Eigenvalues and Eigenvectors: Problems with Solutions

Solution:
The characteristic polynomial has degree [tex]5[/tex], so [tex]A[/tex] is a matrix of order [tex]5[/tex]. Substitute [tex]x=-5[/tex] into the characteristic polynomial:
[tex]\det(-5E-A)=(-5-5)^3(-5+4)^2=(-10)^3\times 1=-1000[/tex]
Since [tex]-5E-A=-(A+5E)[/tex] and multiplying a matrix of order [tex]5[/tex] by [tex]-1[/tex] multiplies its determinant by [tex](-1)^5=-1[/tex], we get
[tex]\det(-5E-A)=-\det(A+5E)\Rightarrow \det(A+5E)=1000[/tex]
Check with the eigenvalues: the eigenvalues of [tex]A[/tex] (with multiplicities) are [tex]5, 5, 5, -4, -4[/tex], so those of [tex]A+5E[/tex] are [tex]10, 10, 10, 1, 1[/tex], and the determinant is their product:
[tex]\det(A+5E)=10^3\times 1^2=1000[/tex]

Solution:
Let [tex]E[/tex] be the identity matrix. In the characteristic determinant the third row has only one non-zero entry, so we expand along it:
[tex]\det(A-\lambda E)=\begin{vmatrix}2-\lambda&-1&1\\-1&2-\lambda&-1\\0&0&1-\lambda\end{vmatrix}=(1-\lambda)\left[(2-\lambda)^2-1\right]=(1-\lambda)^2(3-\lambda)[/tex]
The eigenvalues are [tex]\lambda=1[/tex] (double) and [tex]\lambda=3[/tex]; the largest one is [tex]\lambda=3[/tex]. Its eigenvectors are the non-zero solutions of [tex](A-3E)\vec{x}=\vec{0}[/tex], i.e. of the system
[tex]\begin{cases}-x_1-x_2+x_3=0\\ -x_1-x_2-x_3=0\\ -2x_3=0\end{cases}[/tex]
The third equation gives [tex]x_3=0[/tex], and then the first one gives [tex]x_2=-x_1[/tex]. Hence the eigenvectors are
[tex]\vec{x}=c\left(1, -1, 0\right)^{T},\quad c\ne 0[/tex]
Check:
[tex]A\left(1, -1, 0\right)^{T}=\left(3, -3, 0\right)^{T}=3\left(1, -1, 0\right)^{T}[/tex]
Remark: [tex]\left(1, -1, 1\right)^{T}[/tex] is an eigenvector of the transposed matrix [tex]A^{T}[/tex] for [tex]\lambda=3[/tex] (a typical mistake when rows and columns are mixed up), and [tex]\left(1, 1, 0\right)^{T}[/tex] belongs to the eigenvalue [tex]\lambda=1[/tex].

Solution:
The matrix of the form is the symmetric matrix whose off-diagonal entries equal half the coefficient of [tex]xy[/tex]:
[tex]A=\begin{bmatrix}9&3\\3&1\end{bmatrix}[/tex]
Its characteristic polynomial and eigenvalues ([tex]E[/tex] is the identity matrix):
[tex]\det(A-\lambda E)=(9-\lambda)(1-\lambda)-9=\lambda^2-10\lambda=\lambda(\lambda-10)\Rightarrow \lambda_1=0,\ \lambda_2=10[/tex]
Since [tex]A[/tex] is a real symmetric matrix, an orthogonal change of variables brings the form to the canonical form
[tex]Q=10u^2+0\times v^2=10u^2[/tex]
Hence [tex]Q(x,y)\ge 0[/tex] for all [tex](x,y)[/tex], but [tex]Q[/tex] vanishes at non-zero vectors, for example along the eigenvector [tex]\left(1, -3\right)[/tex] of the eigenvalue [tex]0[/tex]:
[tex]Q\left(1, -3\right)=9-18+9=0[/tex]
Therefore the form is positive semidefinite, but not positive definite. Check: [tex]Q(x,y)=(3x+y)^2[/tex]; also, by Sylvester's criterion, [tex]\Delta_1=9\gt 0[/tex] and [tex]\Delta_2=\det A=0[/tex], so the form is not definite.

Solution:
The scalar matrix [tex]\lambda E[/tex] commutes with every matrix, so
[tex]T^{-1}(\lambda E)T=\lambda\, T^{-1}T=\lambda E[/tex]
Therefore
[tex]B-\lambda E=T^{-1}AT-T^{-1}(\lambda E)T=T^{-1}(A-\lambda E)T[/tex]
The determinant of a product equals the product of the determinants (this also holds for matrices whose entries are polynomials in [tex]\lambda[/tex], since these polynomials form a commutative ring), and these determinants commute with each other as elements of that ring. Hence
[tex]\det(B-\lambda E)=\det(T^{-1})\det(A-\lambda E)\det T=\det(A-\lambda E)\,\det(T^{-1}T)=\det(A-\lambda E)\,\det E=\det(A-\lambda E)[/tex]
Thus [tex]A[/tex] and [tex]B[/tex] have the same characteristic polynomial; in particular, they have the same eigenvalues with the same algebraic multiplicities, the same trace and the same determinant.