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Is it true that the set of all matrices with trace equal to zero a connected and compact subset of the 2*2 matrices over R?

Learnmore
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  • Which topology you put on the space of matrices? – Milly Oct 17 '14 at 04:37
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    The space of $n$-dimensional matrices (over $\mathbb{R}$ or $\mathbb{C}$) is finite-dimensional vector space, so usually one considers the topology induced by any norm (they are all equivalent). – Seirios Oct 17 '14 at 05:45

2 Answers2

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No, definitely not, at least provided you use the standard topology induced by the obvious identification $M_2(\mathbb{R})$ with $\mathbb{R}^4$.

The trace operator $$\text{tr}: M_2(\mathbb{R}) \to \mathbb{R}$$ is linear, and so its kernel (the set of tracefree matrices) is a linear subspace of $M_2(\mathbb{R})$, and is obviously nonempty (it contains $\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}$). In particular, it is an unbounded (but connected) subset of a real vector space and so is not compact.

Travis Willse
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  • Can you explain please why it is connected@travis sir – Learnmore Oct 17 '14 at 05:06
  • All linear subspaces of vector spaces over $\mathbb{R}$ and $\mathbb{C}$ are connected: By linearity they are convex, and so in particular they are path-connected. The space of tracefree, real $2 \times 2$ matrices is just a $3$-dimensional vector space in $M_2(\mathbb{R})$, so as a vector space (via a choice of basis) it's a copy of $\mathbb{R}^3$. – Travis Willse Oct 17 '14 at 05:30
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Only if $n=1$. In general, if $n>1$ let $A=e_1 e_1^T - e_n e_n^T$, then $\operatorname{tr} A = 0$, but $\operatorname{tr} ( \lambda A) = 0$ for all $\lambda$, and $\operatorname{sp} \{ A \}$ is not compact.

copper.hat
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