Review exercise 2.4. Given the matrix

find its echelon form and the dimensions of the column space, nullspace, row space, and left nullspace of .

Answer: We perform elimination on to reduce it to echelon form. In the first step we multiply the first row of by -1 and subtract it from the second row:

and then multiply the first row of by 1 and subtract it from the third row:

Finally we multiply the second row of by 1 and subtract it from the third row:

The resulting matrix

is in echelon form with pivots in columns 1 and 3.

Since has two pivots its rank . The rank of is the same as the rank of so the rank of is also . This is the dimension of the column space of .

The number of columns of is so the dimension of the nullspace of is .

The dimension of the row space of is the same as that of the column space of , so the dimension of the row space of is also .

Finally, the number of rows of is so the dimension of the left nullspace of is .

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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