Derivative of a function whose variable is a matrix

[lus1450]

Homework Statement

Let $f: M_{n \times n} \rightarrow M_{n \times n}$ with $f(X) = X^2$, where $M_{n \times n}$ denotes the vector space of $n \times n$ matrices. Show $f$ is differentiable and find its differential.

Homework Equations

The Attempt at a Solution

So far, I've been looking at the difference quotient in order to "guess" linear transformation $A$ that will satisfy it. We have:
$$\lim_{|h|\to 0} \frac{|f(X+h) - f(X) - Ah|}{|h|} = \lim_{|h|\to 0} \frac{|Xh + hX + h^2 - Ah|}{|h|}$$after a little simplifying.
And I was thinking for a fixed $X$, I have $A(h) = Xh + hX$, as I wanted to get rid of the $Xh + hX$ term in the quotient. However, I'm stuck now since it's $Ah$ and not just $A$. I was thinking of throwing in an $h^{-1}$ to my $A$, but that would make it non-linear. Any suggestions?

 

[Ray Vickson]

Homework Statement

Let $f: M_{n \times n} \rightarrow M_{n \times n}$ with $f(X) = X^2$, where $M_{n \times n}$ denotes the vector space of $n \times n$ matrices. Show $f$ is differentiable and find its differential.

Homework Equations

The Attempt at a Solution

So far, I've been looking at the difference quotient in order to "guess" linear transformation $A$ that will satisfy it. We have:
$$\lim_{|h|\to 0} \frac{|f(X+h) - f(X) - Ah|}{|h|} = \lim_{|h|\to 0} \frac{|Xh + hX + h^2 - Ah|}{|h|}$$after a little simplifying.
And I was thinking for a fixed $X$, I have $A(h) = Xh + hX$, as I wanted to get rid of the $Xh + hX$ term in the quotient. However, I'm stuck now since it's $Ah$ and not just $A$. I was thinking of throwing in an $h^{-1}$ to my $A$, but that would make it non-linear. Any suggestions?

The concept you need is that of the Frechet Derivative; see http://en.wikipedia.org/wiki/Fréchet_derivative or
http://www.maths.lse.ac.uk/Courses/MA409/Notes-Part2.pdf . If $h$ is a matrix, then for $f(x) = x^2$ we have
f(X+h) - f(X) = (X+h)(X+h) - X^2 = hX + Xh + h^2,
so the linear operator $D_X: h \rightarrow M_{n \times n},$ defined as $D_X(h) = Xh + hX$, is the Frechet derivative. Since $D_X$ is linear, it will always be possible to write $D_X(h)$ as $A_X h$ for some matrix $A_X$, if you really want to, but I am not sure if your question really requires that you do so.

 

[lus1450]

Ah okay, thank you! I just realized I was swapping linear transformation and matrix in my thought process, since in my linear algebra class "A" is normally a matrix while in my analysis class, "A" is a linear transformation. As such, it is saying A(h), not A*h. My confusion is cleared up =]

 

[Ray Vickson]

Ah okay, thank you! I just realized I was swapping linear transformation and matrix in my thought process, since in my linear algebra class "A" is normally a matrix while in my analysis class, "A" is a linear transformation. As such, it is saying A(h), not A*h. My confusion is cleared up =]

I take back what I said about it always being possible to write $D_X(h)$ as $A_Xh$ for some matrix $A_X$. I forgot that $h$ is a matrix, not a vector. If it were a vector, that re-write would be possible, but maybe not if it is a matrix. (Actually, $h$ is an $n^2$-dimensional vector, so if we are willing to re-write all matrices as $n^2$-vectors, there would, indeed, be an $n^2 \times n^2$ matrix giving us what we need, but that seems excessive.)