## Linear Algebra and Its Applications, Review Exercise 2.4

Review exercise 2.4. Given the matrix $A = \begin{bmatrix} 1&2&0&2&1 \\ -1&-2&1&1&0 \\ 1&2&-3&-7&-2\end{bmatrix}$

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

Answer: We perform elimination on $A$ to reduce it to echelon form. In the first step we multiply the first row of $A$ by -1 and subtract it from the second row: $\begin{bmatrix} 1&2&0&2&1 \\ -1&-2&1&1&0 \\ 1&2&-3&-7&-2\end{bmatrix}$ $\Rightarrow \begin{bmatrix} 1&2&0&2&1 \\ 0&0&1&3&1 \\ 1&2&-3&-7&-2\end{bmatrix}$

and then multiply the first row of $A$ by 1 and subtract it from the third row: $\begin{bmatrix} 1&2&0&2&1 \\ 0&0&1&3&1 \\ 1&2&-3&-7&-2\end{bmatrix}$ $\Rightarrow \begin{bmatrix} 1&2&0&2&1 \\ 0&0&1&3&1 \\ 0&0&-3&-9&-3\end{bmatrix}$

Finally we multiply the second row of $A$ by 1 and subtract it from the third row: $\begin{bmatrix} 1&2&0&2&1 \\ 0&0&1&3&1 \\ 0&0&-3&-9&-3\end{bmatrix}$ $\Rightarrow \begin{bmatrix} 1&2&0&2&1 \\ 0&0&1&3&1 \\ 0&0&0&0&0\end{bmatrix}$

The resulting matrix $U = \begin{bmatrix} 1&2&0&2&1 \\ 0&0&1&3&1 \\ 0&0&0&0&0\end{bmatrix}$

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

Since $U$ has two pivots its rank $r = 2$. The rank of $A$ is the same as the rank of $U$ so the rank of $A$ is also $r = 2$. This is the dimension of the column space of $A$.

The number of columns of $A$ is $n = 5$ so the dimension of the nullspace of $A$ is $n - r = 5 - 2 = 3$.

The dimension of the row space of $A$ is the same as that of the column space of $A$, so the dimension of the row space of $A$ is also $r = 2$.

Finally, the number of rows of $A$ is $m = 3$ so the dimension of the left nullspace of $A$ is $m - r = 3 - 2 = 1$.

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 .

This entry was posted in linear algebra. Bookmark the permalink.