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let $u_1, \cdots , u_5$ be linearly independent vectors in $\mathbb{R}_m$. Consider the matrix $A = \sum_{i=1}^{5} u_i u_i^T$.

What is the rank of $A$?

Since $A$ is $m\times m$ matrix and there is no column or row that is linearly dependent on another column or row then the rank of $A$ is $m$.

Is it correct? Please advise

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  • $\begingroup$ Therankis5. ${}{}{}{}$ $\endgroup$ – copper.hat Sep 26 '15 at 8:09
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With $U=[u_1\ u_2\ \ldots\ u_5]$ you have $A=UU^T$.

  1. $\text{rank}(U)=\text{rank}(U^T)=5$ $\implies$ $\text{dim}\ker(U^T)=m-5$.
  2. $\ker(A)=\ker(UU^T)=\ker(U^T)$ $\implies$ $\text{dim}\ker(A)=m-5$.
  3. $\text{rank}(A)=m-\text{dim}\ker(A)=m-(m-5)=5$.
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Let $A=\sum_k \alpha_k u_k u_k^T$, with $\alpha_k \neq 0$.

Then $Ax = 0$ iff $\sum_k \alpha_k \langle u_k, x \rangle u_k =0$ iff $\alpha_k \langle u_k, x \rangle = 0$ for all $k$ iff $\langle u_k, x \rangle = 0$ for all $k$ iff $x \in \operatorname{sp} \{ u_1,...,u_5\}^\bot$.

Hence $\ker A = \operatorname{sp} \{ u_1,...,u_5\}^\bot$, $\dim \ker A = m-5$ and so $\dim {\cal R} A = 5$.

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  • $\begingroup$ Why the downvote? $\endgroup$ – copper.hat Sep 26 '15 at 9:11

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