How can I find rank of $A=\sum_{i=1}^4 x_ix_i^T$ without actually finding the matrix $A$?

Question

Suppose $A=\sum\limits_{i=1}^4 x_ix_i^T$ where

$x_1=(1,-1,1,0)^T,x_2=(1,1,0,1)^T,x_3=(1,3,1,0)^T$ and $x_4=(1,1,1,0)^T$.

How can I find the rank of $A$ without explicitly finding the matrix $A$ and then transforming $A$ to an echelon matrix?

I know that $\operatorname{rank}(A)=3$ proceeding the usual way.

We have $\operatorname{rank}(x_ix_i^T)=1$ for each $i$, but that doesn't help me much here. Also, $x_1=2x_4-x_3$. Does that imply there are $3$ linearly independent rows/columns in $A$?

As this question was supposed to be solved in the exam with limited time at hand, I am left to wonder if it can be solved rather quickly.

score 1 · Accepted Answer · answered Jan 03 '19 at 15:13

1

The matrix $A$ can be written as $A=XX^T$ where $X=[x_1\ x_2\ x_3\ x_4]$. Now use the fact that $\operatorname{rank}XX^T=\operatorname{rank}X$, so all you need to do is the find the rank of $X$.

answered Jan 03 '19 at 15:13

A.Γ.

29,518

Thank you. I overlooked this decomposition. – StubbornAtom Jan 03 '19 at 15:26

score 0 · Answer 2 · answered Jan 03 '19 at 14:49

0

Observe that for any vector $c$, $$Ac = \sum_{i=1}^{4} x_ix_i^Tc = \sum_{i=1}^{4} (x^Tc)x_i$$

The image of a linear map is a vector space. That means that for every $Ac$ we can find some set of vectors which spans this set. Can you find such a set (even if the set is linearly dependent)? If you can find such a set, then you just need to count how many linearly independent vectors you need to span the same space (i.e. remove vectors which are linearly dependent).

answered Jan 03 '19 at 14:49

overfull hbox

1,410

How does $x_i x_i^T c$ equal $(x_i^T c) x_i$? – StubbornAtom Jan 03 '19 at 15:04
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$x_i^Tc$ is a scalar, so you can move it around – overfull hbox Jan 03 '19 at 15:14
Oh I see $c$ is a vector. – StubbornAtom Jan 03 '19 at 15:20

How can I find rank of $A=\sum_{i=1}^4 x_ix_i^T$ without actually finding the matrix $A$?

2 Answers2