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user130529
user130529

The $\textrm{trace}$ function is continuous from $\mathcal{M}_2(\mathbb{C})$ to $\mathbb{C}$ and $\{0\}$ is closed in $\mathbb{C}$, hence $\textrm{trace}^{-1}(0)$ is closed. A closed subspace of a Hilbert space is a Hilbert space, hence the space of traceless matrices is a Hilbert space. The subset of Hermitian matrices is not a subspace, but nevertheless is closed as being the reciprocal image of the null matrix of another continuous function from $\mathcal{M}_2(\mathbb{C})$ into itself: $f(M) = M^*M-I$$f(M) = M^*-M$. Your set is the intersection of two closed sets, hence is closed too.

The $\textrm{trace}$ function is continuous from $\mathcal{M}_2(\mathbb{C})$ to $\mathbb{C}$ and $\{0\}$ is closed in $\mathbb{C}$, hence $\textrm{trace}^{-1}(0)$ is closed. A closed subspace of a Hilbert space is a Hilbert space, hence the space of traceless matrices is a Hilbert space. The subset of Hermitian matrices is not a subspace, but nevertheless is closed as being the reciprocal image of another continuous function: $f(M) = M^*M-I$. Your set is the intersection of two closed sets, hence is closed too.

The $\textrm{trace}$ function is continuous from $\mathcal{M}_2(\mathbb{C})$ to $\mathbb{C}$ and $\{0\}$ is closed in $\mathbb{C}$, hence $\textrm{trace}^{-1}(0)$ is closed. A closed subspace of a Hilbert space is a Hilbert space, hence the space of traceless matrices is a Hilbert space. The subset of Hermitian matrices is not a subspace, but nevertheless is closed as being the reciprocal image of the null matrix of another continuous function from $\mathcal{M}_2(\mathbb{C})$ into itself: $f(M) = M^*-M$. Your set is the intersection of two closed sets, hence is closed too.

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user36790
user36790

The $trace$$\textrm{trace}$ function is continuous from $\mathcal{M}_2(\mathbb{C})$ to $\mathbb{C}$ and $\{0\}$ is closed in $\mathbb{C}$, hence $trace^{-1}(0)$$\textrm{trace}^{-1}(0)$ is closed. A closed subspace of a Hilbert space is a Hilbert space, hence the space of traceless matrices is a Hilbert space. The subset of Hermitian matrices is not a subspace, but nevertheless is closed as being the reciprocal image of another continuous function: $f(M) = M^*M-I$. Your set is the intersection of two closed sets, hence is closed too.

The $trace$ function is continuous from $\mathcal{M}_2(\mathbb{C})$ to $\mathbb{C}$ and $\{0\}$ is closed in $\mathbb{C}$, hence $trace^{-1}(0)$ is closed. A closed subspace of a Hilbert space is a Hilbert space, hence the space of traceless matrices is a Hilbert space. The subset of Hermitian matrices is not a subspace, but nevertheless is closed as being the reciprocal image of another continuous function: $f(M) = M^*M-I$. Your set is the intersection of two closed sets, hence is closed too.

The $\textrm{trace}$ function is continuous from $\mathcal{M}_2(\mathbb{C})$ to $\mathbb{C}$ and $\{0\}$ is closed in $\mathbb{C}$, hence $\textrm{trace}^{-1}(0)$ is closed. A closed subspace of a Hilbert space is a Hilbert space, hence the space of traceless matrices is a Hilbert space. The subset of Hermitian matrices is not a subspace, but nevertheless is closed as being the reciprocal image of another continuous function: $f(M) = M^*M-I$. Your set is the intersection of two closed sets, hence is closed too.

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user130529
user130529

The $trace$ function is continuous from $\mathcal{M}_2(\mathbb{C})$ to $\mathbb{C}$ and $\{0\}$ is closed in $\mathbb{C}$, hence $trace^{-1}(0)$ is closed. A closed subspace of a Hilbert space is a Hilbert space, hence the space of traceless matrices is a Hilbert space. The subset of Hermitian matrices is not a subspace, but nevertheless is closed as being the reciprocal image of another continuous function: $f(M) = M^*M-I$. Your set is the intersection of two closed sets, hence is closed too.