Eigenvalues and Eigenvectors: Difficult Problems with Solutions

Solution:
Let [tex]E[/tex] be the identity matrix. The second row of [tex]A-\lambda E[/tex] has only one non-zero entry, [tex]4-\lambda[/tex], so expanding along it gives
[tex]\det(A-\lambda E)=(4-\lambda)\begin{vmatrix}a^2+b^2-\lambda&2ab\\2ab&a^2+b^2-\lambda\end{vmatrix}=(4-\lambda)\left[(a^2+b^2-\lambda)^2-(2ab)^2\right][/tex]
By the difference of squares,
[tex](a^2+b^2-\lambda)^2-(2ab)^2=(a^2-2ab+b^2-\lambda)(a^2+2ab+b^2-\lambda)=\left((a-b)^2-\lambda\right)\left((a+b)^2-\lambda\right)[/tex]
Hence the eigenvalues are
[tex]\lambda_1=4, \quad \lambda_2=(a-b)^2, \quad \lambda_3=(a+b)^2[/tex]
Check with the trace:
[tex]4+(a-b)^2+(a+b)^2=4+2(a^2+b^2)=\operatorname{tr}A[/tex]
The corresponding eigenvectors [tex]\left(0, 1, 0\right)^{T}[/tex], [tex]\left(1, 0, -1\right)^{T}[/tex], [tex]\left(1, 0, 1\right)^{T}[/tex] do not depend on [tex]a[/tex] and [tex]b[/tex]; for example,
[tex]A\left(1, 0, -1\right)^{T}=\left(a^2-2ab+b^2\right)\left(1, 0, -1\right)^{T}=(a-b)^2\left(1, 0, -1\right)^{T}[/tex]
As expected for a real symmetric matrix, all eigenvalues are real and the eigenvectors are mutually orthogonal.

Solution:
Let [tex]E[/tex] be the identity matrix. We expand the characteristic determinant along its third row [tex]\left(-1, 0, -2-\lambda\right)[/tex]:
[tex]\det(A-\lambda E)=\begin{vmatrix}2-\lambda&-1&2\\5&-3-\lambda&3\\-1&0&-2-\lambda\end{vmatrix}=-1\times \begin{vmatrix}-1&2\\-3-\lambda&3\end{vmatrix}+(-2-\lambda)\begin{vmatrix}2-\lambda&-1\\5&-3-\lambda\end{vmatrix}[/tex]
[tex]=-(3+2\lambda)+(-2-\lambda)(\lambda^2+\lambda-1)=-\lambda^3-3\lambda^2-3\lambda-1=-(\lambda+1)^3[/tex]
Hence [tex]\lambda=-1[/tex] is the only eigenvalue, of algebraic multiplicity [tex]3[/tex]. Its eigenvectors are the non-zero solutions of [tex](A+E)\vec{x}=\vec{0}[/tex], i.e. of the system
[tex]\begin{cases}3x_1-x_2+2x_3=0\\ 5x_1-2x_2+3x_3=0\\ -x_1-x_3=0\end{cases}[/tex]
The third equation gives [tex]x_1=-x_3[/tex]; then the first one gives [tex]-3x_3-x_2+2x_3=0[/tex], i.e. [tex]x_2=-x_3[/tex], and the second one, [tex]-5x_3+2x_3+3x_3=0[/tex], holds automatically. So the rank of [tex]A+E[/tex] is [tex]2[/tex], the solution space has dimension [tex]3-2=1[/tex], and with [tex]x_3=-c[/tex] the eigenvectors are
[tex]\vec{x}=c\left(1, 1, -1\right)^{T},\quad c\ne 0[/tex]
Check:
[tex]A\left(1, 1, -1\right)^{T}=\left(-1, -1, 1\right)^{T}=-\left(1, 1, -1\right)^{T}[/tex]
The geometric multiplicity [tex]1[/tex] is less than the algebraic multiplicity [tex]3[/tex], so there is no basis of eigenvectors and [tex]A[/tex] is not diagonalizable. Note that [tex]\left(1, 0, -1\right)^{T}[/tex] is not an eigenvector: [tex]A\left(1, 0, -1\right)^{T}=\left(0, 2, 1\right)^{T}[/tex].

Solution:
The matrix of the form (its off-diagonal entries are half the coefficients of the mixed terms):
[tex]A=\begin{bmatrix}2&2&-2\\2&5&-4\\-2&-4&5\end{bmatrix}[/tex]
Let [tex]E[/tex] be the identity matrix. In [tex]\det(A-\lambda E)[/tex] we add the third row to the second; the new second row is [tex]\left(0, 1-\lambda, 1-\lambda\right)[/tex], and we factor out [tex]1-\lambda[/tex]. Then we subtract the second column from the third and expand along the second row:
[tex]\det(A-\lambda E)=(1-\lambda)\begin{vmatrix}2-\lambda&2&-2\\0&1&1\\-2&-4&5-\lambda\end{vmatrix}=(1-\lambda)\begin{vmatrix}2-\lambda&2&-4\\0&1&0\\-2&-4&9-\lambda\end{vmatrix}[/tex]
[tex]=(1-\lambda)\left[(2-\lambda)(9-\lambda)-8\right]=(1-\lambda)(\lambda^2-11\lambda+10)=-(\lambda-1)^2(\lambda-10)[/tex]
The eigenvalues are [tex]\lambda_1=\lambda_2=1[/tex] and [tex]\lambda_3=10[/tex]. Check: their sum is [tex]12=\operatorname{tr}A[/tex] and their product is [tex]10=\det A[/tex].
For [tex]\lambda=10[/tex] the system [tex](A-10E)\vec{x}=\vec{0}[/tex] gives the eigenvector [tex]\left(1, 2, -2\right)[/tex]. For [tex]\lambda=1[/tex] the matrix [tex]A-E[/tex] has rank [tex]1[/tex], so the eigenvectors fill the plane [tex]x+2y-2z=0[/tex]; in it we choose two orthogonal vectors [tex]\left(2, -1, 0\right)[/tex] and [tex]\left(2, 4, 5\right)[/tex]. Eigenvectors of a symmetric matrix for different eigenvalues are orthogonal, so after normalisation these three vectors are the columns of an orthogonal matrix [tex]Q[/tex]:
[tex]Q=\begin{bmatrix}\frac{2}{\sqrt5}&\frac{2}{3\sqrt5}&\frac13\\-\frac{1}{\sqrt5}&\frac{4}{3\sqrt5}&\frac23\\0&\frac{5}{3\sqrt5}&-\frac23\end{bmatrix},\qquad Q^{T}AQ=\begin{bmatrix}1&0&0\\0&1&0\\0&0&10\end{bmatrix}[/tex]
Therefore the orthogonal substitution [tex]\left(x, y, z\right)^{T}=Q\left(u, v, w\right)^{T}[/tex] brings the form to the canonical form
[tex]f=u^2+v^2+10w^2[/tex]

Solution:
Let [tex]E[/tex] be the identity matrix. Over the complex numbers the characteristic polynomial of [tex]A[/tex] splits into linear factors; let [tex]\lambda_1, \lambda_2, \ldots, \lambda_n[/tex] be the eigenvalues of [tex]A[/tex], each repeated according to its multiplicity:
[tex]\det(A-\lambda E)=(\lambda_1-\lambda)(\lambda_2-\lambda)\ldots(\lambda_n-\lambda)[/tex]
At [tex]\lambda=0[/tex] this gives the following, so all eigenvalues of [tex]A[/tex] are non-zero and the numbers [tex]\frac{1}{\lambda_i}[/tex] make sense:
[tex]\lambda_1\lambda_2\ldots\lambda_n=\det A\ne 0[/tex]
Let [tex]\mu\ne 0[/tex]. Since [tex]A^{-1}A=E[/tex], we can write
[tex]A^{-1}-\mu E=A^{-1}(E-\mu A)=-\mu A^{-1}\left(A-\frac{1}{\mu}E\right)[/tex]
Taking determinants (multiplying a matrix of order [tex]n[/tex] by the number [tex]-\mu[/tex] multiplies its determinant by [tex](-\mu)^n[/tex]) and using [tex](-\mu)\left(\lambda_i-\frac{1}{\mu}\right)=1-\lambda_i\mu[/tex]:
[tex]\det(A^{-1}-\mu E)=(-\mu)^n\det(A^{-1})\prod_{i=1}^{n}\left(\lambda_i-\frac{1}{\mu}\right)=\det(A^{-1})\prod_{i=1}^{n}(1-\lambda_i\mu)[/tex]
Now [tex]1-\lambda_i\mu=\lambda_i\left(\frac{1}{\lambda_i}-\mu\right)[/tex], [tex]\lambda_1\lambda_2\ldots\lambda_n=\det A[/tex] and [tex]\det(A^{-1})\det A=\det E=1[/tex], hence
[tex]\det(A^{-1}-\mu E)=\det(A^{-1})\det A\prod_{i=1}^{n}\left(\frac{1}{\lambda_i}-\mu\right)=\prod_{i=1}^{n}\left(\frac{1}{\lambda_i}-\mu\right)[/tex]
Both sides are polynomials in [tex]\mu[/tex] of degree [tex]n[/tex] that coincide for all [tex]\mu\ne 0[/tex], i.e. for infinitely many values, so they are identically equal. Thus the characteristic roots of [tex]A^{-1}[/tex] are exactly [tex]\frac{1}{\lambda_1}, \frac{1}{\lambda_2}, \ldots, \frac{1}{\lambda_n}[/tex]. The map [tex]\lambda\mapsto\frac{1}{\lambda}[/tex] is one-to-one, so equal factors correspond to equal factors and each root [tex]\frac{1}{\lambda}[/tex] of [tex]A^{-1}[/tex] has the same multiplicity as the root [tex]\lambda[/tex] of [tex]A[/tex].
Remark: the eigenvectors are also the same. If [tex]A\vec{x}=\lambda\vec{x}[/tex], [tex]\vec{x}\ne\vec{0}[/tex], then multiplying by [tex]A^{-1}[/tex] gives [tex]\vec{x}=\lambda A^{-1}\vec{x}[/tex], i.e.
[tex]A^{-1}\vec{x}=\frac{1}{\lambda}\vec{x}[/tex]