# [SOLVED]Existence of Symmetric Matrix

#### Sudharaka

##### Well-known member
MHB Math Helper
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
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
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. 