Linear operators on spaces with a scalar product

When is a unitary $n \times n$ matrix $A$ written as a (multiplicative) commutator $A=X Y X^{-1} Y^{-1}$ with unitary matrices $X, Y$? The condition $\operatorname {det} A=1$ is obviously necessary. Prove that it is also sufficient. In other words, in the group $S U(n)$ every element is a commutator.

By a Jacobi matrix one understands a real matrix of the form

Prove that $\operatorname {Spec}(J)$ is always real and simple.

Does an analogue of Theorem 13 hold when one of the operators $\mathcal{A}, \mathcal{B}$ is Hermitian and the other is an isometry?

Theorem 13: two Hermitian operators $\mathcal{A}, \mathcal{B}$, or two isometries $\mathcal{A}, \mathcal{B}$, on an $n$-dimensional Hermitian space $V$ are simultaneously reducible to diagonal form in some orthonormal basis if and only if they commute.

Let $\mathcal{A}, \mathcal{B}$ be arbitrary commuting linear operators on a vector space $V$ over a field $\mathfrak {K}$. Prove that if each of the operators $\mathcal{A}, \mathcal{B}$ is diagonalizable, then they are simultaneously diagonalizable, i.e., there exists a basis in $V$ consisting of eigenvectors for both $\mathcal{A}$ and $\mathcal{B}$.

Prove that if $\mathcal{A}, \mathcal{B}$ are positive definite linear operators and $\mathcal{A B}=\mathcal{B} \mathcal{A}$, then $\mathcal{A B}$ is also a positive definite operator.

Prove that if ${ }^{t} A=-A$, then $A^{2}$ is a symmetric negative semidefinite matrix. In particular, the nonzero eigenvalues of a skew-symmetric matrix are purely imaginary.

Let $\mathcal{A}$ and $\mathcal{B}$ be Hermitian (symmetric) operators, one of which, say $\mathcal{A}$, is positive definite. Prove that then $\operatorname {Spec}(\mathcal{A B})$ is real.

Let $q(\mathbf{x})$ be a quadratic form in a Euclidean space $\mathbb {E}$ with inner product (|). At which points of the unit sphere $(\mathbf{x} \mid \mathbf{x})=1$ does the form $q$ attain a maximum or minimum? More generally: at which points of the unit sphere does the form $q$ take a stationary value, i.e., all its derivatives at these points in any direction are zero? Prove that the following statement holds.

The quadratic form $q(\mathbf{x})$ takes stationary values exactly at those points of the unit sphere which correspond to eigenvectors of the symmetric operator $\mathcal{F}$, defined by the form $q(\mathbf{x})=(\mathcal{F} \mathbf{x} \mid \mathbf{x})$.

In particular, the maximum of the form $q(\mathbf{x})$ on the unit sphere equals the largest of its canonical coefficients, and the minimum equals the smallest (extremal values of a quadratic form).

Prove the following generalization of Lemma 6.

Theorem 16. Any family of commuting linear operators on a finite-dimensional complex vector space possesses a common eigenvector.

Prove that if

is any set of commuting (pairwise permutable) matrices of order $n$, then there exists a nonsingular matrix $C$ such that the conjugate set

consists of commuting upper triangular matrices.

Let, as usual, $E$ be the identity matrix of order $n ; E_{i j}, 1 \leqslant i, j \leqslant n$, the matrix units. Verify that the family

has cardinality $\left[n^{2} / 4\right]+1$ and consists of linearly independent commuting upper triangular matrices. By $[p / q]$ one understands the integer part of the fraction $p / q$.

Theorem [I. Schur, 1905]. The maximal dimension of a commutative subalgebra of $M_{n}(\mathbb {C})$ equals $\left[n^{2} / 4\right]+1$.

In other words, one must prove that the maximal number of pairwise commuting linearly independent matrices of order $n$ over $\mathbb {C}$ is $\left[n^{2} / 4\right]+1$.

In fact $\mathbb {C}$ can be replaced by any field $\mathfrak {K}$.

Prove the following statement. Let $(V,(* \mid *))$ be a Euclidean vector space of even dimension $n=2 m$, and let $f(\mathbf{x}, \mathbf{y})$ be a nonsingular skew-symmetric form on $V$. Then there exist a decomposition of $V$ into a direct sum $V=V_{1} \oplus V_{2}$ of two $m$-dimensional subspaces and a nonsingular symmetric (with respect to (|)) linear operator $\mathcal{A}: V \rightarrow V$ such that

Here $\mathbf{x}=\mathbf{x}_{1}+\mathbf{x}_{2}, \mathbf{y}=\mathbf{y}_{1}+\mathbf{y}_{2}, \mathbf{x}_{i}, \mathbf{y}_{i} \in V_{i}, i=1,2$.