Problem 1
In the Euclidean space [tex]\mathbb{R}^3[/tex] with the standard orthonormal basis [tex]e_1=\left(1, 0, 0\right),\ e_2=\left(0, 1, 0\right),\ e_3=\left(0, 0, 1\right)[/tex], the vectors [tex]\vec{a}=e_2-2e_3[/tex] and [tex]\vec{b}=e_1-e_2[/tex] are given. The linear operator [tex]f[/tex] is defined by [tex]f(\vec{x})=(\vec{a},\vec{x})\,\vec{b}-(\vec{b},\vec{x})\,\vec{a}[/tex] for every vector [tex]\vec{x}\in\mathbb{R}^3[/tex], where [tex](\vec{u},\vec{v})[/tex] denotes the scalar product of [tex]\vec{u}[/tex] and [tex]\vec{v}[/tex]. Find the matrix of the operator [tex]f^{2012}[/tex] in the basis [tex]e_1,e_2,e_3[/tex].
Solution:
In coordinates [tex]\vec{a}=\left(0, 1, -2\right)[/tex] and [tex]\vec{b}=\left(1, -1, 0\right)[/tex]. The [tex]j[/tex]-th column of the matrix [tex]F[/tex] of [tex]f[/tex] is [tex]f(e_j)=a_j\vec{b}-b_j\vec{a}[/tex], where [tex]a_j[/tex] and [tex]b_j[/tex] are the [tex]j[/tex]-th coordinates of [tex]\vec{a}[/tex] and [tex]\vec{b}[/tex]. Hence (with [tex]\vec{a}[/tex] and [tex]\vec{b}[/tex] as columns)
[tex]F=\vec{b}\,\vec{a}^{T}-\vec{a}\,\vec{b}^{T}=\begin{bmatrix}0&1&-2\\-1&0&2\\2&-2&0\end{bmatrix}[/tex]
The matrix [tex]F[/tex] is skew-symmetric ([tex]F^{T}=-F[/tex]), so the matrix [tex]S=F^2[/tex] of [tex]f^2[/tex] is symmetric: [tex]S^{T}=(F^{T})^2=F^2=S[/tex]. We compute
[tex]S=F^2=\begin{bmatrix}-5&4&2\\4&-5&2\\2&2&-8\end{bmatrix},\qquad F^{2012}=S^{1006}[/tex]
Let [tex]E[/tex] be the identity matrix. The matrix
[tex]S+9E=\begin{bmatrix}4&4&2\\4&4&2\\2&2&1\end{bmatrix}[/tex]
has rank [tex]1[/tex], so [tex]\lambda=-9[/tex] is an eigenvalue of [tex]S[/tex] with a two-dimensional eigenspace, the plane [tex]2x+2y+z=0[/tex]; as [tex]S[/tex] is symmetric, this eigenvalue has multiplicity [tex]2[/tex]. Since [tex]\operatorname{tr}S=-18[/tex], the third eigenvalue is [tex]-18-2\times (-9)=0[/tex], with eigenvector [tex]\left(2, 2, 1\right)[/tex]: indeed [tex]S\left(2, 2, 1\right)^{T}=\vec{0}[/tex] (this vector is orthogonal to [tex]\vec{a}[/tex] and [tex]\vec{b}[/tex]). An orthonormal basis of eigenvectors (as columns) is
[tex]\vec{g}_1=\frac{1}{\sqrt2}\left(1, -1, 0\right)^{T},\quad \vec{g}_2=\frac{1}{3\sqrt2}\left(1, 1, -4\right)^{T},\quad \vec{g}_3=\frac13\left(2, 2, 1\right)^{T}[/tex]
With the orthogonal matrix [tex]Q=\left[\vec{g}_1\ \vec{g}_2\ \vec{g}_3\right][/tex] we have [tex]S=Q\operatorname{diag}(-9,-9,0)\,Q^{T}[/tex], and since [tex]\vec{g}_1\vec{g}_1^{T}+\vec{g}_2\vec{g}_2^{T}+\vec{g}_3\vec{g}_3^{T}=QQ^{T}=E[/tex],
[tex]F^{2012}=S^{1006}=Q\operatorname{diag}\left((-9)^{1006},(-9)^{1006},0\right)Q^{T}=9^{1006}\left(\vec{g}_1\vec{g}_1^{T}+\vec{g}_2\vec{g}_2^{T}\right)=9^{1006}\left(E-\vec{g}_3\vec{g}_3^{T}\right)[/tex]
Here
[tex]\vec{g}_3\vec{g}_3^{T}=\frac19\begin{bmatrix}4&4&2\\4&4&2\\2&2&1\end{bmatrix},\qquad E-\vec{g}_3\vec{g}_3^{T}=\frac19\begin{bmatrix}5&-4&-2\\-4&5&-2\\-2&-2&8\end{bmatrix}[/tex]
therefore
[tex]F^{2012}=9^{1005}\begin{bmatrix}5&-4&-2\\-4&5&-2\\-2&-2&8\end{bmatrix}[/tex]
Check: [tex]S=-9P[/tex], where [tex]P=E-\vec{g}_3\vec{g}_3^{T}[/tex] is the orthogonal projection onto the plane of [tex]\vec{a}[/tex] and [tex]\vec{b}[/tex] and [tex]P^2=P[/tex]. Hence [tex]S^2=-9S[/tex], so [tex]S^k=(-9)^{k-1}S[/tex] and [tex]S^{1006}=(-9)^{1005}\times (-9P)=9^{1006}P[/tex], the same result.
Problem 2
Let [tex]\mathbb{K}[/tex] be a field, and let [tex]A[/tex] and [tex]B[/tex] be arbitrary square matrices of order [tex]n[/tex] with entries from [tex]\mathbb{K}[/tex]. Prove that the matrices [tex]AB[/tex] and [tex]BA[/tex] have the same characteristic polynomial, i.e. [tex]\det(AB-\lambda E)=\det(BA-\lambda E)[/tex], where [tex]E[/tex] is the identity matrix.
Solution:
If one of the matrices, say [tex]A[/tex], is invertible, the claim is immediate: [tex]BA=A^{-1}(AB)A[/tex] is similar to [tex]AB[/tex], and similar matrices have the same characteristic polynomial. In the general case we work with matrices whose entries are polynomials in [tex]\lambda[/tex], i.e. lie in the ring [tex]\mathbb{K}[\lambda][/tex]. This ring is commutative and has no zero divisors; the determinant of a product equals the product of the determinants, and the determinant of a block triangular matrix equals the product of the determinants of its diagonal blocks. Consider the block matrix of order [tex]2n[/tex] ([tex]0[/tex] denotes the zero matrix of order [tex]n[/tex])
[tex]M=\begin{bmatrix}\lambda E&A\\B&E\end{bmatrix}[/tex]
First, we multiply [tex]M[/tex] on the left by a block triangular matrix with determinant [tex]1[/tex]; the result is block triangular:
[tex]\begin{bmatrix}E&-A\\0&E\end{bmatrix}M=\begin{bmatrix}\lambda E-AB&0\\B&E\end{bmatrix}\ \Rightarrow\ \det M=\det(\lambda E-AB)[/tex]
Second, since [tex]\lambda[/tex] is a scalar, [tex]B(\lambda E)=\lambda B[/tex], and
[tex]\begin{bmatrix}E&0\\-B&\lambda E\end{bmatrix}M=\begin{bmatrix}\lambda E&A\\-\lambda B+\lambda B&-BA+\lambda E\end{bmatrix}=\begin{bmatrix}\lambda E&A\\0&\lambda E-BA\end{bmatrix}[/tex]
The left factor has determinant [tex]\det E\times \det(\lambda E)=\lambda^n[/tex], and the right-hand side has determinant [tex]\lambda^n\det(\lambda E-BA)[/tex]. Together with the first result:
[tex]\lambda^n\det M=\lambda^n\det(\lambda E-BA)\ \Rightarrow\ \lambda^n\det(\lambda E-AB)=\lambda^n\det(\lambda E-BA)[/tex]
This is an equality in [tex]\mathbb{K}[\lambda][/tex]. Since [tex]\lambda^n\ne 0[/tex] and [tex]\mathbb{K}[\lambda][/tex] has no zero divisors, we may cancel [tex]\lambda^n[/tex]:
[tex]\det(\lambda E-AB)=\det(\lambda E-BA)[/tex]
Multiplying by [tex](-1)^n[/tex] gives [tex]\det(AB-\lambda E)=\det(BA-\lambda E)[/tex]. This is an equality of polynomials, not only of their values, so the proof works over every field, including finite fields, where equal values at all points of [tex]\mathbb{K}[/tex] would not be enough.
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