Are the spaces of real orthogonal, complex unitary, hermitian or symmetric matrices connected? I want to know which of these are connected and which are not. I think I've to take some continuous map from the set of matrices to $\mathbb R$ or $\mathbb C$ and interpret these matrix sets as inverse images of connected sets to conclude they are connected. What maps do I take? I don't think the determinant map works everywhere, if at all. Can anyone provide a solution?
 A: A continuous map is indeed helpful in proving a set is disconnected.  In fact, consider the determinant over the real orthogonal matrices.  What is the image of this map?  What can you conclude as a result?
Note that any path connected space is connected.  For the symmetric and Hermitian matrices, consider $p:[0,1] \to \Bbb C^{n \times n}$ given by 
$$
t \mapsto (1-t)A + tB
$$
All that's left now is the set of complex unitary matrices.  This set is also path connected, but there is a trickier path here.  Perhaps you can make sense of the map
$$
t \mapsto A^{(1-t)}B^t
$$
Alternatively, note that
$$
A \mapsto e^{iA}
$$
is a continuous onto map from the Hermitian matrices to the complex unitary matrices.

The matrix exponential is probably the quicker way to go here.  We define the exponential map by
$$
\exp(X) = e^X = I + X + \frac 1{2!} X^2 + \frac 1{3!} X^3 + \cdots
$$
We have the following properties: 
$$
\exp \pmatrix{d_1\\&\ddots \\ && d_n} = 
\pmatrix{e^{d_1} \\ & \ddots \\ && e^{d_n}}
$$
and for any invertible $S$ and matrix $A$, we have
$$
\exp(SAS^{-1}) = S \exp(A) S^{-1}
$$
Now, using the spectral theorem, show that if $A$ is Hermitian, then $\exp(iA)$ is unitary.
