Matrix spectral space vs entry space in symmetric matrices Is is known that the space of symmetric matrices $\mathbb{R}_{sym}^{n \times n}$ has $\binom{n}{2}$  dimensions.
And according to the spectral theorem every symmetric matrix $A \in \mathbb{R}_{sym}^{n \times n}$ has a spectral decomposition in terms of 1-rank matrices.
A = $\sum_{i=1}^n \lambda_i v_i v_i^T $
Hence we conclude the dimension space of the symmetric matrices is $n$.
Where is the fallacy of this reasoning ?
Thanks in advance. 

You have correctly stated that for any symmetric $A$, there exist rank $1$ matrices $v_iv_i^T$ such that
$$
A = \sum \lambda _i v_i v_i^T
$$
However, it is impossible to select a fixed set $\{v_1v_1^T,\dots,v_nv_n^T\}$ such that every $A$, there exists a choice of $\lambda_i$ such that $A$ has the above form.  That is, there is no basis for the set of symmetric matrices that consists of $n$ rank-1 matrices.
