Householder similarities reduce a general matrix to upper Hessenberg form and a real symmetric matrix to tridiagonal form

If $n\le 2$, every $n\times n$ matrix is already upper Hessenberg and every real symmetric one is already tridiagonal, so take the identity similarity. Assume $n\ge 3$. For $j=1,\dots ,n-2$, apply [L1] to the tail of column $j$ below the first subdiagonal inside the trailing $(n-j)\times (n-j)$ block. Embedding that reflector into the identity produces a unitary $Q_j$ that annihilates all entries of column $j$ below row $j+1$. Because $Q_j$ acts only on the trailing block, previously created zeros are preserved.

After the $n-2$ steps of 1.1, the product $H=Q_{n-2}^{\ast }\cdots Q_1^{\ast }A{Q}_1\cdots Q_{n-2}$ has zeros below its first subdiagonal, so [L2] says that $H$ is upper Hessenberg.

If $A$ is real symmetric, the same similarity steps remain real orthogonal. Each step that zeros the lower tail in column $j$ also zeros the matching upper tail in row $j$ because symmetry is preserved under orthogonal similarity.

Therefore the final matrix is symmetric and upper Hessenberg, hence tridiagonal by [L2]. This proves the symmetric claim.