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Let $M(n)$ be the space of $n\times n$ real matrices. Let Tr: $M(n)\rightarrow\mathbb{R}$ where Tr($A$) is the trace of matrix $A$. Find the derivative of Tr($A$) in the direction of matrix $B$.

I'm not sure how one would take the derivative of the trace operator to begin with--would I put it in terms of the summation $\sum_{i=1}^na_{i,i}$? And to make it a directional derivative, I multiply the derivative by the matrix $B$?

Any help is appreciated!

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    If you prefer, think of $M(n)$ as some Euclidean space $\Bbb R^N$ (of course, $N = n^2$). Then, the directional derivative is $(\nabla_B \mathrm{Tr})(A) = \left.\frac{d}{dt}\right\vert_0 \mathrm{Tr} (A+tB)$. In this case we need no calculation, though: $\mathrm{Tr}$ is a linear map $M(n) \to \Bbb R$, and so we can identify it with its derivative (at each point), that is, $(\nabla_B \mathrm{Tr})(A) = \mathrm{Tr}(B)$. – 2017-01-30

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The derivative in question is given by

$ \lim_{t \to 0} \frac{Tr(A+tB)-Tr(A)}{t}$

Now we have $Tr(A+tB)-Tr(A)=tTr(B)$

Your turn !

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    This is helpful! We end up taking the limit as $t\rightarrow\infty$ of $\frac{tTr(B)}{t} = Tr(B)$, so our answer is just $Tr(B)$? – 2017-01-30
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    Yes, just $Tr(B)$. – 2017-01-31
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To, find the directional derivative of $ Trace (A) $ along the matrix $B$ , we first define a function $ f(X)=Tr(AX) $ , where $ X $ is a matrix in $ M(\mathbb{R}^{n}) $ . $$ $$ Then $ D_{B}f(X) $= $ \lim_{t \rightarrow 0} \frac{f(X+tB)-f(X)}{t}= \lim_{t \rightarrow 0}\frac{Tr(A(X+tB)-Tr(AX)}{t} $ , $$ $$or, $ D_{B}f(X)=\lim_{t \rightarrow 0 } \frac{Tr(AX)+t Tr(AB)-Tr(AX)}{t}=\lim_{ t \rightarrow 0}\frac{t Tr(AB)}{t}=Tr(AB)=Tr(B^{T}A^{T}) $ , $$ $$or, $ D_{B}f(X)=Tr(B^{T}A^{T}) $, $$ $$ , or, $D_{B}f(I)=Tr(B^{T}A^{T})$ , where I is the identity matrix in $ \mathbb{R}^{n}$ , $$ $$ , or,$ D_{B}Tr(A)=Tr(B^{T}A^{T}) $ , since $ f(I)=Tr(A I)=Tr(A)$.