Matrix Operation
Trace Norm
- $A := (a_{i}^{j})_{i,j} \in \mathbb{R}^{n \times n}$,
- $B := (b_{i}^{j})_{i,j} \in \mathbb{R}^{n \times n}$,
We denote $i$-th leargest eignen value by $\lambda_{i}(A)$.
Proposition 1
- $x \in \mathbb{R}^{n}$,
(1)
\[\mathrm{tr} \left( A \right) = \sum_{i=1}^{n} \lambda_{i}(A)\](2)
\[x^{\mathrm{T}} A x = A \bullet (x x^{\mathrm{T}}) = \mathrm{tr} \left( A (x x^{\mathrm{T}}) \right)\](3)
\[\begin{eqnarray} \mathrm{tr} \left( A + B \right) & = & \mathrm{tr}(A) + \mathrm{tr}(B) \nonumber \end{eqnarray}\](4)
\[\begin{eqnarray} c \in \mathbb{R}, \ \mathrm{tr} \left( c A \right) & = & \mathrm{tr}(cA) \nonumber \end{eqnarray}\]proof
(1)
TODO
(2)
\[\begin{eqnarray} x^{\mathrm{T}} A x & = & \sum_{i=1}^{n} x^{i} a_{i}^{j} x_{j} \nonumber \\ \mathrm{tr} \left( A x x^{\mathrm{T}} \right) & = & \mathrm{tr} \left( (a_{j}^{i})_{i,j} (x_{j}x^{i})_{i,j} \right) \nonumber \\ & = & \mathrm{tr} \left( ( \sum_{k=1}^{n} a_{k}^{i} x_{j} x^{k} )_{i,j} \right) \nonumber \\ & = & \sum_{l=1}^{n} \sum_{k=1}^{n} a_{k}^{l} x_{l} x^{k} \nonumber \end{eqnarray}\]$\Box$