Why Householder transformation can not be chosen to be an identity matrix?

Question

For Householder transformation, we know that $H = I-uu^T$, where $\|u\|_2=\sqrt{2}$. When it acts on any vector $x$, $Hx$ and $x$ is symmetric with respect to $span(u)^T$. But I have read a monography "Stewart. Matrix Algorithm I: Basic Decomposition, 1998, SIAM". It is written as follows on Page 257:

Combining these two observations we get Algorithm 1.1 — a program to generate a Householder transformation. Note that when $x = 0$, any $u$ will do. In this case the program housegen returns $u = 2e_1$, where $e_1$ is the first column of an identity matrix. This choice does not make $H$ the identity, but the identity with its $(1,1)$-element changed to $-1$. (In fact, it is easy to see that a Householder transformation can never be the identity matrix, since it transforms $u$into $-u$.)

My question is that I donot understand why the Householder matrixcannot be an identity matrix. Because when $x=0$, it means that we donot need to do any transformation, i.e., we can take $H = I$ identity matrix. But the author said we cannot take identity matrix. Where do I misunderstand? Thanks very much.

Answer 1

By definition, the Householder transformation is a reflection about a plane or hyperplane. The plane is described by its unit normal vector u and the transformation is then H = I - 2uu'. If u is a unit vector, 2uu' cannot be zero and thus H cannot be the identity matrix.

You can find plenty of non-Householder transformations H that, for certain arguments, give the same result as some Householder transform. H = I for x = 0 is one such example. Yet f(x) = x is not the same as f(x) = x² even though f(0) = 0 is true for both of them.

Answer 2

Short answer:

When he says that when $x=0$, any $u$ will do, he means any $u$ that satisfies $\|u\|_2 = \sqrt{2}$. Otherwise, it wouldn't be a Householder reflector.

So yes, you have found a transformation that "introduces" zeros in $x$, but you haven't proven that that transformation is a Householder reflector (spoiler: it isn't ;) ).