The drawbacks with Jacobi's method are rather obvious, with perhaps the most negative feature being the fact that we cannot tell * a priori* how many transformations are needed. Can we do better? The answer to this is yes and is given by a clever algorithm outlined by Householder. It was ranked among the top ten algorithms in the previous century. We will discuss this algorithm in more detail below.

The first step consists in finding an orthogonal matrix \( \mathbf{S} \) which is the product of \( (n-2) \) orthogonal matrices $$ \mathbf{S}=\mathbf{S}_1\mathbf{S}_2\dots\mathbf{S}_{n-2}, $$ each of which successively transforms one row and one column of \( \mathbf{A} \) into the required tridiagonal form. Only \( n-2 \) transformations are required, since the last two elements are already in tridiagonal form.