Question

The sum of \(r\) rank one matrices has rank \(r\). Can anyone point me towards a proof?

Answer 1

I am confused, I don't think this is true. I can think of 2 rank 1 matrices when added together give a rank 2 matrix.

Answer 2

I edited the Post...

Answer 3

fixed @kainui

Answer 4

The proof I wrote has a problem that I just found....

Here it is

suppose \(A\) is the sum of \(r\) rank one linearly independent matrices (the book left out the linearly independent part but for sure it is necessary as we can take any rank one matrix and add it to itself and the result is a rank one matrix). Then we have \[A=\sum_{i=1}^rv_iu_i^T\]where the \(v_i\) are linearly independent. We can use the Gram-Schmidt Process to find a basis \(\{v_i\}_i^n\) (the first \(r\) \(v_i\) are the same as before.\\

Consider the matrix \[B=\underset{i=r+1}{\sum^n}v_iv_i^T\] then we have \[(A+B)=\sum_{i=1}^nv_iu_i^T\] where \(u_i=v_i\) for \(r+1\le i \le n\).

Now suppose \((A+B)x=0 \) for some vector \(x\). Then we have \[\sum_{i=1}^nv_i(u_i^Tx)=0.\] Since our \(v_iu_I^T\) are linearly independent, the sum of any \(i\) of them can't be \(0\). So it must be that \(v_i^Tx=o \ \forall i\). Since the \( v_i=u_i\) form a basis, it must be that \(x=0\) and so \((A+B)\) has full rank \(n\). Since rank\((X+Y)\le\)rank\((X)+\)rank\((Y)\)we have that \(n\le\)rank\((A)\)+rank\((B)\), also rank\((B)\le n-r\). So we have \(n\le \text{rank}(A)+n-r\) so that \(r\le \text{rank}(A)\) and for sure \(\text{rank}(A)\le r\) as it is the sum of rank one matrices. So \(\text{rank}(A)=r\).

The problem is this line 'Since the \( v_i=u_i\) form a basis, it must be that \(x=0\)'

Because only the last \(n-r\) are a part of the basis. It may be an easy fix but I can't see it.

Answer 5

here is my reasoning, I use the Dirac's notation.
Since we have:
\[\left| {{v_i}} \right\rangle \left\langle {{v_i}} \right|\left. x \right\rangle = 0\quad \forall i = 1,...,n\]
then I multiply from the left by \(\left\langle x \right|\), so I get:
\[\left\langle x \right.\left| {{v_i}} \right\rangle \left\langle {{v_i}} \right|\left. x \right\rangle = 0\]
and, then:

\[{\left| {{x_i}} \right|^2} = 0\]
being:
wherein:
\[{{x_i}}\]
is the \(i\)-th component of \(\left| x \right\rangle \) along \({\left| {{v_i}} \right\rangle }\)
namely the operator:

Answer 6

is the projection operator along the subspace generated by \[{\left| {{v_i}} \right\rangle }\]