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The characteristic polynomial isIts roots give the two eigenvaluesTheir sum is zero, agreeing with the trace of the matrix.

The characteristic polynomial isFor $\lambda =2$, the eigenvector equation reduces to $x=y$; for $\lambda =8$, it reduces to $x=-y$. Choose the following unit vectors consistently for the remaining parts:Each eigenspace consists of all scalar multiples of its displayed vector, including zero; an eigenvector itself must be nonzero.

The characteristic polynomial, using the convention $\det (\lambda I-A)$, isIts zeros give the eigenvalues, including algebraic multiplicity:Their product is $45$ and their sum is $-1$, agreeing with the determinant and trace of the matrix.

Direct matrix multiplication givesSubtracting verifies $\boxed{A^2-4A-5I=0}$. This is also the Cayley-Hamilton theorem for the characteristic polynomial ${\lambda }^2-4\lambda -5$.

The characteristic polynomial is $(1-\lambda )(4-\lambda )-4=\lambda (\lambda -5)$. Solving each eigenvalue equation gives the eigenvalues and eigenvectorsFor unit-length representatives divide each displayed direction by $\sqrt{5}$. The two eigendirections are orthogonal, consistently with the matrix being a symmetric matrix.

The characteristic polynomial of the homogeneous equation is $r^2+7r+12=(r+3)(r+4)$. Thus the complementary function is $C_1{e}^{-3x}+C_2{e}^{-4x}$. The forcing resonates with the first term, so a resonant exponential particular solution is $y_p=Kx{e}^{-3x}$. Substitution gives $y_p^{\prime \prime }+7y_p^{\prime }+12y_p=K{e}^{-3x}$, hence $K=2$.

The full solution is $y=(C_1+2x)e^{-3x}+C_2{e}^{-4x}$. At zero, the two initial conditions give $C_1+C_2=1$ and $2-3C_1-4C_2=0$. Solving yields $C_1=2$, $C_2=-1$, so

The characteristic polynomial is $(1-\lambda {)}^2-4=(\lambda -3)(\lambda +1)$. For $\lambda =3$ the components are equal; for $\lambda =-1$ they are opposite. After normalization, the eigenvalues and associated eigenvectors areAn overall sign change of either normalized eigenvector is equivalent.

Yes. The matrix is a real symmetric matrix, so the spectral theorem for real symmetric matrices provides an orthonormal basis of real eigenvectors and hence a real diagonal form. Explicitly its characteristic polynomial is ${\lambda }^2-7\lambda -4$, with distinct real eigenvalues

The characteristic polynomial is ${\lambda }^2-2\lambda \cos \theta +1$, with eigenvalues $\cos \theta \pm i\sin \theta$. A real diagonal form can have only real eigenvalues, so it requires $\sin \theta =0$. Conversely, when $\theta =k\pi$, the rotation matrix is already $(-1{)}^{k}I$. Thus

The characteristic polynomial isIts roots give the eigenvalues

The matrix is symmetric under interchange of the first and third coordinates. For an antisymmetric vector $(1,0,-1)$, direct multiplication gives $A(1,0,-1{)}^{\mathsf{T}}=-6(1,0,-1{)}^{\mathsf{T}}$. For a symmetric vector $(p,q,p)$, the linear transformation reduces toThe reduced characteristic polynomial is $\lambda (\lambda -12)$. At $\lambda =0$, its equations give $q=2p$; at $\lambda =12$, they give $q=-p$. Normalizing the three eigenvectors givesTheir norms are one. Their unnormalized pairwise dot products are $1-1=0$, $1-1=0$ and $1-2+1=0$, proving mutual orthogonality. The eigenvalues sum to six and have product zero, agreeing with the matrix trace and determinant from part (a).

The characteristic polynomial factors as $(r+1)(r+2)$, so the homogeneous solution is $A{e}^{-x}+B{e}^{-2x}$. The forcing $e^{-x}$ already occurs in that solution, so a constant multiple of $e^{-x}$ cannot be a particular solution. Use $y_p=Cx{e}^{-x}$ instead. Its derivatives are $C(1-x)e^{-x}$ and $C(x-2)e^{-x}$, so substitution gives $C{e}^{-x}$ and fixes $C=1$. ThusThe extra factor of $x$ accounts for the repeated forcing exponent in the method of undetermined coefficients.

A left eigenvector satisfies $(M-\lambda I{)}^T\mathbf{x}=0$. Here $M$ is a symmetric matrix, so its left and right eigenspaces have the same components. The characteristic polynomial factors asThe eigenvalues and all corresponding nonzero row eigenvectors areFor example, in the $xy$ block, equal components give the eigenvalue $3$ and opposite components give $-1$, while the $z$ coordinate is unaffected.

For the $C$ obtained above, the characteristic polynomial isThus the eigenvalues are $5$ and $-5$. For $\mu =5$, $(C-5I)\mathbf{v}=0$ gives $v_2=-3v_1$; for $\mu =-5$, it gives $v_1=3v_2$. Normalizing the eigenvectors yieldsBoth are unit vectors, their dot product is zero, and either overall sign is allowed for each eigenvector.

The characteristic polynomial of $A$ isIts distinct eigenvalues are $3$ and $-3$. For $3$, the eigenvector relation gives $y=-2x$, so choose $(1,-2{)}^T$. For $-3$, it gives $x=-2y$, so choose $(2,-1{)}^T$. Put these independent eigenvectors into columns:Then $AB=BC$ column by column. Also $\det B=3\ne 0$, so this choice supplies the diagonalization of a matrix $B^{-1}AB=C$. Scaling columns or interchanging their order gives other valid choices, with the matching order in $C$.

The vector $(1,-1,0{)}^T$ is an eigenvector with eigenvalue $4$. The complementary subspace of vectors with first two components equal is preserved by $A$: a vector $(s,s,t{)}^T$ maps to $(4s+2t,4s+2t,4s+2t{)}^T$. There, the eigenvalue equations reduce to the $2\times 2$ matrix $\left(\begin{array}{cc}4& 2\\ 4& 2\end{array}\right)$, whose characteristic polynomial is $\lambda (\lambda -6)$. For $\lambda =6$, $t=s$; for $\lambda =0$, $t=-2s$.

Normalising by the Euclidean norm, a complete orthonormal basis of eigenvectors isSign changes of any of these eigenvectors are equally valid. The eigenvalues also check part (a): their sum is $10$ and the sum of their squares is $52$.

Let $p(t)=\det (tI-A_{\alpha })$ be the real characteristic polynomial. It is a monic cubic, so $p(t)\to +\infty$ as $t\to +\infty$. At zero,The intermediate value theorem therefore gives some $t_0>0$ with $p(t_0)=0$. A root of the characteristic polynomial is an eigenvalue, hence $A_{\alpha }$ has a positive real eigenvalue for every real $\alpha$. This uses the general fact that an odd-dimensional real matrix with positive determinant has a positive eigenvalue.

The first coordinate direction is already an eigenvector of $A$ with eigenvalue $4$. On the last two coordinates, the matrix is $\left(\begin{array}{cc}2& 4\\ 4& 2\end{array}\right)$, with characteristic polynomialThus the three eigenvalues are $4,6,-2$. For $\lambda =6$, the last two components are equal; for $\lambda =-2$, they are opposite. Normalising by the Euclidean norm gives the complete listEach sign-reversed vector is an equally valid choice. The three displayed eigenvectors are mutually orthogonal, as expected for a real symmetric matrix with distinct eigenvalues.

The characteristic equation has the repeated polynomial root $m=-2$, because both $e^{-2x}$ and $x{e}^{-2x}$ occur as independent solutions. Hence its characteristic polynomial isComparing constant terms gives

An eigenvalue makes the shifted matrix singular, so its characteristic polynomial vanishes:Thus the characteristic equation is