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Let $B$ be the matrix whose $i$th column consists of the coordinates of ${\beta }_i$ in the basis $(e_j)$. Define the upper-triangular matrix $U$ and lower-triangular matrix $L$ byThe first identity in part b says $I=BU$, so $B=U^{-1}$. The second says that the matrix $C$ of $\sigma (c)$ is $C=BL=U^{-1}L$. Since $U$ has diagonal entries one, $\det U=1$, and thereforeThus $\det (tU-L)$ is the characteristic polynomial of the Coxeter element in its geometric representation.

Using the convention ${\chi }_T(x)=\det (xI-A)$, the characteristic polynomial isSince $x^2+1>0$ for every $x\in \mathbb{R}$, the only eigenvalue over the real numbers is $1$. Solving $(A-I)v=0$ givesThus the only eigenspace over the real numbers is the axis of the rotation.

Over the complex numbers, the characteristic polynomial splits into three distinct roots of a polynomial, $1,i,-i$. Direct multiplication gives the eigenvectorsThey are linearly independent: the last two have two-dimensional determinant $2i\ne 0$ in the first two coordinates, and the first has nonzero third coordinate. ThusThis is a basis of eigenvectors of ${\mathbb{C}}^3$, proving that $T^{\prime }$ has a diagonalizable matrix.

For the given matrix $A$, the characteristic polynomial isFor $\lambda =4$, the eigenvector equation gives $v_2=v_1$; for $\lambda =-1$, it gives $3v_1+2v_2=0$. Thus the eigenvalues and their corresponding eigenspaces areFor any eigenvector $v$ with eigenvalue $\lambda$, the vector-valued function $e^{\lambda x}v$ satisfies $Y^{\prime }=AY$, since its derivative is $\lambda e^{\lambda x}v=e^{\lambda x}Av$. The two displayed eigenvectors are linearly independent, so the general solution of the linear system of ordinary differential equations is

Choose an ordered basis $\mathcal{B}$ of $V$ and let $A=[T{]}_{\mathcal{B}}$. Define the characteristic polynomial byIf a different basis gives $B=P^{-1}AP$ for an invertible matrix $P$, multiplicativity of the determinant givesThus the definition is independent of the chosen basis. The alternate convention $\det (A-xI)$ differs by the constant factor $(-1{)}^{\dim V}$ and gives the same roots.

By the spectral theorem for real symmetric matrices, there is an orthogonal matrix $Q$ such thatFor clarity, the diagonalization can also be justified recursively. The characteristic polynomial has a complex root of a polynomial by the fundamental theorem of algebra. By ii its eigenvalue is real; taking a nonzero real or imaginary part of an associated eigenvector in ${\mathbb{C}}^n$ gives a real eigenvector, which can be normalized. Its orthogonal complement is invariant, because $⟨Ax,z⟩=⟨x,Az⟩=\lambda ⟨x,z⟩=0$. Restriction to this linear subspace is again symmetric, and induction on dimension gives an orthonormal basis of eigenvectors.

All ${\lambda }_i\ge 0$ by ii. The permitted invariance of determinant and matrix trace under change of basis, applied to the diagonal matrix, yields

We use the monic polynomial convention for the characteristic polynomial of an $n\times n$ matrix:It has degree $n$, since the only contribution to the coefficient of $t^n$ in the determinant is the product of the $n$ diagonal terms $t$. The alternative convention $\det (A-t{I}_n)=(-1{)}^n{\chi }_A(t)$ has the same polynomial roots.

Put $c=\cos \theta$, $s=\sin \theta$. The characteristic polynomial is $(c-\lambda {)}^2-s^2$, givingWhen $s=0$, these eigenvalues coincide and $A=cI=\pm I$.

Direct matrix multiplication givesConsequently $\boxed{A^2-6A+5I=0}$, the Cayley-Hamilton theorem for its characteristic polynomial $t^2-6t+5$. Rearranging gives $A(6I-A)=5I$, so

Call the given matrix $M$. Its middle coordinate decouples, immediately giving the eigenvalue $3$ with normalized eigenvector $(0,1,0{)}^T$. The remaining characteristic polynomial comes from the $xz$ block:For $\lambda =-4$, the first row of $(M+4I)v=0$ gives $x=\sqrt{3}z$; for $\lambda =8$, the first row of $(M-8I)v=0$ gives $z=-\sqrt{3}x$. After vector normalization, one possible complete set isEach has unit Euclidean norm. The dot products with $v_2$ vanish because the other two middle components are zero, and $v_1\cdot v_3=(-\sqrt{3}+\sqrt{3})/4=0$. This explicitly verifies orthonormality, also predicted by the spectral theorem for real symmetric matrices with distinct eigenvalues. Changing any eigenvector's sign gives an equally valid normalized choice.

The characteristic polynomial is $(5-\lambda )(2-\lambda )-4={\lambda }^2-7\lambda +6=(\lambda -1)(\lambda -6)$. For $\lambda =1$, the eigenvector equation gives $y=2x$; for $\lambda =6$, it gives $x=-2y$. The normalized eigenvectors can therefore be chosen asBoth have unit Euclidean norm, and their dot product is $(-2+2)/5=0$, explicitly verifying orthogonality. This also follows from the spectral theorem for a real symmetric matrix with distinct eigenvalues.

True. Orthogonality gives $A_{ki}{A}_{kj}={\delta }_{ij}$ and $A_{ik}{A}_{jk}={\delta }_{ij}$, so $A^T=A^{-1}$. Thus $C_{ij}=A_{ki}{B}_{k\ell }{A}_{\ell j}$ is a similar matrix to $B$. If $B_{ij}{v}_j=\lambda v_i$, set $w_i=A_{ji}{v}_j$. ThenThe transformed eigenvector is nonzero because $A$ is invertible. Applying the inverse transformation gives the reverse correspondence, and similar matrices also have the same characteristic polynomial, including algebraic multiplicities of all eigenvalues.

The characteristic polynomial is $\det (\lambda I-M)=\lambda ({\lambda }^2-2)$. Solving the three eigenvector equations givesEach eigenspace consists of the scalar multiples of its displayed nonzero vector. Their pairwise dot products vanish: $v_{+}\cdot v_{-}=1-2+1=0$ and $v_{\pm }\cdot v_0=1-1=0$. This also agrees with the orthogonal diagonalization of a real symmetric matrix.

The upper-left block is a planar rotation matrix. Its real characteristic polynomial is $(\lambda -\cos \theta {)}^2+{\sin }^2\theta$, which is strictly positive for real $\lambda$ because the angle excludes integer multiples of $\pi$. Hence any real eigenvector has $x=y=0$. The remaining equation is $5z=\lambda z$, givingThere are infinitely many nonzero solutions, spanning one independent eigenvector direction.

Differentiate $z^{\prime }=yz$ without dividing by $z$:Thus the linearization of a Riccati equation givesIts characteristic polynomial is $(\lambda -a)(\lambda -b)$, explaining both the distinct-root exponentials and the repeated-root factor $A+Bx$. Nonzero solutions reconstruct $y=z^{\prime }/z$ between their zeros; the identically zero $z$ also satisfies the linear equation.

Computing the characteristic polynomial with the convention $\det (\lambda I-H)$ givesThus the three eigenvalues areThey can also be checked directly on the respective eigenvectors $(1,0,-1{)}^T$, $(1,-1,1{)}^T$, and $(1,2,1{)}^T$.

With $\nu =0$, the characteristic polynomial factors into $(1-\lambda )[{\lambda }^2-\lambda -({\mu }^2+4)]$, up to the overall sign convention. The eigenvalues are $1$ and $(1\pm \sqrt{17+4{\mu }^2})/2$. Integer values require an odd integer $m$ satisfying $(m-2\mu )(m+2\mu )=17$. The positive integer factorization $17=1\cdot 17$ forces $m=9$ andThe resulting eigenvalues and corresponding nonzero eigenvectors areAny nonzero scalar multiple is equally valid. The factorization also shows that no other positive integer $\mu$ works.

The characteristic polynomial is $(1/2-\lambda )(-1/2-\lambda )-2={\lambda }^2-9/4$. Its two roots give the eigenvalues

The characteristic polynomial isThus the eigenvalues and corresponding eigenvectors over $\mathbb{C}$ areAll nonzero scalar multiples of these vectors are also eigenvectors. Distinct eigenvalues make them linearly independent. Put them in the columns ofThe equations $A{v}_{\pm }={\lambda }_{\pm }{v}_{\pm }$ are exactly $AB=BC$, giving the diagonalizationThere is no diagonalization over $\mathbb{R}$, since the matrix has no real eigenvalues.

Compute the characteristic polynomial using $\det (\lambda I-M)$:Its three distinct real roots are $1$ and $3\pm \sqrt{15}$. For $\lambda =1$, the first row of $(M-I)v=0$ gives $z=3y$ and the second then gives $x=y$, so take $v_1=(1,1,3{)}^T$.

For either root of ${\lambda }^2-6\lambda -6=0$, take $z=1$ and solve the row equations to get $v=(\lambda -8,5-\lambda ,1{)}^T$. Substitution makes the first two row residuals opposite multiples of the quadratic equation and the third zero. Thus a complete choice isEvery nonzero scalar multiple is another valid eigenvector in the corresponding one-dimensional eigenspace.