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### Prove: if $\text{rank}(A)=\text{rank}(A^2)$ then $Ax=0$ and $A^2x=0$ have the same set of solutions.

$$n-\dim(\ker A^2)=\text{rank}~ A^2=\text{rank}~ A=n-\dim(\ker A)$$ we get $$\dim(\ker A^2)=\dim(\ker A)\tag{1}$$ We also know: $$x\in\ker A\Rightarrow Ax=0\Rightarrow A^2x=0\Rightarrow x\in \ker A^2$$...

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### Proof of $\operatorname{rank} \left( X’ V X \right) = \operatorname{rank} (X)$

Here is a proof of the result. I assume that in this context, $V$ being "non-ngeative definite" implies that it is also symmetric (or Hermitian if we're dealing with complex matrices). We ...
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### Rank of partitioned matrix

To be precise, you need to prove that $\operatorname{col}(X_1) \cap \operatorname{col}(X_2) = 0$, not $\emptyset$. And yes, this is implied by $X_1 a \ne X_2 b$ for $a, b \ne 0$, since every nonzero ...
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### Is the set of matrices with rank at most $2$ closed?

I'm going to address only your posted question, regarding whether there is a problem with your argument. Yes, there is a big problem. It is true that any individual row operation is a continuous ...
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### When is ${\rm rank}(A+B)={\rm rank}(A)+{\rm rank}(B)$ true?

Note that the rank is equal to the dimension of the column space (and row space). $rank(A + B) = \dim col(A + B)$ $\leq \dim (col(A) + col(B))$ $\leq \dim col(A) + \dim col(B)$ $= rank(A) + rank(B)$ ...
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### Prove: if $\text{rank}(A)=\text{rank}(A^2)$ then $Ax=0$ and $A^2x=0$ have the same set of solutions.

One can easily check that $$\rm Im(A)=Im(A^2)\Longleftrightarrow Ker(A)=Ker(A^2)$$ Make use of the formula $$\rm dim(Ker(A))+dim(Im(A))=n$$

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