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The characteristic polynomial isFor $\lambda =2$, the eigenspace is spanned only by $(1,0,0{)}^T$, so its geometric multiplicity is one although its algebraic multiplicity is two. Hence $M$ is not diagonalisable.

If $B=S^{-1}AS$, thenso the characteristic polynomials, eigenvalues, and algebraic multiplicities agree. The converse is false: $I_2$ and $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ have the same characteristic polynomial, but the first is diagonalisable and the second is not, so they cannot be similar.

The Cayley-Hamilton theorem states that a square matrix satisfies its characteristic polynomial. If a $2\times 2$ matrix is diagonalisable, $A=S\mathrm{diag}({\lambda }_1,{\lambda }_2)S^{-1}$; applying $p(t)=(t-{\lambda }_1)(t-{\lambda }_2)$ to the diagonal matrix gives zero and hence $p(A)=0$.

If $B^k=0$, every eigenvalue $\lambda$ satisfies ${\lambda }^k=0$, so all eigenvalues are zero and the characteristic polynomial is $t^n$. Cayley--Hamilton then gives $B^n=0$.

The characteristic polynomial of the homogeneous equation isso $y_h=C_0+(C_1+C_{2}x)e^{2x}$. For a particular integral put $y=e^{2x}v$. ThenTaking $v^{\prime \prime }=x/2-1/4$ gives $v=x^3/12-x^2/8$ after discarding homogeneous terms. Therefore

The recurrence shows that $T_n$ is a polynomial with integer coefficients. Part (a) givesThus $\cos (\pi /n)$ is a root of a nonzero integer polynomial and is algebraic.

Put $z=2\cos (\pi /n)$. The recurrencedefines monic integer polynomials and gives $S_k(z)=2\cos (k\pi /n)$. Hence $z$ is an algebraic integer. If $\cos (\pi /n)$ is rational, then the rational algebraic integer $z$ is an integer. Since $-2\le z<2$, this leaves the values corresponding toThis is the rational cosine of an integral submultiple of pi result. For every $n\ge 4$, the value lies strictly between $1/2$ and $1$, so it is irrational.

On $|z|\ge \rho$, put $w=1/z$, so $|w|\le 1/\rho$. The Taylor polynomialsconverge uniformly there to $e^w$. MoreoverFor all sufficiently large $n$, the uniform error is below half this lower bound, so $|f_n(z)|>0$ whenever $|z|\ge \rho$. Every zero therefore lies in $|z|<\rho$.

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.

False. At eigenvalue $2$, algebraic multiplicity two and minimal-polynomial exponent two force one block $J_2(2)$. At eigenvalue $1$, the blocks have total size five and largest size two, so there are only two possible partitions,Hence there are exactly two conjugacy classes satisfying the given polynomial data, represented byandThere cannot be three mutually non-conjugate examples.

True. Suppose $A,B$ are real and $AP=PB$ for an invertible complex matrix $P=X+iY$, with $X,Y$ real. Taking real and imaginary parts givesThe polynomial $q(t)=\det (X+tY)$ is not identically zero, since $q(i)=\det P\ne 0$. Choose a real $t$ outside its finite zero set. Then $Q=X+tY$ is real and invertible, and $AQ=QB$. Thus $A$ and $B$ are conjugate over $\mathbb{R}$, as stated by complex similarity of real matrices implies real similarity.

Integrating the ordinary differential equation four times gives a quartic polynomial. It is useful to use the linearity of the equation and split the solution into a gravity part and a force part:wheresatisfies the clamped and torque-free boundary conditions with $F=0$, whilesatisfies the homogeneous equation and contributes the endpoint force. Direct differentiation verifiesThusthe clamped-free beam under uniform load and endpoint force.

Substitute $y=y_0+y_F$ into the energy functional and integrate the resulting polynomial:The term linear in $F$ cancels. HenceAs required, $E_0$ is independent of the force and $\mathcal{E}$ is a quadratic function of it.

For an $n\times n$ matrix $A$, the characteristic polynomial isThe Cayley-Hamilton theorem states that ${\chi }_A(A)=0$.

Over $\mathbb{C}$, choose a basis in which $A$ is upper triangular, with diagonal entries ${\lambda }_1,\dots ,{\lambda }_n$. For the standard invariant flag $V_j=⟨e_1,\dots ,e_j⟩$,The factors $A-{\lambda }_{j}I$ commute, so applying their product in descending order sends $V_n$ successively into $V_{n-1},\dots ,V_0=0$. Hencewhich proves the theorem.

Direct expansion gives the commutator product rule:

Put $C=[B,A]$. Since $C$ commutes with $A$, repeated use of the product rule givesBy linearity, for every polynomial $\phi$,

Let $D(X)=[B,X]$ and suppose $f(A)=0$. For $k=1$,Assume inductively thatBoth $u$ and $C$ are polynomials in, or commute with, $A$, so $u{C}^m=C^{m}u=0$. Apply the derivation $D$ to $u{C}^m=0$ and multiply on the left by $C^m$:Since $D(u)=f^{(k+1)}(A)C$, this saysThe induction is complete. Taking $f={\chi }_A$ and $k=n$ givesThus $[B,A]$ is nilpotent, which is the Jacobson lemma for a commuting commutator.

Suppose $f$ is prime in $F[X]$ and divides $gh$ in $R[X]$. It divides one factor in $F[X]$, say $g=fq$ for $q\in F[X]$. Write $q=a{q}_0$ with $q_0\in R[X]$ primitive and $a\in F$. The product $f{q}_0$ is primitive. Sincehas all coefficients in $R$, a denominator of $a$ would divide every coefficient of the primitive polynomial $f{q}_0$; it must therefore be a unit. Thus $a\in R$, so $q\in R[X]$ and $f$ divides $g$ in $R[X]$. Hence $f$ is prime in $R[X]$.