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I must determine if a each of the following is a subspace of the vector space consisting of all complex $2\times 2$ matrices.

  1. All matrices with real diagonals.
  2. All matrices for which the sum of the diagonal entries is zero.
  3. All singular complex matrices.
  4. All matrices whose determinant is real.
  5. All matrices of the form $\left[\begin{array}{cc}a&b\\{a_1}&{b_2}\end{array}\right]$ where $a,b \in C$

Are there steps I can take in proving or disproving each of these? I've gotten nowhere with this problem.

I think that only number two is a subspace, but I have no way of proving or disproving any of them.

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  • $\begingroup$ What have you tried? Do you know what is meant by a "subspace" in this case? Do you have any guesses towards the right solutions here? $\endgroup$ – Omnomnomnom Oct 20 '14 at 21:31
  • $\begingroup$ I think that only number two is a subspace, but I have no way of proving or disproving any of them. $\endgroup$ – jerry2144 Oct 20 '14 at 21:36
  • $\begingroup$ Can we assume that $a_1,b_2$ are not both zero? $\endgroup$ – Omnomnomnom Oct 20 '14 at 21:40
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In this context, a set $S$ subspace iff for all $x,y \in S$ and $a \in \Bbb C$, we have

  • $x + y \in \Bbb C$
  • $a x \in \Bbb C$

So, for example, I know that the first set (call it $S_1$) is not a subspace because $M = \pmatrix{1&0\\0&1} \in S$, but $$ i M = \pmatrix{i&0\\0&i} \notin S $$

On the other hand, I know that the second set (call it $S_2$) is a subspace. If $M,N \in S_2$ and $a \in \Bbb C$, then the sum of the diagonal of $M+N$ will be zero, as will be the sum of the diagonal of $a M$.


You are correct in your guess that $S_2$ is the only subspace. The key to "disproving" the rest is to find an element (or elements) of the set that break the rules of a subspace.

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