Matrix spectral space vs entry space in symmetric matrices

Question

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.

Answer 1

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.

Answer 2

The dimension space of $n$-symmetric matrices is $\frac{n(n+1)}{2}$. By the spectral theorem every symmetric matrix, with real coefficient, is diagonalizable; in other words every symmetric matrix is similar to a diagonal matrix. But this not means that the dimension of $n$-symmetric matrices in $n$.

The set of diagonal matrices is a vector subspace of symmetric matrices, and is very easy to find a symmetric matrix such that is not diagonal.

Answer 3

The linear subspace of symmetric matrices is actually of dimension $n(n+1)/2$. The linear subspace of diagoanl matrices is of dimension $n$. The similar transformation (spectral decomposition) maps $n(n+1)/2$ space to $n$ space. You have confused 2 different spaces with a single space.