+i⟩=(|0⟩+i|1⟩)/\sqrt{2}$ up to a global phase, giving $P_Y(+)=1$. At $\omega t=\pi$, it is $|-⟩$ up to a global phase, giving $P_X(+)=0$. Both retain equal energy measurement probabilities.

At $t=2\pi /\omega$, the time-evolution operator is $U(t)=-I$, so $|\psi (t)⟩=-|+⟩$. The physical quantum state ray, its density operator, and every measurement's probabilities have returned. The representative vector returns exactly only at $t=4\pi /\omega$, when $U(t)=I$. The minus sign is a global phase for this isolated evolution; comparing an evolved branch with an unevolved branch through a controlled unitary gate can make it a relative quantum phase, as explained in global phase of an uncontrolled quantum gate is unobservable.

Suppose a unitary matrix $U$ is symmetric under ordinary transpose, and write its entrywise real and imaginary parts as $A$ and $B$. They are real symmetric matrices. Expanding $U^{†}U=I$ gives $A^2+B^2+i(AB-BA)=I$, proving the displayed identities. Commuting real symmetric matrices admit simultaneous diagonalization: diagonalize $A$, then diagonalize the restriction of $B$ to each invariant eigenspace of $A$. A common real orthonormal eigenbasis givesIf $A$ has simple eigenvalues, every eigenvector of $A$ is already an eigenvector of $B$, because its one-dimensional eigenspace is invariant. The eigenvalues of $A$ and $B$ are paired as the real and imaginary parts of the same eigenvalue of $U$.

If a real linear operator $A$ on an $n$-dimensional vector space has $n$ distinct real eigenvalues, any real square root $B$ commutes with $A$ and preserves each one-dimensional eigenspace. There $B$ is multiplication by a real $b_i$, hence ${\lambda }_i=b_i^2\ge 0$. Conversely, if all the eigenvalues are nonnegative, choosing $b_i=\sqrt{{\lambda }_i}$ on an eigenbasis supplies a real square root. Without simple spectrum the necessity fails: a real quarter-turn squares to $-I$.

Suppose $D$ has $n$ distinct eigenvalues over a field and $C$ commutes with $D$. Each eigenspace of $D$ is one-dimensional. Since commuting maps preserve eigenspaces, $C$ acts as a scalar on each of them. The eigenvectors of $D$ give a common eigenbasis and hence simultaneous diagonalization. For real matrices, the result gives a real eigenbasis only if the eigenvalues of $D$ are real; otherwise it applies over $\mathbb{C}$.

A real nonnegative normalized filter $w$ defines $g_Z=\int w(t)e^{itH}{h}_Z{e}^{-itH}dt$. In an energy eigenbasis, it multiplies each matrix element by $\hat{w}(E_i-E_j)$. A Fourier cutoff at the spectral gap removes couplings between a unique ground state and excitations. The filtered terms still sum to $H$, because the total Hamiltonian is unchanged by its own unitary time evolution. Each term commutes with the ground projector, but need not be minimized there.

For $u=(1,\dots ,1{)}^T$, $J_n=u{u}^T$ maps $v$ to $({\sum }_i{v}_i)u$. Therefore $u$ is an eigenvector with eigenvalue $n$, while the zero-sum vector subspace is its kernel. For $n\ge 2$ a basis of the kernel is $e_1-e_n,\dots ,e_{n-1}-e_n$. These vectors and $u$ form an eigenbasis, proving diagonalization and characteristic polynomial $t^{n-1}(t-n)$. Both algebraic multiplicity and geometric multiplicity are one at $n$ and $n-1$ at zero. At $n=1$ the sole eigenvalue is 1; the zero-sum space has dimension zero and contributes no additional root.