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Let $A_1 \in \mathbb C^{n \times m}$, where $2 \leq m < n$, be a rank-$1$ matrix and $$A_2 = A_1 D$$ where $D$ is a full rank diagonal matrix. Prove that

$$\begin{bmatrix} A_1 \\ A_2\end{bmatrix}$$

is rank-$2$.

This might be a stupid question, but I'm really lost.

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The statement is false.

Let $A_1 = \begin{bmatrix} 1 \\ 1\end{bmatrix}$. Check that it is rank $1$.

$A_1D \in C^{2 \times 1}$

$\begin{bmatrix} A_1 \\ A_2 \end{bmatrix} \in C^{4 \times 1},$

Hence $$\operatorname{rank}\left( \begin{bmatrix} A_1 \\ A_2 \end{bmatrix}\right)= 1$$

Edit:

Let $A_1 = \begin{bmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 0 \end{bmatrix}$ and $D = I_2$. Hence $A_1 = A_2$.

Hence $$\operatorname{rank} \left( \begin{bmatrix} A_1 \\ A_2 \end{bmatrix}\right)= 1$$

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  • $\begingroup$ Thanks.. I will edit the question accordingly. $\endgroup$ Aug 2, 2017 at 8:29
  • $\begingroup$ I think my example still work for your edited question. $\endgroup$ Aug 2, 2017 at 8:32
  • $\begingroup$ sorry a typo i did $\endgroup$ Aug 2, 2017 at 8:33

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