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[SOLVED] Existence of Symmetric Matrix

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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 $U.$
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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. :)