Path connectedness and convexity of some sets of matrices I have no idea of proving the following statements:


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*$GL_n(\mathbb C) = \{ M \in M_n(\mathbb C): \det(M) \neq 0 \} $ is open and path connected in $M_n(\mathbb C)$. Is it convex?

*$SL_n(\mathbb R) = \{ M \in GL_n(\mathbb R): \det (M) = 1 \}$ is closed and path connected in $M_n(\mathbb R)$ (Hint: using the fact that the polynominal function: $z \to \det((1-z)A + zB)$ has a finite number of roots). Is it convex?

*The set of diagonalisable matrices of $M_n(\mathbb R)$ is path connected. Is it open or closed? Is it convex?
What is the general idea when dealing with the path connectedness of these sets of matrices?
 A: A Givens rotation $G_{ij}(\theta)$ (see here) satisfies $\det G_{ij}(\theta) = 1$ and is path connected to $I$ by the map
$t \mapsto  G_{ij}(t\theta)$. Givens rotations can be used to
introduce zeros into a matrix to get the QR decomposition (see
here). Note that the rotations are real.
In particular, given a square matrix $A$, there are Givens rotations
$\Gamma_k$ such that $\Gamma_1 ... \Gamma_m A = R$, where $R$ is
upper triangular. It follows that $\det A = \det R = \prod_k [R]_{kk}$. If $A$ is invertible, we see that $[R]_{kk} \neq 0$ for all
$k$.
Since each rotation $\Gamma_k$ has a path $\gamma_k$ such that
$\gamma_k(0) = I$ and $\gamma_k(1) = \Gamma_k$, we see that there
is a path $A \to  \Gamma_m A \to \cdots \to \Gamma_1 ... \Gamma_m A = R$.
If we write $R = D+U$, where $U$ is strictly upper triangular,
we see that the path $t \mapsto R -tU$ connects $R$ to a diagonal
matrix and $\det R = \det D$.


*

*Since $\mathbb{C}\setminus \{0\}$ is path connected, for  every $z \neq 0$ there is a path $\xi_z$ in $\mathbb{C}\setminus \{0\}$ such that $\xi_z(0) = z, \xi(1) = 1$, hence the path
$t\mapsto \operatorname{diag}(\xi_{[D_{11}]}(t),...,\xi_{[D_{nn}]}(t))$ connects the diagonal matrix $D$ to $I$. It follows from the above that any $A \in GL_n(\mathbb C)$ is path
connected to $I$ and hence $GL_n(\mathbb C)$ is path connected.
Since $M \mapsto  \det M$ is continuous, we see that $GL_n(\mathbb C)=\det^{-1} ( \{0\}^c)$ is open. Since $I, -I \in GL_n(\mathbb C)$ and $0 \notin GL_n(\mathbb C)$, we see that it is not convex.

*We see that $SL_1(\mathbb R) = \{1\}$ which is trivially path
connected, so suppose $n \ge 2$. Let $\delta_k(t) = (1-t+t {1 \over |D_{kk}|}) D_{kk}$ for $k=1,...,n-1$ and note that the map $t \mapsto \operatorname{diag}(\delta_1(t),...,\delta_{n-1}(t), { 1 \over \delta_1(t) \cdots \delta_{n-1}(t)})$ connects
$D$ to the matrix $\operatorname{diag}(...,\operatorname{sgn} D_{kk},...)$ which has $\pm 1$s on the diagonal. If $A \in SL_n(\mathbb R)$ then $\det D = \det A =1$, hence an
even number of the $D_{kk}$ are negative. By pairing off the
negative $D_{kk}$ and using an appropriate Givens rotation (with $\theta = \pi$) we see that there is a path from $D$ to $I$.
Since $M \mapsto  \det M$ is continuous, we see that $SL_n(\mathbb R)=\det^{-1} ( \{1\})$ is closed. Since
$J=\operatorname{diag}(-1,-1,1,1,...), I \in SL_n(\mathbb R)$, but
${1 \over 2} (I+J)$ is not invertible, we see that $SL_n(\mathbb R)$
is not convex.

*If $A$ is diagonalisable, so is $\lambda A$ for any $ \lambda$, hence there is a path to $0$, and so the space of diagonalisable
matrices is path connected. If $n=1$ then clearly the space of diagonal $1 \times 1$ matrices (that is, $\mathbb{R}$) is open, closed and convex, hence we may assume 
that $n \ge 2$. Let $A= \begin{bmatrix} 0 & 1   \\
0& 0  \\
  & & I \end{bmatrix}$, and
$A_n= \begin{bmatrix} 0 & 1   \\
0& {1 \over n} \\
& & I\end{bmatrix} $. Then $A_n \to A$, but $A$ is not
diagonalisable and the $A_n$ are, hence the set of diagonalisable
matrices is not closed.
Now let $B= \begin{bmatrix} 0 & 0   \\
0& 0  \\
  & & I \end{bmatrix}$, and
$B_n= \begin{bmatrix} 0 & {1 \over n}   \\
0& 0 \\
& & I\end{bmatrix} $. Then $B_n \to B$, but $B$ is
diagonalisable and the $B_n$ are not, hence the set of diagonalisable
matrices is not open.
If we let $C_1= \begin{bmatrix} 0 & 1   \\
0& 1  \\
  & & I \end{bmatrix}$,
$C_2= \begin{bmatrix} 1 & 1   \\
0& 0  \\
  & & I \end{bmatrix}$ then we see that $C_1,C_2$ are diagonalisable
but ${1 \over 2} (C_1+C_2)$ is not.
