# Which of the following subsets of $M_n(\mathbb{R})$ are compact (NBHM)

Following is a list of problems from an exam for admission into Ph. D program. I have just compiled all previous questions on compactness of certain subsets of matrices and i tried to work out . I would be thankful if some one can check the solutions and please suggest if there are any better ways to do and if the solution is wrong please let me know what could be done for them.

1. $\{ A\in M_n(\mathbb{R}) : \text { A is real symmetric Matrix with eigenvalues$|\lambda|\leq 2$}\}$

Solution: We see that any real symmetric matrix is diagonalizable. So, I believe each element in $S$ is similar to some $\begin{bmatrix} a& 0\\0&b\end{bmatrix}$ where $|a|\leq 2, |b|\leq 2$ in which case we have boundedness. I am not so certain about the closedness I believe it is closed though i can not write it in detail.

2. $\{ A\in M_n(\mathbb{R}) : \text { A is diagonalizable Matrix with eigenvalues$|\lambda|\leq 2$}\}$

Solution: We see this is same as above set $S$ and so should be compact.

3. $\{\text {unitary matrices in$M_2(\mathbb{C}$)}\}$

Solution: My justification for this makes no sense.. I would edit this once i got some clear idea..

4. $\{ A\in M_n(\mathbb{R}) : \text { det (A)} =1\}$

Solution: We can consider $\begin{bmatrix} 1& n\\0&1\end{bmatrix}$ to be in given set and $n$ can be as large as i wish.. so this set is not bounded so is not compact.

5. $\{\text {Trace (A) :$A\in M_n(\mathbb{R})$is orthogonal}\}\subset \mathbb{R}$

Solution: I have no idea. only thing i could see is that determinant is $\pm 1$ but i do not see any properties regarding Trace.

6. $\{ A\in M_n(\mathbb{R}) : \text { A is invertible diagonal Matrix}\}$

Solution: We can consider $\begin{bmatrix} n& 0\\0&n\end{bmatrix}$ for $n\neq 0$to be in given set and $n$ can be as large as i wish.. so this set is not bounded so is not compact.

7. $\{ A\in M_n(\mathbb{R}) : \text { A is upper triangular Matrix}\}$

Solution: We can consider $\begin{bmatrix} 1& n\\0&1\end{bmatrix}$ to be in given set and $n$ can be as large as i wish.. so this set is not bounded so is not compact.

Thank you for sparing your valuable time in checking my solutions. Once somebody confirm that every thing is fine and no modifications needed i would delete this question so please post only comments (unless you believe it is so beautiful and worth being an answer :)).

Thank you :)

• Maybe it makes sense to give the questions a number. If someone has checked a question then he can refer to it more easily. Success when it concerns your exam. – drhab Jan 20 '14 at 9:37
• Yes.. I will edit this:) – user87543 Jan 20 '14 at 9:39
• It seems to me that the matrix you give in 7 is not unitary. $|det(A)|=1$ is a necessary but not sufficient condition for A being unitary. – Vincent Pfenninger Jan 20 '14 at 9:55
• @VincentPfenninger : Oh yes... I should have been more careful... I would try saying something more sensible and edit that... Thank you – user87543 Jan 20 '14 at 9:58
• $5$ and $6$ are not same. The closedness in $5$ is same as eigenvalues being continuous function of orthogonal matrices. check that result if it is true. – Vishal Gupta Jan 20 '14 at 10:01

## 4 Answers

Your Argument looks okay to me for $7,6,4,1,2$

$5$. Look $O(n,\mathbb{R})$ is compact in $M_n(\mathbb{R})$, Now $Trace:M_n(\mathbb{R})\to\mathbb{R}$ is a linear map so continuous map, so image of compact set under continuos map is compact.

$3$. $UU^*=U^*U=I\Rightarrow \|Ux\|=\|x\|$, so set of all unitary matrices are bounded.

now $U\to U,U\to U^*$ are continous and so $f:U\to UU^*$ is continous from $M_n(\mathbb{C})\to$ itself. now $U(n,\mathbb{C})=f^{-1}(I)$ so closed also.

• +1 for trace map...you may delete your statement saying $7$ and $6$ are correct.. then you may get another upvote :D – user87543 Jan 26 '14 at 2:53
• what about $3$ ;) – Marso Jan 26 '14 at 2:57
• this is perfect.... :D once you delete those statements about $4,6,7$ I would accept it :D May be you can write "Rest is fine" as P.S :P – user87543 Jan 26 '14 at 3:00
• done!{}{}{}{}{}}{{}{}{} – Marso Jan 26 '14 at 3:03
• @ElAngelExterminador You mean to say $\text{Trace}:O(n,R) \to \mathbb{R}$ is continuous?? – crskhr Jan 18 '15 at 2:59

A few comments:

(2) The matrix $\begin{bmatrix}a & 0 \\ 0 & a\end{bmatrix}$ is positive semidefinite if and only if $a\geq 0$ (you write $a\in\mathbb{R}$). Also, this set is closed. In particular, a matrix $$\begin{bmatrix}a & b \\ c & d\end{bmatrix}$$ is positive semidefinite if and only if it is symmetric and has non-negative eigenvalues. We can check that both eigenvalues are non-negative by checking that the trace and determinant are both nonnegative. Hence, the above matrix is positive semidefinite if and only if it satisfies the equations $$b=c,\qquad a+d\geq 0,\qquad ad-bc\geq 0.$$

(5) A matrix in this set doesn't have to be similar to $\begin{bmatrix}a & 0 \\ 0 & a\end{bmatrix}$, since the two eigenvalues don't have to be the same. In any case, bounding the eigenvalues isn't necessarily sufficient for bounding the matrix.

(7) This argument is not correct. The matrix $\begin{bmatrix}1 & n \\ 0 & 1\end{bmatrix}$ is unitary if and only if $n=0$.

(Hint for 9) What do you know about the eigenvalues of an orthogonal matrix?

• Eigen values of orthogonal matrix are complex numbers with modulus $1$... – user87543 Jan 20 '14 at 10:06
• @PraphullaKoushik Well no, in general they are complex numbers with unit modulus. – Jim Belk Jan 20 '14 at 10:06
• Yes Yes.. I was in hurry... I actually meant to say modulus $1$ – user87543 Jan 20 '14 at 10:07
• @PraphullaKoushik So how are the eigenvalues related to the trace? What does this tell us? – Jim Belk Jan 20 '14 at 10:08
• trace is sum of eigenvalues... I am not sure how to make use of this... i only know boundedness... :O – user87543 Jan 20 '14 at 10:09

I don't think the set from $6$ is the same as the set from $5$. A symmetric matrix is diagonalizable by a orthonormal matrix, this is a stronger condition to just being diagonalizable. For a $2\times 2$ matrix, take the matrix $$\begin{bmatrix}0& 1 \\ 1& b\end{bmatrix}\begin{bmatrix} 1& 0\\0&2\end{bmatrix}\begin{bmatrix}0& 1 \\ 1& b\end{bmatrix}^{-1}= \begin{bmatrix}0& 1 \\ 1& b\end{bmatrix}\begin{bmatrix} 1& 0\\0&2\end{bmatrix}\begin{bmatrix}-b& 1 \\ 1& 0\end{bmatrix} = \begin{bmatrix}2& 0 \\ b& 1\end{bmatrix}$$ and set $b$ to be as large as you want.

For $7$, the matrix $U=\begin{bmatrix} 1& n\\0&1\end{bmatrix}$ is not unitary. $UU^*=\begin{bmatrix} 1+n^2& n\\n&1\end{bmatrix}\neq I$

For $9$, if $A$ is orthogonal, its columns all have the norm of $1$, so you have boundedness

• Yes Yes... this is a valid point.. I actually meant $(6)$ is contained in $(5)$ and as $(5)$ is comapct hoping to conlcude $6$ is compact (I have to show $6$ is closed but could not make it out) – user87543 Jan 20 '14 at 9:55
• My point exactly. $6$ is NOT contained in $5$ as it contains some non-symmetric matrices which are diagonalizable. – 5xum Jan 20 '14 at 9:59
• Oh my bad.. I was confused... Thanks for the clarification... :) – user87543 Jan 20 '14 at 10:01
• I added an example of a matrix in $6$ which is not bounded. – 5xum Jan 20 '14 at 10:16
• Oh.. Thank you.. That helps me.. :) – user87543 Jan 20 '14 at 10:20

your 2 ans is wrong..bcoz u can take upper triangular matrix with distinct entries on diagonal ,then it is diagonalizable but it is not bounded if u take one of d non diagonal entry large

• A friendly tip from your friendly community moderator: Your answer is ok mathematically, but it is also difficult to read. It is probably a good idea to edit your answer and expand the argument filling in some of the details. Brevity is a virtue in math most of the time, but if you roam around our site you will notice that this style is common in comments only. Also, your style is very much ok when exchanging SMSs with friends, but here many prefer "school English". No need to save characters! Also members who don't have English as their first language may have trouble understanding. – Jyrki Lahtonen Dec 26 '14 at 15:18
• as Mr.Jyrki Lahtonen said please edit your answer accordingly... thank you for your interest in my quetsion :) – user87543 Dec 26 '14 at 17:09