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True. By Jordan normal form, the characteristic polynomial records the total size of the blocks for each eigenvalue, while the minimal polynomial records the largest block for each eigenvalue. In dimension three this determines every block partition: for algebraic multiplicity three the possibilities $3$, $2+1$, and $1+1+1$ have largest block sizes three, two, and one; multiplicities one and two are equally immediate. Thus the two matrices have the same Jordan form and are conjugate. This is the characteristic and minimal polynomials determine similarity in dimension three phenomenon.

For ordered bases $\mathcal{B}=(v_1,\dots ,v_n)$ of $V$ and $\mathcal{C}=(w_1,\dots ,w_m)$ of $W$, the matrix representation of a linear map $\gamma :V\to W$ is the matrix whose $j$th column consists of the $\mathcal{C}$-coordinates of $\gamma (v_j)$. Thus

The operators $\alpha ,\beta :V\to V$ are conjugate linear operators when there is an isomorphism $s:V\to V$ such thatIn one basis their matrices therefore satisfy $[\beta ]=[s{]}^{-1}[\alpha ][s]$, so they are similar matrices.

For invertible $\beta$, the mapis linear, and its inverse is $A\mapsto \beta A{\beta }^{-1}$; hence it is a linear isomorphism of $\mathrm{End}(V)$. If ${\beta }^{\prime }=s^{-1}\beta s$, define ${\Phi }_s(A)=s^{-1}As$. A direct substitution givesso ${\varphi }_{{\beta }^{\prime }}$ and ${\varphi }_{\beta }$ are conjugate. This is the conjugation operator on an endomorphism space.

It remains to compute the Jordan normal form over $\mathbb{C}$. By the preceding conjugacy, we may put $\beta$ in Jordan form.

Ifthen the matrix units $E_{ij}$ are eigenvectors becauseThusThis includes the scalar case $\lambda =\mu$, when ${\varphi }_{\beta }$ is the identity.

Otherwise $\beta$ has one size-two Jordan block. Multiplying $\beta$ by a nonzero scalar does not change ${\varphi }_{\beta }$, and conjugating within its Jordan class allows us to use $J=I+N$ with $N=E_{12}$. Put $D={\varphi }_J-I$. Direct multiplication givesHence $D^3=0$, $\mathrm{rank}D=2$, and $\mathrm{rank}{D}^2=1$. The nilpotent Jordan blocks of $D$ therefore have sizes three and one. Adding the identity givesThese two cases are the Jordan normal form of conjugation on two-by-two matrices.

The Jordan normal form theorem says that every complex square matrix is similar to a direct sum of Jordan blocks, uniquely up to their order. For an eigenvalue $\lambda$, if its blocks have sizes $n_1,\dots ,n_r$, then its algebraic multiplicity and geometric multiplicity areand its contribution to the minimal polynomial is $(t-\lambda {)}^{{\max }_j{n}_j}$. Thuswhere $s_{\lambda }$ is the largest $\lambda$-block size.

For the given matrix,Moreover,HenceThe eigenvalue $-1$ therefore has one Jordan block of size two, so

A Jordan block is $J_m(\alpha )=\alpha I+N$, where $N$ has ones on the superdiagonal and zeros elsewhere. Since $N^2=0$ for $m=2$, Taylor expansion givesThe Jordan normal form is the block-diagonal matrix, unique up to block order, to which $A$ is similar and whose blocks are Jordan blocks.

A block of size at least two for eigenvalue $\lambda$ contributes a generalized eigenvector in $\ker (A-\lambda I{)}^2\setminus \ker (A-\lambda I)$. Thus the displayed equality for every $\lambda$ is equivalent to every block having size one, which is diagonalizability.

The transpose of each Jordan block is similar to it by reversing its basis. Hence $A^T$ has the same Jordan form as $A$.

Write $A=PJ{P}^{-1}$. Let $R$ be block diagonal with the reversal matrix on each Jordan block. Both $R$ and $RJ$ are symmetric and $R$ is invertible. Thereforewhere $C,D$ are symmetric and $C$ is nonsingular.

Choose a shift $\mu$ which is not an eigenvalue. Inverse iteration applies power iteration to $B=(A-\mu I{)}^{-1}$, implemented by a linear system solve rather than by forming an explicit matrix inverse:A single LU decomposition of $A-\mu I$ can be reused for all the solves at this fixed shift. An eigenpair $A{u}_j={\lambda }_j{u}_j$ transforms intoThus an eigenvalue ${\lambda }_{\ast }$ uniquely closest to $\mu$ becomes the dominant eigenvalue of $B$. For a diagonalizable matrix, the argument in the preceding subpart gives convergence of diagonalizable fixed-shift inverse iteration, provided the starting coefficient in the target eigenspace is nonzero, with power iteration direction error $O(q_{\mu }^k)$, whereFor any prescribed simple eigenvalue, taking a sufficiently close but unequal shift makes the numerator small while the denominator stays close to its positive separation from the remaining eigenvalues. This makes $q_{\mu }$ small and gives very fast convergence per iteration. The Rayleigh quotient of $v_k$ estimates ${\lambda }_{\ast }$ directly. If ${\beta }_k$ estimates the associated eigenvalue of $B$, the corresponding estimate is $\mu +{\beta }_k^{-1}$.

The selection principle also applies when the other eigenvalues have nontrivial blocks in the Jordan normal form: their contributions acquire powers of $k$, multiplying strictly smaller geometric factors. A simple uniquely closest eigenvalue still attracts the direction when the Riesz projection of $v_0$ onto its generalized eigenspace is nonzero. For nonreal target eigenvalues, allow complex shifts and arithmetic; a real shift cannot distinguish a conjugate pair of equal distances. Setting $\mu ={\lambda }_{\ast }$ makes the solve singular, and an exactly zero starting target component prevents convergence to that target in exact arithmetic. When ${\lambda }_{\ast }<\mu$ is real, account for the sign behavior of fixed-shift inverse iteration.

Let $x_{\ast }=A^{-1}b$, $e_k=x_k-x_{\ast }$ and $G=M^{-1}N$. Subtract the exact fixed-point equation to obtain $e_k=G^k{e}_0$. Hence convergence for every initial vector is equivalent to $G^k\to 0$. If this holds, it holds after complexification; applying it to a complex eigenvector gives ${\lambda }^{k}v\to 0$, so $|\lambda |<1$. Conversely, the complex Jordan normal form expresses each power as a finite sum of terms $\left(\genfrac{}{}{0ex}{}{k}{r}\right){\lambda }^{k-r}$ from its blocks. Every term tends to zero when all eigenvalues have modulus below one. ThusThe convergence criterion for a stationary iteration needs complex triangularization or Jordan normal form. The paper's assertion that every real matrix is similar to a real upper-triangular matrix is false when nonreal eigenvalues occur; complex triangularization supplies the intended argument. In the Gauss-Seidel method, with $A=D+L+U$ split into diagonal, strictly lower and strictly upper parts, $\boxed{M=D+L,N=-U}$.

For $n\ge 4$, eliminating the ghost values gives the symmetric pentadiagonal matrixAll other off-diagonal entries vanish. The first two equations areand the last two areFor $n=3$ the matrix is $\left(\begin{array}{ccc}7& -4& 1\\ -4& 6& -4\\ 1& -4& 7\end{array}\right)$, for $n=2$ it is $\left(\begin{array}{cc}7& -4\\ -4& 7\end{array}\right)$, and for $n=1$ it is $(8)$ because both ghost contributions meet the same unknown. The diagonal entries are positive in every case, so the lower triangular matrix $D+L$ is invertible and the Gauss-Seidel iteration is well defined.

An eigenvalue $\lambda$ of $G=-(D+L{)}^{-1}U$ makes $H(\lambda )=\lambda (D+L)+U$ singular. Let ${\ell }_i$ and $r_i$ be the sums of absolute lower and upper entries in row $i$. The row's diagonal-dominance margin isInterior full rows have $d_i=6$, ${\ell }_i=r_i=5$, giving margin $|\lambda |-5$. Boundary rows have larger margins; for instance the first gives $7|\lambda |-5$, and the second gives at least $2|\lambda |-5$. Therefore $|\lambda |>5$ would make $H(\lambda )$ a strictly diagonally dominant matrix, hence nonsingular, a contradiction.

At $|\lambda |=5$ it is weakly diagonally dominant with strict boundary rows. Its nonzero nearest-neighbour entries make it an irreducibly diagonally dominant matrix. The nonsingularity theorem for that class again contradicts singularity. For the small matrices above all rows are already strict at this value. HenceThere is also a stronger conclusion. For $T=\mathrm{tridiag}(-1,2,-1)$,including the $n=1$ overlap. Since $T$ is positive definite, $A$ is a symmetric positive-definite matrix. The Gauss-Seidel convergence from diagonal dominance and positive definiteness theorem consequently gives the stronger bound $\rho (G)<1$.

Consider the diagonal matrix $D=2I_2$ and a Jordan blockBoth have characteristic polynomial $(t-2{)}^2$, trace $4$, determinant $4$, and eigenvalue $2$ with algebraic multiplicity two. But $\ker (D-2I)$ is the whole plane, whereas $(J-2I)(a,b{)}^T=(b,0{)}^T$ vanishes only when $b=0$. Their eigenspaces have different dimensions, so eigenvectors under similarity rules out matrix similarity. There is an even shorter proof: $S^{-1}(2I)S=2I$ for every invertible matrix $S$.

The missing information is how the linear operator acts beyond its eigenvectors. Here $D-2I=0$, while $J-2I\ne 0$ but $(J-2I{)}^2=0$. Their minimal polynomials are respectively $t-2$ and $(t-2{)}^2$. Over a field where the characteristic polynomial splits, the full sizes of the Jordan blocks, encoded by the Jordan normal form, determine the matrix similarity class.

For a finite square matrix, powers tend to zero exactly when every eigenvalue has modulus less than one. In a complex Jordan normal form, the entries of a block power are sums of factors $\left(\genfrac{}{}{0ex}{}{k}{j}\right){\lambda }^{k-j}$, which tend to zero when $|\lambda |<1$; a zero-eigenvalue block becomes zero after finitely many powers. Conversely, an eigenvector gives $G^{k}v={\lambda }^{k}v$, so convergence to zero excludes $|\lambda |\ge 1$. This criterion underlies the convergence theorem for a stationary iterative method for a linear system.