# Show derivative of matrix map is surjective

Suppose I have the map h:Mn(R)→Sn(R)h\colon M_n(\mathbb{R})\to S_n(\mathbb{R}), X↦XXTX\mapsto XX^T, how would I compute its derivative to show that it’s surjective? (Here, Sn(R)S_n(\mathbb{R}) are the symmetric matrices.)

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You’re right, sorry i changed it
– eeser boy
Oct 20 at 20:51

1

I noticed you have 8 questions, a lot of them answered, but none of them accepted. Please take the time to accept answers which were useful to you.
– Aloizio Macedo
Oct 20 at 21:27

3.1 on math.toronto.edu/mgualt/courses/MAT1300F-2016/docs/…
– Unit
2 days ago

@Unit that doesn’t have any solutions, why do you give me that website?
– eeser boy
2 days ago

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1

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h(X+H)=(X+H)(X+H)T=(X+H)(XT+HT)=XXT+XHT+HXT+HHT.h(X+H)=(X+H)(X+H)^T=(X+H)(X^T+H^T)=XX^T+XH^T+HX^T+HH^T.

It follows that the derivative is given by h′X=X(⋅)T+(⋅)XTh’_X=X(\cdot)^T+(\cdot) X^T.

It is not surjective is general (for instance, h′0=0h’_0=0). Don’t you want to restrict to some particular subset to compute the derivative?

If you want to show II is a regular value, let XX be in h−1(I)h^{-1}(I). Now, let AA be a symmetric matrix. Verify that h′X(12AX)=A.h’_X(\frac{1}{2}AX)=A. This shows surjectivity.

yes I want to show 1 is a regular value
– eeser boy
Oct 20 at 20:57