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Here is a clipping from Milnor's Morse Theory. Since this question about linear algebra, I will present my question below so that no prior knowledge of the materials in the book required to answer this question.

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The only relevant thing here is the matrix of size $(\lambda+v)^2$ at the bottom. Here $\Big((E^{\tau}_0)_{**}(W_i,W_j)\Big)$ is a negative definite matrix of size $\lambda^2$ and for convenient, let's call this matrix $M$. The $(\lambda+v)^2$-matrix become $$ H_c= \begin{pmatrix} M & cA\\\ cA^t & -4I+c^2B \end{pmatrix} $$ where $A$ and $B$ are some fixed matrix of size $\lambda \times \nu$ and $v^2$ respectively, $M$ is a negative definite matrix of size $\lambda^2$, and $c$ is an arbitrary real number.

Now the claim is that we can pick $c$ such that $H_c$ is a negative definite matrix. I.e., there is $c$ such that for any $x \neq 0$, $x^tH_cx <0$. I've tried this by writting $x = (a \quad b)^t$ where $a$ and $b$ are column vectors of lenght $\lambda$ and $\nu$ respectively and then do the block matrix multiplication $$ \begin{pmatrix} a & b \end{pmatrix}^t H_c \begin{pmatrix} a \\ b \end{pmatrix} $$ and see if i can adjust $c$ to make the whole expression negative no matter how big or small $a$ and $b$ are. But i think it's messy and i want a clean or general way to get the result. I hope somebody could help me with this. Thank you.

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Negative definite matrices form an open subset of all symmetric matrices, since they are just the symmetric matrices whose eigenvalues are all negative and the eigenvalues vary continuously with the entries of a matrix. Now $H_0=\begin{pmatrix} M & 0 \\ 0 & -4I\end{pmatrix}$ is clearly negative definite since $M$ and $-4I$ are. So, by continuity, $H_c$ is also negative definite for all $c$ sufficiently close to $0$.

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  • $\begingroup$ Thank you for the answer. Can you explain a bit why eigenvalues vary continuously ? I'm so bad at this. I would also appreciate if you can give a nice reference to learn linear algebra related to differential geometry. $\endgroup$ – Si Kucing Jul 28 at 22:47
  • $\begingroup$ The eigenvalues are the roots of the characteristic polynomial, so you are reduced to showing that the roots of a monic polynomial vary continuously with its coefficients. For that, see math.stackexchange.com/questions/63196/…. $\endgroup$ – Eric Wofsey Jul 28 at 23:09
  • $\begingroup$ Thank you so much Eric. $\endgroup$ – Si Kucing Jul 28 at 23:15
  • $\begingroup$ Last question if you don't mind. Can we conclude that negative definite matrices is open by considering continous function $(v,A) \mapsto v^tAv$ as pointed out here ? $\endgroup$ – Si Kucing Jul 29 at 1:11
  • $\begingroup$ Not in any trivial way, since you need a condition for all $v$ at once. You can make it work by restricting only to unit vectors $v$ and using the fact that the set of unit vectors is compact. $\endgroup$ – Eric Wofsey Jul 29 at 1:13

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