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let $M$ be a $n \times n$ matrix of rank $k (\neq n)$ if $\lambda \neq 0$ is an eigenvalue of $M$ with corresponding unit column vector $u$. with $Mu=\lambda u$,then which of the following is\are true?.

  • 1). $rank(M-\lambda uu^{*})=k-1$

  • 2). $rank(M-\lambda uu^{*})=k$

  • 3).$rank(M-\lambda uu^{*})=k+1$

  • 4).$(M-\lambda uu^{*})^{n}$$=M^{n}-\lambda^{n}uu^{*}$

solution I tried

in the given question

rank$(M-\lambda uu^{*})$$=$rank$(M-Muu^{*})$

taking $M$ common we get rank$[M(I-uu^{*})]$

further that I don't know how to proceed

please help!

Thank you

nmasanta
  • 9,222
gaurav saini
  • 763

1 Answers1

3

Let $\lambda_i$ be the non-zero eigenvalues of $M.$ And, $u_i$ be the unit eigenvectors corresponding to $\lambda_i$ (Choose $u_i$ to be mutually orthogonal and assume $\lambda_1=\lambda$ and $u_1=u$.)

Now write the spectral decomposition of $M$. That is write $M=\sum_i \lambda_i u_i u_i^*.$ This immediately gives the eigendecomposition of the matrix $M-\lambda uu^*$, that is, $$M-\lambda uu^*=\sum_{i=2}^{k} \lambda_i u_iu_i^*.$$

From the spectral decomposition it is clear that $rank(M-\lambda uu^*)=k-1.$

A straightforward computation and the fact that $u_i$ are orthonormal also yields that $$(M-\lambda uu^*)^n=M^n -\lambda^n uu^*.$$

Raghav
  • 3,026
  • From Spectral theorem how can we say that rank is $k-1$? and how can we show that $u_i u^{*}_i$ is the orthogonal projection? – TheStudent May 03 '20 at 05:54
  • Each $u_iu_i^*$ is a $1$ dimensional eigenspace. The sum written above shows that the range is precisely the direct sum of $k-1$ subspaces each of dimension $1.$ – Raghav May 03 '20 at 22:05