2

$1$. Let $F:M_2(\mathbb{R}) \to M_2(\mathbb{R})$ given by $F(X)=X^TX$.

$2$.$F:M_2(\mathbb{R}) \to S_2(\mathbb{R})$ given by $F(X)=X^TX$ where $S_2(\mathbb{R})=$ {$X \in M_2(\mathbb{R}): X^T=X$}.

Does $O$ and $I$ are regular values of 1 and 2?


My thoughts:-

after calculating the derivative I get $D_vF(X)=v^tX+X^Tv$.

taking $X=I$ it becomes $D_vF=v^t+v$.

sothe range set is the set of symmetric matrix.

For 1. no value will be a regular value as the given range is $M_2(\mathbb{R})$.

For 2.$O$ is not a regular point since $D_OF=O$.But not sure about $I$.But I think it will be a regular value but cant find proper justification.

1 Answers1

1
  1. Instead of using the space of all matrices as your target, use the space of symmetric matrices as the target.

  2. Compute kernel of the derivative and it's dimension.

  3. Conclude that the derivative is surjective using rank+nullity theorem.

  4. Conclude that the preimage of 1 is a submanifold.

  5. (Bonus) Repeat steps 1-4 for matrices of arbitrary dimension.

Moishe Kohan
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