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Let $M_{m\times n}(\mathbb{R})$ be the set of all $m\times n $ matrices with real entries. Which of the following matrices are correct.

  1. There exists a $A\in M_{2\times 5}(\mathbb{R})$ such that the dimension of the null space of $A$ is 2.
  2. There exists a $A\in M_{2\times 5}(\mathbb{R})$ such that the dimension of the null space of $A$ is 0.
  3. There exists a matrix $B\in M_{5\times 2}$ such that $AB$ is the identity matrix.
  4. There exists a $A\in M_{2\times 5}(\mathbb{R})$ whose null space is $\{(x_1,x_2,x_3,x_4,x_5)\in \mathbb{R^5}: x_1=x_2,x_3=x_4=x_5\}$

Since $\text{rank}A\leq$min$(m,n)$ so rank $A\leq2$. By rank nullity theorem we can conclude that $\text{nullity}A\geq3$. Hence option 1 and 2 are false. Option 3 is true, easy to check. But I am confused with the option 4. I don't know how to reason with that. Can anybody help me with any ideas or hints? Thanks.

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  • $\begingroup$ Can you write down a basis for the subspace in 4? Then you'll know its dimension. $\endgroup$ – André 3000 May 2 '17 at 4:42
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    $\begingroup$ Ya got it. The basis is $\{(1,1,0,0,0), (0,0,1,1,1)\}$ so the dimension is 2. Hence option 4 is false. $\endgroup$ – Kushal Bhuyan May 2 '17 at 5:08
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Only 3 is true as dimensions of null space should be atleast 3. In last option dimension of null space is 2.

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It's the number of free variables you have to count, that's the dimension of the nullspace.

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