Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Really basic question on notation...if $u,v$ are column vectors and $A$ is a matrix (we're in $\mathbb{R}^{d}$ here), is it customary to write $$A+uv^{T}:=A+uv^{T}I,$$ or...? I've honestly never seen the shorthand before until today.

share|cite|improve this question
Thanks for all the answers. I was evidently up way too late doing math, and kept thinking of $uv^{T}$ as a scalar (e.g. a dot product) just from the habit of seeing $v^{T}u$ much more often, and that was confusing me since adding a scalar to a matrix makes no sense (unless one just defines it by entry-wise addition like we do scalar multiplication, but I am not aware of this being done anywhere). – Taylor Martin Jan 27 '13 at 23:26
up vote 1 down vote accepted

$$A+uv^{T}$$ This is fine. As $d\times 1$ multiplied with $1 \times d$ gives $d \times d$, the result is correct, and it is written this way all the time. This is why it is important to have dimensions and order of operations correct so there is no confusion, but this is fine. The $I$ does not do anything to make it more obvious, though it does not hurt. To be even more obvious, it is sometimes necessary to write the dimension for $I$ as in $I_d$ or sometimes $I_{d \times d}$, but all that is completely unnecessary here.

share|cite|improve this answer

There is no reason to multiply with $I$. Note that $v^T$ is a row vector, hence we multiply a $d\times 1$ matrix with a $1\times d$ matrix and obtain a $d\times d$ matrix. Contrast this with $v^Tu$, wich is a scalar (or $1\times 1$ matrix).

share|cite|improve this answer

AFAIK, it's done all the time in quantum physics, as the general "outer product" of a ket and a bra. Such an outer product of vectors is a linear operator, and if $u,v$ lie in the same space as the domain(=co-domain) of $A$ it makes perfect sense to add $A$ and $uv^T$.

share|cite|improve this answer

It's not simply a notation, nor a definition. That's an actual equality between the "two" entities. Remember that the identity matrix $I$ is the identity to the product between matrices. This means that for any matrix $M$ such that the product $MI$ is defined, we'll have $MI=M$.

Since $A\in \mathcal{M}_ d(\mathbb{R}) \cong \mathbb{R}^{d\times d}$ and $u$ and $v$ are column vectors, we have $u,v\in \mathbb{R}^{d\times 1}$. Therefore $v^T \in \mathbb{R}^{1\times d}$ and we'll have $uv^T\ \in \mathbb{R}^{d\times d}$. Since the product $uv^T$ is defined, by the above paragraph it will follow that $uv^T=uv^TI$.

You might be interested in the Sherman-Morrison formula.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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