1

If $A$ is a n by n matrix such that $rank(A)=1$ and $Null(A)=Col(A)^\perp$ Prove $A=A^\perp$

I already proved that $dim(Col(A)^\perp)=1$ and $dim(Null(A))=n-1$, but then what comes next?

I am also thinking that if A is orthogonally diagonalizable, then $A=A^\perp$, but how to prove this?

Tito Eliatron
  • 6,678
Logan
  • 252
  • 1
    You may find useful to decompose the matrix $A = uv^T$ for some vectors $u,v$: https://math.stackexchange.com/questions/1545118/a-rank-one-matrix-is-the-product-of-two-vectors

    you want to conclude that $u=v$

     – FeedbackLooper Nov 09 '20 at 18:56
  • I'm still very lost, but thank you. – Logan Nov 09 '20 at 18:59
  • Do you get the first step? that rank 1 matrix can be decomposed as $A=uv^T$? – FeedbackLooper Nov 09 '20 at 19:00
  • em..., not really, let me think about it – Logan Nov 09 '20 at 19:01
  • $\operatorname{Null}(A) = \operatorname{Col}(A)^\perp = \operatorname{Null}(A^\top)$. For the last equality see https://en.wikipedia.org/wiki/Row_and_column_spaces – V.S.e.H. Nov 09 '20 at 19:04
  • 1
    you can start by noting that if $A$ has rank 1, thus it has just 1 non zero eigenvalue $\lambda$. So if you expand it in eigenvalues-eigenvectors $A = P^{-1}\text{diag}(\lambda,0,0,\dots,0)P$. So when you perform the multiplication $P^{-1}\text{diag}(\lambda,0,0,\dots,0)P$ you end-up multiplying one row of $P$ with one column of $P^{-1}$ (times $\lambda$). Think about this, and the last part of the proof is in the answer of Trevor Gunn or in the comment of bodil. – FeedbackLooper Nov 09 '20 at 19:05
  • 1
    @RodrigoAldana I think that $A$ has rank 1 shouldn't really matter. What if we change to the assumption that $\operatorname{rank}(A) = 2$? :) – V.S.e.H. Nov 09 '20 at 19:11
  • You are totally right. The key fact (as you pointed out) was that $\text{Null}(A)=\text{Col}(A)^\bot$. I was just trying to give some more intuition to @Logan about his problem. Thanks for pointing that out! – FeedbackLooper Nov 09 '20 at 19:25

1 Answers1

2

Since $A$ has rank 1, it is of the form $A = uv^\top$. Then $\operatorname{Null}(A) = \{v\}^\perp$ and $\operatorname{Col}(A)^\perp = \{u\}^\perp$. Since $\operatorname{Null}(A) = \operatorname{Col}(A)^\perp$, it follows that $u$ is parallel to $v$.

So now $A = c(uu^\top)$ for some scalar $c$ and it follows that $A$ is symmetric.

Sera Gunn
  • 27,981