The general principle of convergence says that a real sequence converges exactly when it is Cauchy. Convergent sequences are Cauchy by the triangle inequality. Conversely, a Cauchy sequence is bounded, so Bolzano–Weierstrass gives a convergent subsequence $a_{n_k}\to a$; the Cauchy property then forces the entire sequence to converge to $a$.

A sequence $(a_n)$ is Cauchy when, for every $\epsilon >0$, some $N$ satisfies $|a_m-a_n|<\epsilon$ whenever $m,n\ge N$.

The vectors $(1,0,1{)}^T,(0,1,0{)}^T,(1,0,-1{)}^T$ are eigenvectors of $B$ with eigenvalues $2,1,0$. Applying the exponential to these eigenspaces gives

Write $A=S\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)S^{-1}$. The power series gives $e^A=S\mathrm{diag}(e^{{\lambda }_1},\dots ,e^{{\lambda }_n})S^{-1}$. Therefore

If $Av=\lambda v$, then $p(A)v=p(\lambda )v$, proving one inclusion. Conversely, if $\mu$ is an eigenvalue of $p(A)$, factor $p(z)-\mu =c{\prod }_j(z-{\lambda }_j)$. If none of the ${\lambda }_j$ were eigenvalues of $A$, every $A-{\lambda }_{j}I$ would be invertible, making $p(A)-\mu I$ invertible, a contradiction. Thus $\mu =p(\lambda )$ for some eigenvalue $\lambda$ of $A$.

A complex matrix is diagonalisable when it is similar to a diagonal matrix, equivalently when the vector space has a basis of its eigenvectors.

For the transpose shift, $(Cu{)}_1=0$ and $(Cu{)}_i=u_{i-1}$ for $i\ge 2$. If $Cu=\lambda u$, the first equation and the subsequent recurrence force every component to vanish, both for $\lambda =0$ and for $\lambda \ne 0$. Thus $C$ has no eigenvalues, illustrating that a bounded infinite-dimensional operator need not have one.

The eigenvalue equation is $u_{i+1}=\lambda u_i$, soThis is a nonzero square-summable vector exactly when $c\ne 0$ and $|\lambda |<1$. Hence every point of the open unit disc is an eigenvalue, with the displayed one-dimensional eigenspace.

With ${\chi }_A(t)=\det (tI-A)$, direct expansion givesThus the eigenvalues are $3,1,-1,-2$. Kernel calculation gives eigenspace dimensions $1,2,1,1$, respectively.

An eigenvalue $\lambda$ is a scalar for which $\ker (A-\lambda I)$ contains a nonzero vector; that kernel is its eigenspace. The characteristic polynomial has degree $n$, so the fundamental theorem of algebra supplies at least one complex root and hence an eigenvalue. Its eigenspace dimension can be any integer from $1$ to $n$.

The line/plane decomposition givesSubstituting the values in part (a) and simplifying yields

Let $N=n{n}^T$ and $K_{ij}={\epsilon }_{ijk}{n}_k$. Since $K^2=N-I$, inversion separately on $\mathbb{R}n$ and $n^{\perp }$ givesprovided $(\alpha +\beta )({\alpha }^2+{\gamma }^2)\ne 0$.

On the line spanned by $n$, $A$ multiplies by $L=\alpha +\beta =a+b+c$. On $n^{\perp }$, it is the composition of a rotation with a dilation by $r=\sqrt{{\alpha }^2+{\gamma }^2}$. Consequently every area in that plane is multiplied by $r^2={\alpha }^2+{\gamma }^2$.

The distinguished unit vector is $n=(1,1,1)/\sqrt{3}$. Comparing diagonal, symmetric off-diagonal, and antisymmetric parts givesso $A_{ij}=\alpha {\delta }_{ij}+\beta n_i{n}_j+\gamma {\epsilon }_{ijk}{n}_k$.

Dotting $c={\sum }_j{\gamma }_j{a}_j$ with $e_i$ gives $e_i\cdot c={\gamma }_i{\lambda }_i$, proving ${\gamma }_i=(e_i\cdot c)/{\lambda }_i$. AlsoCombining these with $V=|\Delta |/6$ gives the displayed face formula, with the orientation chosen so its signed right-hand side is positive.

The $i$th face through $v$ contains the two edge directions $e_j$ with $j\ne i$, so $a_i\cdot e_j=0$ for those $j$. The one-dimensional space with these two orthogonality conditions is spanned by $f_i$, hence $a_i={\lambda }_i{f}_i$ for some nonzero real ${\lambda }_i$.

The final face contains $v_i=v+e_i$, so $c\cdot (v+e_i)+d=0$. Therefore $e_i\cdot c=-(c\cdot v+d)$.

Given $\epsilon >0$, choose a partition of $[a,d]$ whose upper-minus-lower sum is below $\epsilon$, and refine it by inserting $b,c$. The contribution from subintervals lying in $[b,c]$ is nonnegative and no larger than the total difference. Restricting the refined partition to $[b,c]$ therefore gives upper-minus-lower sum below $\epsilon$, so the criterion proves integrability there.