If $A$ is $m \times n$ matrix with rank$(A)= 1$. How do we show that $$A^2 =\lambda A$$ for some $\lambda$?
How do we show this? All I could show is that if rank$(A)=1$ then rank$(A^2)=1.$
Am I heading in the right direction?
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8http://math.stackexchange.com/questions/991346/matrices-of-rank-1-show-that-a2-c-cdot-a-for-some-scalar-c – vudu vucu Oct 26 '15 at 14:09
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$A$ should be $n\times n$, otherwise $A^2$ is not well defined. – Ofir Schnabel Oct 27 '15 at 09:02
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$A$ must be square to build $A^2$. Having rank one implies $A=xy^T$ where $x,y$ are (column) vectors. Now $$ A^2=xy^Txy^T=x(\underbrace{y^Tx}_{\lambda})y^T=x\,\lambda\, y^T=\lambda\, xy^T=\lambda A. $$
A.Γ.
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you must have $m=n$, let $u$ be the generator of the image of $A$, $A(u)=cu$, for every $x$ in $\mathbb{R}^n$, there exists $d(x)\in \mathbb{R}$ such that $A(x)=d(x)u$, $A^2(x)=A(d(x)u)=d(x)A(u)=d(x)cu=cd(x)u=cA(u)$, thus $A^2=cA$.
Giovanni
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Tsemo Aristide
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if $n=m$ and $rank(A)=1$ then $dim(Im(A))=1$ thus $dim(Ker(A))=n-1$ and so, there exists a base $b=\{x_1,...,x_n\}$ of $\mathbb{R}^n,\ Ax_1=y_1$ and $\forall i \in \{2..n\}\ Ax_i=0 $.
We can write $\forall x \in \mathbb{R}^n\ Ax=y= \sum_{i=1}^n{\lambda_iAx_i}$ as $b$ is a base.
As $\forall i \in \{2..n\} \ Ax_i=0,\ Ay= \sum_{i=1}^n{\lambda_i A^2x_i} = \lambda_1A^2x_1$.
As $Ax_1 = \sum_{i=1}^{n}{\gamma_ix_i},\ Ay=\lambda_1\gamma_1Ax_1$.
In the same way $x= \sum_{i=1}^n{\lambda_ix_i}$ as $b$ is a base so $Ax=\lambda_1Ax_1$.
Finally $\forall x, A^2x=\gamma_1 (\lambda_1 Ax_1) = \gamma_1 Ax $. Thus $A^2=\lambda A$ with $\lambda = \gamma_1$ .
scam00
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