#Marks for some Theorems
1. (Gershgorin Circle Theorem) Let $A$ be a comlex $n \times n$ matrix, with entries ${a_{ij}}$. For $i \in \left\{ {1, \ldots ,n} \right\}$ let ${R_i} = \sum\nolimits_{j \ne i} {\left| {{a_{ij}}} \right|} $ be the sum of the absolute values of the non-diagonal entries in the $i$-th row. Let $D\left( {{a_{ii}},{R_i}} \right)$ be the closed disc centered at ${{a_{ii}}}$ with radius ${{R_i}}$. Such a disc is called a Gershgorin disc. Then every eigenvalue of $A$ lies within at least one of the Gershgorin discs $D\left( {{a_{ii}},{R_i}} \right)$.
2. (Hadamard's inequality) If $N$ is the matrix having columns $v_i$, then \[\left| {\det \left( N \right)} \right| \le \prod\limits_{i = 1}^n {{{\left\| {{v_i}} \right\|}_2}},\]and equality is achieved if and only if the vectors are orthogonal or at least one of the columns is 0.
The proof of Hadamard's inequality is quiet interesting. (Wikipedia) The result is trivial if the matrix is sigular, so assume the columns of $N$ are linearly independent. By dividing each column by its length, it can be seen that the result is equivalent to the special case where each column has length 1, in other words if $e_i$ are unit vectors and $M$ is the matrix having the $e_i$ as columns then $\left| {\det \left( M \right)} \right| \le 1$, and equality is acheived if and only if the vectors are an orthogoanl set, that is when the matrix is unitary. The general result now follows: \[\left| {\det \left( N \right)} \right| \le \left( {\prod\limits_{i = 1}^n {{{\left\| {{v_i}} \right\|}_2}} } \right)\left| {\det \left( M \right)} \right| \le \left( {\prod\limits_{i = 1}^n {{{\left\| {{v_i}} \right\|}_2}} } \right).\] For the positive definite case, let $P = {M^ * }M$ and let the eigenvalues of $P$ be ${\lambda _1},{\lambda _2}, \ldots ,{\lambda _n}$. By assumption, each entry in the diagonal of $P$ is 1, so the trace of $P$ is $n$. Applying the inequality of arithmetic and geometric means, \[\det P \le \prod\limits_{i = 1}^n {{\lambda _i}} \le {\left( {\frac{1}{n}\sum\limits_{i = 1}^n {{\lambda _i}} } \right)^n} = {\left( {\frac{1}{n}{\rm{tr}}P} \right)^n} = {1^n} = 1,\] so \[\det \left( M \right) = \sqrt {\det P} \le 1.\] If there is equality then each of the $\lambda_i$'s must all be equal and their sum is $n$, so they must all be 1. The matrix $P$ is Hermitian, therefore diagonalizable, so it is the identity matrix - in other words the columns of $M$ are an orthonormal set and the columns of $N$ are an orthogonal set.
