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Question:

Let $A_{n\times n}$ be the vector space of all real $n\times n$ matrices. If I define a map $$g:A_{n\times n}\rightarrow A_{n\times n}$$ such that: $$g\left ( X \right )=X^{2}$$

In this case, what is the derivative of $g$?

I am thinking whether the formula: $g'\left ( X \right )=2XX{'}$ works. But I have no idea how the derivative of a matrix is defined. please help?

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@Arturo Magidin: I meant $X^{'}$. –  M.Krov Mar 2 '12 at 5:48
    
Is g really linear? –  NKS Mar 2 '12 at 5:49
    
@NKS: Of course not; lapsus on my part. –  Arturo Magidin Mar 2 '12 at 5:50
    
@m_p2009: How are you even defining the derivative of $g$ in the first place? Before you can try to come up with a formula, what is the definition? –  Arturo Magidin Mar 2 '12 at 5:51
    
Are you taking $X$ to be a matrix-valued function $(a,b)\to A_{n\times n}$ so that we have $$\frac{d}{dt}g(X(t))=2X(t)X'(t)\;\; ?$$ –  anon Mar 2 '12 at 6:45

2 Answers 2

up vote 2 down vote accepted

If we consider $g$ as a function from $\mathbb{R}^{n^2}\to \mathbb{R}^{n^2}$, then its derivative makes sense and the procedure to find the derivative goes in the usual way.

So, find $g(X+H)-g(X)$ and recognize the linear term and the remainder term which when divided by $\|H\|$ goes to $\bf{0}$ as $H\to \bf{0}$.

You would see that for any matrix $X\in A_{n\times n}$, we have $g'(X)H=XH+HX$.

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Again, I can only write here, since I don't have enough points.

As Ashok said, you may consider your product as a product from $\mathbb R^{n^2}$ to itself, or your matrices maybe can be made to be elements of a Frechet space, whereby we can define the Gateaux or Frechet derivatives as the linear maps that best describe the change of the function, i.e., the derivative would then be a linear map L , so that:

$$\lim_{||h||->0}\frac{||f(x+h)-f(x)-L(x,h)||_2}{||h||_1}=0$$

is the derivative.

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