Vector Spaces and Linear Operators: Very Difficult Problems with Solutions

Solution:
Every homogeneous polynomial of degree [tex]k[/tex] is a linear combination of the monomials
[tex]x_1^{a_1}x_2^{a_2}\ldots x_n^{a_n},\qquad a_i\in\mathbb{Z},\ a_i\ge 0,\qquad a_1+a_2+\ldots+a_n=k[/tex]
Distinct monomials are linearly independent (a polynomial is the zero polynomial only if all its coefficients are zero), so these monomials form a basis. Hence the dimension equals the number of solutions of [tex]a_1+\ldots+a_n=k[/tex] in non-negative integers.
Count them by "stars and bars": a solution corresponds to a row of [tex]k[/tex] stars and [tex]n-1[/tex] bars, where the bars cut the stars into [tex]n[/tex] groups of sizes [tex]a_1,\ldots,a_n[/tex]. Such a row has [tex]n+k-1[/tex] places and is determined by choosing the [tex]k[/tex] places of the stars, so the dimension is
[tex]\binom{n+k-1}{k}=\frac{(n+k-1)!}{k!\,(n-1)!}[/tex]
Check: for [tex]n=2,\ k=2[/tex] the basis is [tex]x_1^2,\ x_1x_2,\ x_2^2[/tex] and [tex]\binom{3}{2}=3[/tex]. Note that [tex]\binom{n+k}{k}[/tex] is the dimension of the space of all polynomials of degree at most [tex]k[/tex], not of the homogeneous ones.

Solution:
The conditions are homogeneous linear equations in the [tex]n^2[/tex] values of [tex]f[/tex], so the solutions form a subspace [tex]L[/tex]. Call the [tex](n-2)^2[/tex] points [tex](k,l)[/tex] with [tex]k,l\in\{2, \ldots, n-1\}[/tex] interior and the remaining points boundary. The number of boundary points is
[tex]n^2-(n-2)^2=4n-4[/tex]
Let [tex]R:L\to\mathbb{R}^{4n-4}[/tex] send [tex]f[/tex] to its values at the boundary points; [tex]R[/tex] is linear. We show that [tex]R[/tex] is an isomorphism.
The map [tex]R[/tex] is injective. Let [tex]f\in L[/tex] vanish at the boundary points and let [tex]M[/tex] be the largest value of [tex]f[/tex]. Suppose [tex]M\gt 0[/tex]; then it is attained at an interior point [tex](k,l)[/tex]. There [tex]f(k,l)[/tex] is the average of four values, each [tex]\le M[/tex], so all four are equal to [tex]M[/tex]; in particular [tex]f(k+1,l)=M[/tex]. Repeating this, we move to the right with the value [tex]M[/tex] until we reach the boundary point [tex](n,l)[/tex], where [tex]f=0\lt M[/tex], a contradiction. So [tex]M\le 0[/tex], i.e. [tex]f\le 0[/tex]; applying this to [tex]-f[/tex] gives [tex]f\ge 0[/tex], hence [tex]f\equiv 0[/tex].
The map [tex]R[/tex] is surjective. For prescribed boundary values, move them to the right-hand side: the conditions become a square linear system of [tex](n-2)^2[/tex] equations in the [tex](n-2)^2[/tex] interior values. For zero boundary values it has only the zero solution (by the injectivity), so its determinant is non-zero and the system is solvable for every right-hand side.
Hence [tex]R[/tex] is an isomorphism and
[tex]\dim L=4n-4[/tex]
Note that the four corners [tex](1, 1)[/tex], [tex](1, n)[/tex], [tex](n, 1)[/tex], [tex](n, n)[/tex] occur in no condition at all, but they still have to be counted (forgetting them gives [tex]4n-8[/tex]).

Solution:
On the complex vector space [tex]M_n(\mathbb{C})[/tex] of square matrices of order [tex]n[/tex] define the following function, where [tex]Y^*=\overline{Y}^{T}[/tex] is the conjugate transpose of [tex]Y[/tex]:
[tex]\langle X,Y\rangle=\operatorname{tr}(XY^*)=\sum_{i,j=1}^{n}x_{ij}\overline{y_{ij}}[/tex]
This is the standard inner product of [tex]\mathbb{C}^{n^2}[/tex] applied to the entries of the matrices, so it is an inner product: it is linear in [tex]X[/tex], [tex]\langle Y,X\rangle=\overline{\langle X,Y\rangle}[/tex], and [tex]\langle X,X\rangle=\sum_{i,j}|x_{ij}|^2\gt 0[/tex] for [tex]X\ne 0[/tex]. By the Cauchy-Schwarz inequality, which holds in every inner product space,
[tex]|\langle A,B\rangle|^2\le\langle A,A\rangle\times \langle B,B\rangle[/tex]
Since [tex]A[/tex] and [tex]B[/tex] are Hermitian, [tex]A^*=A[/tex] and [tex]B^*=B[/tex], hence
[tex]\langle A,B\rangle=\operatorname{tr}(AB^*)=\operatorname{tr}(AB),\qquad \langle A,A\rangle=\operatorname{tr}(A^2),\qquad \langle B,B\rangle=\operatorname{tr}(B^2)[/tex]
Substituting these into the Cauchy-Schwarz inequality gives the required inequality:
[tex]\left|\operatorname{tr}(AB)\right|^2\le\operatorname{tr}(A^2)\times \operatorname{tr}(B^2)[/tex]
(Here [tex]\operatorname{tr}(A^2)=\sum_{i,j}|a_{ij}|^2\ge 0[/tex] is real. Equality holds exactly when [tex]A[/tex] and [tex]B[/tex] are linearly dependent.)