Homotopy of Involutory Matrices? I want to construct a homotopy from the matrix 
$$
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & -1 & 0 & 0\\
0 & 0 & -1 & 0\\
0 & 0 & 0 & 1\\
\end{bmatrix}
$$
to the identity matrix within the space of involutory matrices ($A^{2} = I$). Is this even possible?
Also, how do I determine the number of path connected components of this space?
 A: No such path exists. Probably the shortest proof is to observe that the trace of a matrix is a continuous function, that the trace of a matrix satisfying $A^2 = 1$ takes on a discrete set of possible values, and that it's $0$ for your matrix but $4$ for the identity matrix. 
Just for fun, let's interpret the title question to ask: what is the homotopy type of the space of (real) involutory matrices? As we've seen, this space has $5$ connected components depending on the number of eigenvalues which are $1$ vs. $-1$; these can be analyzed separately. The space of involutory matrices with a given set of eigenvalues has the property that every pair of such matrices is conjugate to a diagonal matrix, hence it is acted on transitively by $\text{GL}_4(\mathbb{R})$. The stabilizer of this action depends on the eigenvalues:


*

*For the eigenvalues $1, 1, 1, 1$ and $-1, -1, -1, -1$ the stabilizer is $\text{GL}_4(\mathbb{R})$ and the space is a point, since scalar matrices are central.

*For the eigenvalues $1, 1, 1, -1$ and $1, -1, -1, -1$ the stabilizer is $\text{GL}_3(\mathbb{R}) \times \text{GL}_1(\mathbb{R})$ and the space $\text{GL}_4(\mathbb{R})/(\text{GL}_3(\mathbb{R}) \times \text{GL}_1(\mathbb{R}))$ is diffeomorphic to the space of lines in $\mathbb{R}^4$, or equivalently the real projective space $\mathbb{RP}^3$. This is a relatively well-understood space from the point of view of homotopy theory; in particular, its fundamental group is $\mathbb{Z}_2$ and it has universal cover the $3$-sphere $S^3$. . 

*For the eigenvalues $1, 1, -1, -1$ the stabilizer is $\text{GL}_2(\mathbb{R}) \times \text{GL}_2(\mathbb{R})$ and the space $\text{GL}_4(\mathbb{R})/(\text{GL}_2(\mathbb{R}) \times \text{GL}_2(\mathbb{R}))$ is diffeomorphic to the space of planes in $\mathbb{R}^4$, or equivalently the Grassmannian $\text{Gr}_2(\mathbb{R}^4)$. This is again a relatively well-understood space from the point of view of homotopy theory; its fundamental group is $\mathbb{Z}_2$ and it has universal cover the space of oriented $2$-planes in $\mathbb{R}^4$, which as it turns out is diffeomorphic to $S^2 \times S^2$. (This reflects an exceptional isomorphism and doesn't generalize.)

A: Let $M(t)$ be a path through the involutory matrices such that $M(0) = \text{diag}(1,-1,-1,1)$ and $M(1)$ is the identity matrix.
Let $p_t(x) = \det(xI - M(t))$, written as
$$
p_t(x) = x^4 + a_3(t)x^3 + a_2(t)x^2 + a_1(t)x + a_0(t)
$$
Clearly, $p_0(x) = (x-1)^2(x+1)^2 = x^4 - 2x^2 + 1$, while $p_1(t) = (x-1)^4$.  Note that $a_3(0) \neq a_3(1)$.
Now, note that there are finitely many possibilities for the characteristic polynomial of an involutory matrix.  However, $a_3(t)$ depends continuously on $t$.  So, $a_3(t)$ is a non-constant continuous function on $[0,1]$ that attains finitely many values.  This is a contradiction.
Thus, no such path exists.
In fact, the path components are precisely the sets of matrices with identical characteristic polynomials.  So, there will be $5$ path components in this space.
A: If we're talking about the reals, then presumably there's no way to do this. Why? Because if $A^2 = I$, then $A$ satisfies $x^2 - 1 = 0$, whose only roots are $+1$ and $-1$. So the domain of $A$ splits as a direct sum of the $+1$ and $-1$ eigenspaces. For your matrix, these each have dimension 2; for the identity, the $+1$ eigenspace has dimension 4. Suppose you had a path $t \mapsto M(t)$ from $M(0) = A$ to $M(1) =I$. Consider $H = \{ t | M(t) \text{ has a nontrivial -1 eigenspace} \}$. Let $h$ be the least upper bound of $H$. What's the $-1$ eigenspace of $M(h)$ look like? Trivial or not? For $t< h$ you've got $-1$ as an eigenvalue twice, and for $t > h$ (at least near $h$) you've got it as an eigenvalue at most once. That seems implausible, at the very least. I'm guessing this can be converted to an actual proof. 
