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Use the characteristic polynomial convention $p(\lambda )=\det (\lambda I-P)$, with $I$ the identity matrix. Put $u=\lambda -1/2$. ThenThe required characteristic polynomial equation for this matrix is thereforeUsing $\det (P-\lambda I)$ instead negates the cubic polynomial and leaves the eigenvalues unchanged.

Define the characteristic polynomial using $p(\lambda )=\det (\lambda I-P)$, where $I$ is the identity matrix. Expanding the determinant along the first row givesThus the characteristic equation isThe convention $\det (P-\lambda I)$ changes the sign of this cubic polynomial but gives the same roots.

The factorization of the characteristic polynomial shows that $\lambda =1$ is an eigenvalue. To find its normalized eigenvector, solve $(P-I)v=0$ together with $v_C+v_H+v_B=1$. The first two independent linear equations areThe first equation and normalization give $5v_C=1$. The second gives $v_H-v_B=v_C/2=1/10$, while $v_H+v_B=4/5$. Consequently,All entries are positive and sum to one. This probability vector is the stationary distribution.

For $1=(1,1,1{)}^T$, the entries of $P^{T}1$ are the column sums of $P$. Henceso $1$ is an eigenvalue of $P^T$, with eigenvector $1$.

The determinant is unchanged by matrix transposition, soThus $P-I$ has a nontrivial null space, givingThe same argument with $\lambda I$ shows that a matrix and its transpose have the same characteristic polynomial. It does not say that they have the same eigenvectors: here $P1=(0.9,1.2,0.9{)}^T\ne 1$.

The linear ordinary differential equation has differential operator $(D-3{)}^2$. Put $y=e^{3x}v$ and use the product rule twice:Substitution cancels the $v$ and $v^{\prime }$ terms, leaving $v^{\prime \prime }=4$. Two integrations give $v=2x^2+C_{1}x+C_2$. Hence the general solution isThe two arbitrary constants account for the order of the differential equation; the $x^2$ factor in the particular term reflects the repeated root $3$ of the characteristic polynomial.

For $A$, the triangular matrix formula gives the characteristic polynomialThe three eigenvalues are distinct, so part (d) already proves the existence of an eigenvector basis. Explicit choices, ordered by eigenvalues $1,3,2$, areTheir column matrix has determinant one, and direct multiplication gives respectively one, three and two times the indicated vector.

For $B$, expansion along the last row givesAgain the three eigenvalues are distinct. An explicit eigenvector basis, ordered by eigenvalues $4,2,3$, isIts column matrix has determinant $-2\ne 0$. In particular, $B$'s diagonal entries are not its eigenvalues: it is not a triangular matrix.

Finally $C=I_3$, so every nonzero vector is an eigenvector with eigenvalue one. Choose the standard basis $e_1,e_2,e_3$. These are linearly independent even though their eigenvalues coincide. Distinct eigenvalues are sufficient for an eigenvector basis, and are not necessary.

Since $B$ is upper triangular, its eigenvalues are its diagonal entries. Equivalently, its characteristic polynomial is $\lambda (\lambda -3/8)$. Hence

The characteristic polynomial of this constant-coefficient differential equation is $r^2-r-2=(r-2)(r+1)$. Its distinct roots give

The initial values are $p_0=1$ and $p_1=1/2$. The characteristic polynomial of the linear recurrence iswith roots $3/4$ and $-1/4$. Hence $p_n=a(3/4{)}^n+b(-1/4{)}^n$. The initial conditions give $a+b=1$, $3a-b=2$, so $a=3/4$, $b=1/4$. Therefore

The characteristic polynomial of the homogeneous linear recurrence is $(r-2{)}^2$, so its repeated-root solutions are $(A+Bn)2^n$. A constant particular solution is one. Thus $u_n=1+(A+Bn)2^n$. The initial conditions give $A=-1$ and $B=1$, henceThe factor $n$ is required by the repeated root of a polynomial; omitting it would lose one independent solution.

The characteristic polynomial is $r^2-3r+2=(r-1)(r-2)$. Therefore $u_n=A+B{2}^n$. The initial values give $A+B=1$ and $A+2B=0$, yielding $A=2$, $B=-1$. HenceThe two constants are fixed by the two initial conditions of this second-order linear recurrence.

The iteration matrix has eigenvalues $0,-1/2,1/2$: expanding its characteristic polynomial down the first column leaves the triangular two-by-two block with diagonal entries $-1/2,1/2$. The distinct eigenvalues ensure diagonalizability. Thus its spectral radius is $1/2<1$.

The previous bound gives $\Vert f_n\Vert \le 2^{-n}\Vert f_0\Vert \to 0$, so the Gauss-Seidel iteration converges for every initial guess under the stated diagonalization. Its limit is the exact solution $(1,1,-1{)}^T$. Since $e_n=S{f}_n$, a fixed matrix also sends this transformed error to zero; its operator norm can account for any amplification in the original coordinates.

The characteristic polynomial isThus the eigenvalues are 1 and 11. Corresponding eigenvectors are $(3,-1{)}^T$ and $(1,3{)}^T$. Their dot product is zero and each has Euclidean norm $\sqrt{10}$. Normalizing givesThe columns are orthonormal, so $P^{T}P=I$. The eigenvector equations give $AP=PD$ and hence $P^{T}AP=D$, proving the claimed orthogonal diagonalization of a real symmetric matrix.

The characteristic polynomial is $\det (\lambda I-A)={\lambda }^2-\lambda -1$. Its distinct eigenvalues are $\alpha =(1+\sqrt{5})/2$ and $\beta =(1-\sqrt{5})/2$. For either eigenvalue $\lambda$, the vector $(\lambda ,1{)}^{\top }$ is an eigenvector, since $A(\lambda ,1{)}^{\top }=(\lambda +1,\lambda {)}^{\top }=\lambda (\lambda ,1{)}^{\top }$.

ChooseThe determinant of $S$ is $\alpha -\beta =\sqrt{5}\ne 0$, soIts columns being eigenvectors gives $AS=SD$, and left multiplication by $S^{-1}$ proves $S^{-1}AS=D$.

The characteristic polynomial isIts two distinct real roots are the eigenvaluesTheir product is $-1$, so $\psi =-1/\phi$. Equivalently, ${\phi }^2=\phi +1$ implies $1/\phi =\phi -1$. The positive eigenvalue is the golden ratio.

The decomposition into $\mathrm{span}\{u\}$ and the zero-sum vector subspace gives characteristic polynomial $t^{n-1}(t-n)$. For $n\ge 2$, $n$ has algebraic multiplicity and geometric multiplicity one; zero has both multiplicities $n-1$. At $n=1$, only the eigenvalue 1 occurs, with both multiplicities one. Referring to a multiplicity zero for the absent root zero is merely a bookkeeping convention, not an assertion that it is an eigenvalue.

The interior grid is $x=1,2,3$, with fixed-end displacements $u_0=u_4=0$. HenceUse the monic characteristic polynomial convention $p(\lambda )=\det (\lambda I-A)$, where $I$ is the identity matrix. Expansion of the determinant along the first row givesThe eigenvalues are the roots of a polynomial $p(\lambda )=0$. With the alternative determinant convention $\det (A-\lambda I)$, the characteristic polynomial is $-p(\lambda )$ because the matrix has odd size; its roots of a polynomial are unchanged.

By polynomial division, write $f(t)=q(t){\chi }_A(t)+r(t)$ with $\mathrm{deg}r<n$. Because ${\chi }_A$ is a monic polynomial, this division works even over a commutative ring. Evaluating at $A$ makes the multiple of ${\chi }_A(A)$ vanish, leaving $f(A)=r(A)$. This explains the reduction in two-coefficient reduction of matrix powers without multiplying out every intermediate matrix.

Over a field, the minimal polynomial $m_A$ divides ${\chi }_A$ and gives the strongest such universal reduction: use the polynomial remainder modulo $m_A$ instead. For the scalar matrix $2I_2$, the characteristic polynomial is $(t-2{)}^2$, while the minimal polynomial is only $t-2$. Every matrix power is then a scalar multiple of $I_2$.

For a nonzero positive-weight eigenform at a good prime, the two oldforms $f(\tau ),f(p\tau )$ are independent. At level multiplied by $p$, the Hecke operator is the bad-prime operator, with matrix $\left(\begin{array}{cc}\lambda & 1\\ -\chi (p)p^{k-1}& 0\end{array}\right)$. Distinct roots $\alpha ,\beta$ of its characteristic polynomial give eigenforms $f-\beta f(p\tau )$ and $f-\alpha f(p\tau )$. Weight-zero constants are an exception to the independence assertion.

A principal minor is the determinant of a square submatrix selected using the same index set for rows and columns. For a three-by-three matrix, the coefficient of $\lambda$ in its monic characteristic polynomial is the sum of its three principal minors of order two.