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

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.

• 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. ;) Jan 19 '20 at 9:04

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 ;) ).