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  1. Consider the set of all $n×n$ matrices with real entries as the space $\mathbb{R}^{n^2}$. Which of the following sets are compact?
    (a) The set of all orthogonal matrices.
    (b) The set of all matrices with determinant equal to unity.
    (c) The set of all invertible matrices.

  2. In the set of all $n×n$ matrices with real entries, considered as the space $\mathbb{R}^{n^2} $, which of the following sets are connected?
    (a) The set of all orthogonal matrices.
    (b) The set of all matrices with trace equal to unity.
    (c) The set of all symmetric and positive definite matrices.


determinant mapping is continuous. so 1.(c) is not true as the image is not compact. but not sure for others. and 2.(a) is not true as the image is not connected. but not sure for others.

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1 Answer 1

1) a) It is bounded and closed, thus compact. It is bounded because $||A||_F^2=trace(A^TA)=n$. It is closed because all orthogonal matrices can be thought of as preimage of $I$ under the map $A^TA$.

b) It is unbounded, thus not compact. Take the diagonal matrix $D=diag(n,1/n,1,...,1)$. It's determinant is 1, but norm is unbounded.

c) User argument in (b)

2) a) don't know, but I think its not connected. Will update.

b) Take any two matrices $A$ and $B$ with unit trace. Define the path $f(t)=tA+(1-t)B$ between $A$ and $B$. Every matrix in the path has trace 1, thus they are path-connected and hence connected.

c) Set of all symmetric positive definite matrices is a convex set. Thus, there should be a path connecting any two matrices in that set. Thus it is connected. You can use the argument in b) as well.

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The set of orthogonal matrices can be partitioned into matrices with determinant +1 and -1, both are open as they are the inverse images of $(-\infty, 0)$ and $(0,\infty)$ of the determinant function. –  ACARCHAU Jan 9 '13 at 4:54

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