When computing the eigenvalues of the symmetric matrix $M\in\mathbb{R}^{n\times n}$ the best you can do with Householder reflector is drive $M$ to a tridiagonal form. As was mentioned in a previous answer because $M$ is symmetric there is an orthogonal similarity transformation which results in a diagonal matrix, i.e., $D=S^TMS$. It would be convenient if we could find the action of the unknown orthogonal matrix $S$ strictly using Householder reflectors by computing a sequence of reflectors and applying $H^T$ from the left to $M$ and $H$ from the right to $M$. However this is not possible because of the way the Householder reflector is designed to zero out columns. If we were to compute the Householder reflector to zero out all the numbers below $M_{11}$ we find
$$
M=\left(\!\!{\begin{array}{ccccc}
	* &* & * & *&* \\
	* &* & * & *&*  \\
	* &* & * & *&* \\
	* &* & * & *&* \\
	* &* & * & *&* \\
\end{array}}\!\!\right)\rightarrow
H^T_1M=\left(\!\!{\begin{array}{ccccc}
	* &*' & *' & *'&*' \\
	0 &*' & *' & *'&*'  \\
	0 &*' & *' & *'&*' \\
	0 &*' & *' & *'&*' \\
	0 &*' & *' & *'&*' \\
\end{array}}\!\!\right).
$$
But now the entries $M_{12}-M_{1n}$ have been altered by the reflector $H^T_1$ applied on the left. Thus when we apply $H_1$ on the right it will no longer zero out the first row of $M$ leaving only $M_{11}$. Instead we will obtain
$$
H^T_1M=\left(\!\!{\begin{array}{ccccc}
	* &*' & *' & *'&*' \\
	0 &*' & *' & *'&*'  \\
	0 &*' & *' & *'&*' \\
	0 &*' & *' & *'&*' \\
	0 &*' & *' & *'&*' \\
\end{array}}\!\!\right)\rightarrow
H^T_1MH_1=\left(\!\!{\begin{array}{ccccc}
	* &*'' & *'' & *''&*'' \\
	*' &*'' & *'' & *''&*''  \\
	*' &*'' & *'' & *''&*'' \\
	*' &*'' & *'' & *''&*'' \\
	*' &*'' & *'' & *''&*'' \\
\end{array}}\!\!\right).
$$
Where not only did we not zero out the row but we may destroy the zero structure we just introduced with the reflector $H^T_1$.

However, when you opt to drive $M$ to a tridiagonal structure you will leave the first row untouched by the action of $H^T_1$, so
$$
M=\left(\!\!{\begin{array}{ccccc}
	* &* & * & *&* \\
	* &* & * & *&*  \\
	* &* & * & *&* \\
	* &* & * & *&* \\
	* &* & * & *&* \\
\end{array}}\!\!\right)\rightarrow
H^T_1M=\left(\!\!{\begin{array}{ccccc}
	* &* & * & *&* \\
	*' &*' & *' & *'&*'  \\
	0 &*' & *' & *'&*' \\
	0 &*' & *' & *'&*' \\
	0 &*' & *' & *'&*' \\
\end{array}}\!\!\right).
$$
Thus when we apply the same reflector from the right we obtain
$$
H^T_1M=\left(\!\!{\begin{array}{ccccc}
	* &* & * & *&* \\
	*' &*' & *' & *'&*'  \\
	0 &*' & *' & *'&*' \\
	0 &*' & *' & *'&*' \\
	0 &*' & *' & *'&*' \\
\end{array}}\!\!\right)\rightarrow
H^T_1MH_1=\left(\!\!{\begin{array}{ccccc}
	* &*' & 0 & 0&0 \\
	*' &*'' & *'' & *''&*''  \\
	0 &*'' & *'' & *''&*'' \\
	0 &*'' & *'' & *''&*'' \\
	0 &*'' & *'' & *''&*'' \\
\end{array}}\!\!\right).
$$

Applied recursively this allows us to drive $M$ to a tridiagonal matrix $T$. You can complete the diagonalization of $M$ efficiently, as was mentioned previously, using Jacobi or Givens rotations both of which are found in the Golub and Van Loan book *Matrix Computations*. The accumulated actions of the sequence of Householder reflectors and Jacobi or Givens rotations allows us to find the action of the orthogonal matrices $S^T$ and $S$ without explicitly forming them.