Here I put in order some common properties of matrices to help remembering.

1. If ${M = A^\top A}$ is a real symmetric matrix, ${M}$ has non-negative eigenvalues and orthogonal eigenvectors, which can be demonstrated as follows.Let ${\lambda_i}$ be the ${i^{th}}$ eigenvalue and ${v_i}$ be the corresponding eigenvector.

$\displaystyle \begin{array}{rcl} &M v_i = \lambda_i v_i \\ \Rightarrow &A^\top A v_i = \lambda_i v_i \\ \Rightarrow &v_i^\top A^\top A v_i = \lambda_i v_i^\top v_i \\ \Rightarrow &\| A v_i \|^2 = \lambda_i \| v_i \|^2 \\ \Rightarrow &\lambda_i = \dfrac{\| A v_i \|^2}{\| v_i \|^2} \geq 0 \end{array}$

$\displaystyle \begin{array}{rcl} & M v_i = \lambda_i v_i \\ \Rightarrow & v_j^\top M v_i = \lambda_i v_j^\top v_i \\ \Rightarrow & v_j^\top M^\top v_i = \lambda_i v_j^\top v_i \\ \Rightarrow & (M v_j)^\top v_i = \lambda_i v_j^\top v_i \\ \Rightarrow & \lambda_j v_j^\top v_i = \lambda_i v_j^\top v_i \\ \Rightarrow & (\lambda_j - \lambda_i) v_j^\top v_i = 0 \\ \Rightarrow & v_j^\top v_i = 0 \end{array}$

2. The trace ${tr(A)}$ of a matrix ${A}$ is the sum of eigenvalues ${\lambda_i}$, i.e.,

$\displaystyle tr(A) = \sum_i{\lambda_i}$

and it is invariant w.r.t a change of basis (the trace is only defined for a square matrix). From another point of view, let ${\lambda_1, \lambda_2, \ldots, \lambda_k}$ be the distinct eigenvalues of ${A}$, and ${\mu_A(\lambda_i)}$ be the corresponding algebraic multiplicity of ${\lambda_i}$ (the algebraic multiplicity of ${\lambda_i}$ is the multiplicity or degree of ${\lambda_i}$ in the characteristic polynomial of ${A}$). We have

$\displaystyle tr(A) = \sum_i^k{\mu_A(\lambda_i)\lambda_i}.$

3. Corresponding to the above property, the determinant ${det(A)}$ of a matrix ${A}$ is the product of eigenvalues ${\lambda_i}$, i.e.,

$\displaystyle del(A) = \prod_i {\lambda_i}.$

4. Geometrically, if we interpret a matrix ${A}$ as an affine transformation, the determinant ${det(A)}$ is the change in volume. The trace can be interpreted as the infinitesimal change in volume, as the derivative of the determinant.
5. The eigenvalues of the ${k^{th}}$ power of ${A}$, i.e., the eigenvalues of ${A^k}$, for any positive integer ${k}$, are ${\lambda_i^k}$. Therefore, we have

$\displaystyle tr(A^k) = \sum_i{\lambda_i^k}.$

Similarly, if ${A}$ is invertible, then the eigenvalues of ${A^{-1}}$ are ${\dfrac{1}{\lambda_i}}$, and thus

$\displaystyle tr(A^{-1}) = \sum_i{\lambda_i^{-1}}.$

6. To be continued…