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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

$0$ .