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.
- There exists a $A\in M_{2\times 5}(\mathbb{R})$ such that the dimension of the null space of $A$ is 2.
- There exists a $A\in M_{2\times 5}(\mathbb{R})$ such that the dimension of the null space of $A$ is 0.
- There exists a matrix $B\in M_{5\times 2}$ such that $AB$ is the identity matrix.
- 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.
It's the number of free variables you have to count, that's the dimension of the nullspace.