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Michael Giudici
Michael Giudici
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Let $\Gamma$ be the graph whose vertices are the 2-dimensional subspaces of $\mathbb{R}^2$$\mathbb{R}^n$ and two will be adjacent if they intersect in a 1-dimensional subspace. Then $PGL(n, \mathbb{R})$ is in the automorphism group of $\Gamma $. I think it is the full automorphism group but that will need checking.

Now any real matrix is similar to its Jordan normal form, which for real matrices is block diagonal with blocks either lower triangular or having size 2x2. Thus it fixes a vertex of $\Gamma$.

Let $\Gamma$ be the graph whose vertices are the 2-dimensional subspaces of $\mathbb{R}^2$ and two will be adjacent if they intersect in a 1-dimensional subspace. Then $PGL(n, \mathbb{R})$ is in the automorphism group of $\Gamma $. I think it is the full automorphism group but that will need checking.

Now any real matrix is similar to its Jordan normal form, which for real matrices is block diagonal with blocks either lower triangular or having size 2x2. Thus it fixes a vertex of $\Gamma$.

Let $\Gamma$ be the graph whose vertices are the 2-dimensional subspaces of $\mathbb{R}^n$ and two will be adjacent if they intersect in a 1-dimensional subspace. Then $PGL(n, \mathbb{R})$ is in the automorphism group of $\Gamma $. I think it is the full automorphism group but that will need checking.

Now any real matrix is similar to its Jordan normal form, which for real matrices is block diagonal with blocks either lower triangular or having size 2x2. Thus it fixes a vertex of $\Gamma$.

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Michael Giudici
Michael Giudici
  • 2.2k
  • 14
  • 18

Let $\Gamma$ be the graph whose vertices are the 2-dimensional subspaces of $\mathbb{R}^2$ and two will be adjacent if they intersect in a 1-dimensional subspace. Then $PGL(n, \mathbb{R})$ is in the automorphism group of $\Gamma $. I think it is the full automorphism group but that will need checking.

Now any real matrix is similar to its Jordan normal form, which for real matrices is block diagonal with blocks either lower triangular or having size 2x2. Thus it fixes a vertex of $\Gamma$.