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.


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.

  • $\begingroup$ Thanks very much for you reply. I got it. ;). by the way, could you please write the equations in math mode for more accurate expressions, eg., $H=I-2*u*u'$ mode. Thanks. ;) $\endgroup$ – sunshine Jan 19 at 9:04

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 ;) ).

  • $\begingroup$ get it, thanks very much. ;). $\endgroup$ – sunshine Jan 19 at 9:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.