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Problems #6,16, pg.237 are graded.

#6, pg.237: If a $7 \times 5$ matrix $A$ has $\mathrm{rank \hspace{2pt}} 2$, find $\mathrm{dim Nul} A$, $\mathrm{dim Row} A$, and $\mathrm{rank} A^T$.

Solution: By the rank-nullity theorem, we know that for any matrix $A$, $$\mathrm{dim Col} A = \mathrm{dim Row} A = \mathrm{dim Row} A^T=\mathrm{dim Col}A^T=\mathrm{rank}A,$$ and so we conclue that $$\mathrm{dim Row} A = 2$$ and $$\mathrm{rank A^T} \stackrel{\mathrm{def}}{=} \mathrm{dim Col A^T} = 2.$$ Also by the rank-nullity theorem, $$\mathrm{rank} \hspace{2pt} A + \mathrm{dim Nul}A = 5,$$ so we compute $$\mathrm{dim Nul}A = 5-\mathrm{rank} A = 5-2 = 3.$$

#16, pg.237: Is $A$ is a $7 \times 5$ matrix, what is the smallest possible dimension of $\mathrm{Nul} A$?

Solution: The rank-nullity theorem tells us that $$\mathrm{dim Nul}A = 5-\mathrm{rank}A.$$ What are the possible ranks that a $7 \times 5$ matrix may have? In this case the answer is $5$. Let us explain why: recall that $\mathrm{rank}A=\mathrm{dim Col}A=\mathrm{dim Row}A$. We have a characterization of $\mathrm{dim Col}A$ as "the number of pivot columns of $A$" and a characterization of $\mathrm{dim Row}A$ as "the number of nonzero rows of $A$" when $A$ is an echelon form. Since $A$ has $7$ rows, there could be at most $7$ nonzero rows. Since $A$ has $5$ columns, there could be at most $5$ pivot columns. This restricts the rank to having values $0,1,2,3,4,$ or $5$. So we see that the possible values for $\mathrm{dim Nul}A$ are $5,4,3,2,1,$ or $0$.

Therefore the smallest possible dimension of $\mathrm{dim Nul}A$ is $0$.