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.


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*$\{ 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.

*$\{ 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.

*$\{\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..

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

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.

*$\{\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.

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

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.

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

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 :)

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
