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.

FOR 1 (a) may be true as determinant mapping is continuous and it maps to the compact set{1,-1} but it is only a necessary condition.and (c) is not true as determinant mapping is continuous and it maps to a non compact set.do not know about (b).but i think it is not true. FOR2 (a) is not correct.do not know about (b) & (c)

  • $\begingroup$ NBHM exam problem $\endgroup$
    – Srijan
    Apr 20 '13 at 4:53
  1. Compactness

    (a) This is true. First, the set of these matrices is bounded by $\sqrt{n}$, that is $||A|| \leq \sqrt{n}$. One way to see this is to first show that for orthogonal matrices, $||A||_2 = 1$, and then show that $||A|| \leq \sqrt{n}||A||_2$. To show that it is closed is not that difficult either. Let $\{A^k\}_{k\in\mathbb{N}}$ be a sequence of orthogonal matrices with limit $A$. Due to the nature of the norm $A$ is the component-wise limit, that is $A_{i,j} = \lim_{k\to\infty} A^k_{i,j}$. It should now be easy to see that $A$ is also orthogonal. Another option for showing that the set is closed is to observe (or show) that the map $A\to A^T A$ is continuous and the set of orthogonal matrices is the preimage of $I$.

    edit: Actually, showing that $||A||\leq \sqrt{n}$ is even easier, namely, almost by definition we have, $||A|| = \sqrt{\mathrm{tr}\left(A^T A\right)}$, but $A^T A = I$, so $||A|| = \sqrt{\mathrm{tr}\left(I\right)} = \sqrt{n}$.

    (b) This is false as this set is unbounded. For example, take the matrix with $a_{1,1} = M$, $a_{2,2} = \frac{1}{M}$, $a_{k,k}=1$ and zeros of of diagonal. The norm of this matrix is greater than $M$, so we get a family of matrices whose determinant is $1$ and whose norm cannot be bounded.

    (c) The same argument as the previous case.

    1. Connectedness

    (a) This set is not connected. For example, the set of matrices with determinant $1$ is a closed set as is the set of matrices with determinant $-1$ and they are disjoint and they cover the set of orthogonal matrices.

    (b) If $A$ and $B$ are matrices with trace $1$, then $f(t) = t A + (1-t) B$ is a path from $A$ to $B$ and it is very easy to see that for each $t$, $f(t)$ is a matrix with trace $1$. The set is thus path connected and therefore connected.

    (c) Similar argument as for the previous case works. For $A$ and $B$ positive definite, for each $t\in [0,1]$, $f(t)$ is a symmetric positive definite matrix, which is easy to see straight from the definition. So this space is also path connected.

  • $\begingroup$ Bizak how do you know that matrices with determinant 1(or -1) are open set in $\mathbb{R}^n²$???in reffrence to question 2 part 1 $\endgroup$ Mar 19 '21 at 10:22
  • $\begingroup$ @Ibs I think I wrote that the set of those matrices is closed, not open. It is closed because taking a determinant is a continuous function, so preimages of closed sets are closed. $\endgroup$ Mar 26 '21 at 17:08
  • $\begingroup$ @Bizak but to show that set is disconnected you need open disjoint sets,......how can you prove it with taking closed disjoint sets? $\endgroup$ Mar 27 '21 at 12:24
  • $\begingroup$ @Ibs The two closed sets cover the entire set (and are not empty) which is one of the equivalent ways of stating that a set is not connected. Briefly, because they cover the set, and are disjoint, then their complements are two open sets, which cover the space. $\endgroup$ Mar 28 '21 at 11:41
  • $\begingroup$ Bizjak; yes I got it...thanks $\endgroup$ Mar 28 '21 at 17:41

Let $E:=\Bbb R^{n^2}$.

  1. (a) We have $O_n:=F^{-1}(\{0\})$ where $F\colon E\to E$ is defined as $F(A):=A^tA-I$. We have to show that $F$ is continuous, but it follows from the fact that $M\colon E^2\to E$, $M(A,B)=A\cdot B$ and $Ad\colon E\to E$, $Ad(M):=M^t$ are continuous. We have $\lVert Ax\rVert=\lVert x\rVert$ for all $x$ where $\lVert\cdot \rVert$ is the Euclidian norm, which proves boundedness.

    (b) The map $\det\colon E\to \Bbb R$ is continuous, which gives closeness. But it's not bounded, as $\pmatrix{a&0&0\\0&a^{—1}&0\\ 0&0&I_{n-2}}$ shows.

    (c) The sequence $\{\frac 1k I_n\}$ of invertible matrices converges to the null matrix, which is not invertible.

  2. (a) We can write $$O_n=(O_n\cap \det^{—1}(-\infty,0))\sqcup (O_n\cap \det^{—1}(0,+\infty)),$$ hence as a disjoint union of two open sets for the induced topology.

    (b) The set of such matrices is arcwise connected.

    (c) If $A$ and $B$ are symmetric positive definite, then so is $tA+(1-t)B$ for $0\leq t\leq 1$.

  • $\begingroup$ 2 (b) gives what result? It's not compact because the diagonal matrices of the form $(n,1/n,1,\ldots,1)$ are unbounded. $\endgroup$
    – tomasz
    Sep 15 '12 at 18:04
  • $\begingroup$ @tomasz I was going to edit but the site was in reading mode. Now I've added the missing steps. $\endgroup$ Sep 15 '12 at 19:40

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