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…