MHB The Existence of Symmetric Matrices in Subspaces

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A 4-dimensional subspace \(U\) of \(3 \times 3\) matrices must contain a symmetric matrix. The dimension of the subspace of symmetric matrices, \(\mathcal{S}\), is 6. If the intersection of \(U\) and \(\mathcal{S}\) were zero-dimensional, it would lead to a contradiction via the Grassmann theorem, as the combined dimension would exceed the total dimension of the space. Therefore, there must exist at least one non-zero symmetric matrix in \(U\). This confirms the existence of symmetric matrices within the specified subspace.
Sudharaka
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Hi everyone, :)

Here's a question I am stuck on. Hope you can provide some hints. :)

Problem:

Let \(U\) be a 4-dimensional subspace in the space of \(3\times 3\) matrices. Show that \(U\) contains a symmetric matrix.
 
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I suppose you mean a symmetric matrix different from $0.$ If $\mathcal{S}$ is the subspace of the symmetric matrices, then $\dim \mathcal{S}=\dfrac{3(3+1)}{2}=6.$ If $\dim (\mathcal{S}\cap U)=0,$ then by the Grassmann theorem $\dim (U+\mathcal{S})=6+4-0=10>9=\dim \mathbb{K}^{3\times 3}$ (contradiction). So, there exists a symmetric and non null matrix belonging to $U.$
 
Fernando Revilla said:
I suppose you mean a symmetric matrix different from $0.$ If $\mathcal{S}$ is the subspace of the symmetric matrices, then $\dim \mathcal{S}=\dfrac{3(3+1)}{2}=6.$ If $\dim (\mathcal{S}\cap U)=0,$ then by the Grassmann theorem $\dim (U+\mathcal{S})=6+4-0=10>9=\dim \mathbb{K}^{3\times 3}$ (contradiction). So, there exists a symmetric and non null matrix belonging to $U.$

Yes indeed, it should be different from the zero matrix. Thanks very much for your reply. I understand it fully. :)
 
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