Are the spaces of real orthogonal, complex unitary, hermitian or symmetric matrices connected?

Question

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?

Answer 1

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.

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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.