Let $A$ be an m×n matrix with $m$<$n$ and rank$(A) =m$. If $B=AA^T$ , $C=A^TA$ and the eigenvalues and corresponding eigenvectors of $B$ are known, find the non-zero eigenvalues and corresponding eigenvectors of $C$

My thinking: Since rank$(A)=m$, rank$(B)=m$, rank$(C) =m$. Hence $C$ has $m$ non zero eigenvalues. $A$ is not a square matrix. If an eigenvalue of $B$ is $\alpha$ and the corresponding eigenvector is $x$ then $Bx=\alpha x$. I can't proceed further. Please help me. Thank you in advance...


We have,

$$Bx = \alpha x \\ (AA^\top)x = \alpha x \\ A^\top (AA^\top) x = A^\top \alpha x \\ (A^\top A) A^\top x= \alpha A^\top x \\ CA^\top x = \alpha A^\top x $$ Let $A^\top x = q \ \in \mathbb{R}^{n}$. Then we have,

$$Cq = \alpha q$$ This shows that the eigenvectors of C are the eigenvectors of B multiplied by $A^\top$. What about the eigenvalues? We have $m$ distinct eigenvalues of B, but C has $n$ eigenvalues. Well as you guessed correctly, C is of rank(m), so it has $n-m$ zero eigenvalues, and the other $m$ eigenvalues are same as B.

  • $\begingroup$ What are the eigenvectors corresponding eigen value 0? $\endgroup$ May 6 at 17:51
  • $\begingroup$ To find the zero eigenvectors you are solving Cq = 0, so q is the null-space of C. $\endgroup$
    – orchi_d
    May 6 at 18:02

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