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I don't know if it suppose to be an easy or a hard question but here it is:

Let $A$ a $n\times n$ matrix such that $\operatorname{rank}(A)$ =1

$\mathbf{1.}$ Proof that there exists a matrix $B$ similar to $A$ when the first $n-1$ columns of $B$ are $0$.

$\mathbf{2.}$ Proof that there exists a matrix $C$ similar to $A$ when all the rows of $C$, except the first one, are $0$.

gal rehavam
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1 Answers1

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Think of rank $1$ matrix as $A=uv^T$.
See this question for the explanation.

Then it is possible to choose (via rotation what means that transformation matrix is orthogonal) coordinate frame where one of the vectors would have coordinates, $a,0,0,\dots,0$, let it be $v_r=Rv$.

The second vector $u_r$ can have any coordinates in this frame and it is transformed $u_r=Ru$.

Obviously matrix equal to $u_rv_r^T=Ru (Rv )^T=R(uv^T)R^T$ is similar to $uv^T$.

Now calculate $u_rv_r^T$, you should obtain matrix with only one-non zero column.

For the second case change vector with one non-zero coordinate.

Widawensen
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