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Do all matrices of the order $m\times n$ and rank $3$ form a vector space?

If I take linear combination of two such matrices, then the resulting matrix should also have the rank $3$. But how can I show this? (Or prove it wrong?) And does this apply to a matrix of order less than or equal to $m$?

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    What is the rank of $A-A$? – 2017-02-18
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    Look at the origin... – 2017-02-18

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The answer is no, since $\forall A\in\mathbb{R}^{m\times n} \mathrm{rank}(A-A)=\mathrm{rank}(0)=0$

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    Wouldn't the answer change if it was $\le 3$? – 2017-02-18
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    If $A$ is in your vector space and $\mathrm{rank}(A)>0$ then, on the one hand, $-A$ has the same rank and, on the other hand, $-A$ also is another member of your vector space. since $A$ and $-A$ are both members of your space, their sum $A-A=0$ must be an element of that space, too. But $\mathrm{rank}(0)=0$. conclusion: such a vectorspace can not exist (except for $\mathrm{rank}(A)=0$, **which indeed would define a vector space.**) – 2017-02-18
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    @AdityaDev, as to the question in your comment above, note that, given enough room, $\begin{bmatrix}1&0&0&0&0&0\\0&1&0&0&0&0\\0&0&1&0&0&0\\0&0&0&0&0&0\\0&0&0&0&0&0\\0&0&0&0&0&0\\\end{bmatrix}+\begin{bmatrix}0&0&0&0&0&0\\0&0&0&0&0&0\\0&0&0&0&0&0\\0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1\\\end{bmatrix}=\begin{bmatrix}1&0&0&0&0&0\\0&1&0&0&0&0\\0&0&1&0&0&0\\0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1\\\end{bmatrix}$ – 2017-02-18