Source: videos addedÂ .I would re-ask my question in the more general context of the standard. It would be useful to me if these definitions also hold for linear operations that cannot be (exactly) modelled by sums or differences of training examples.

5

Since linear functions are invariant under scale, the answer is yes.

For $n \times n$ matrices $A$ and $B$, the following holds:

$$\mbox{trace}(A+B)=\mbox{trace}(A)+\mbox{trace}(B)$$

$$\mbox{trace}(A-B)=\mbox{trace}(A)-\mbox{trace}(B)$$

$$\mbox{trace}(A \circ B)=\mbox{trace}(A)*\mbox{trace}(B)$$

In your example of $A=\begin{pmatrix}1&0\\0&0\end{pmatrix}$ and $B=\begin{pmatrix}1&1\\0&0\end{pmatrix}$, we have
$$\mbox{trace}(A \circ B)=\mbox{trace}(\begin{pmatrix}1&1\\0&0\end{pmatrix} \begin{pmatrix}1&1\\0&0\end{pmatrix})=\mbox{trace}(\begin{pmatrix}1&1\\0&0\end{pmatrix}^2)=\mbox{trace}(\begin{pmatrix}1&1\\0&0\end{pmatrix}^2)=\mbox{trace}(A^2)=\mbox{trace}(A)+\mbox{trace}(A)=2+0=2$$

In your example of $A=\begin{pmatrix}1&0\\0&0\end{pmatrix}$ and $B=\begin{pmatrix}1&1\\0&0\end{pmatrix}$, we have
\mbox{trace}(

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