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Let the $n \times n$ complex matrix $A$ have rank 1. Prove: $A^2 = c\cdot A$ for some scalar $c$.

What I know is that all matrices having rank 1 have rows based on a scalar multiple of the other. This should probably be enough to figure out a very simple proof but I've been out of academic pursuits for 15 years and am having a hard time shaking the dust off.

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OP is right: it’s enough to figure out a rather simple proof. This “rows based on a scalar multiple of the other” feature can be expressed in $A = C\,R$ form where $R$ is an 1 × n matrix (for example, any of non-zero rows of $A$) and $C$ is an n × 1 matrix. Then, $$ A^2 = C\,R\,C\,R $$ (remember that matrix product is associative even for non-square matrices).

Look at the middle multiplication operation (of 3): what is “$R\,C$”? Matrix product of a row (at the left) and a column (at the right) gives 1 × 1 matrix. Then, let $c:=R\,C$, or more precisely, let c be the only element of said 1 × 1 matrix. (This kind of multiplication is known as the dot product.)

Obviously, matrix multiplication by an 1 × 1 matrix yields a scalar multiplication, so $$ A^2 = C\,c\,R = c\cdot C\,R = c\cdot A$$

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If $A$ has rank$~1$, there is some nonzero vector $\def\v{\mathbf v}\v$ such that $\def\x{\mathbf x}A\x$ is a scalar multiple of $\v$ for all vectors$~\x$ (which multiple it is depends on $\x$ of course). In particular this holds for $\v$ itself, say $A\v=c\v$, where $c$ is a scalar. Now if $\x$ is an arbitrary vector and $A\x=d\v$ (with again $d$ a scalar), then one computes $$ A^2\x=A(d\v)=dA\v=dc\v=cd\v=cA\x \qquad\text{so } A^2\x=cA\x\text{ independently of $d$.} $$ Since this is true for all $\x$, one has $A^2=cA$.

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If $A$ has rank one, then all columns are multiples of one of the columns and so $A$ can be written as $A = u v^*$, where $u,v$ are non-zero vectors.

Then $A^2 = u v^* u v^* = (v^* u) u v^* = c A$, where $c=v^* u$.

To see why we can write $A = u v^*$:

We know that $\dim {\cal R} A = 1$, so ${\cal R} A =\operatorname{sp} \{ u\} $ for some unit length $u$. In particular $Ax = \phi(x) u$, where multiplying by $u^*$ we see $\phi(x) = u^* Ax$. Letting $v = A^* u$, we have $Ax = v^* x u = u v^* x$ which gives $A=u v^*$.

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Let denote $A=(c_1\; c_2\;\cdots c_n)$ where $c_i$ is the $i$-th column of $A$. Since $A$ has the rank $1$ then without loss of generality we have $c_1\ne0$ and $c_i=\alpha_i c_1,\; i\ge2$ hence

$$A=c_1(1\;\alpha_2\cdots \alpha_n)=:CR$$

Now we have

$$A^2=(CR)(CR)=C\underbrace{(RC)}_{\text{scalar}}R=(RC)CR=RC\times A$$ and $$\operatorname{tr}(A)=\operatorname{tr}(CR)=\operatorname{tr}(RC)=RC$$ hence

$$A^2=\operatorname{tr}(A)A$$

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