## Linear Algebra and Its Applications, Review Exercise 2.5

Review exercise 2.5. Given the matrices

$A = \begin{bmatrix} 0&0&1 \\ 0&0&1 \\ 1&1&1 \end{bmatrix} \qquad B = \begin{bmatrix} 0&0&1&2 \\ 0&0&1&2 \\ 1&1&1&0 \end{bmatrix}$

find their ranks and nullspaces.

Answer: We can use elimination to reduce $A$ to echelon form. We first exchange the first and third rows:

$\begin{bmatrix} 0&0&1 \\ 0&0&1 \\ 1&1&1 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&1&1 \\ 0&0&1 \\ 0&0&1 \end{bmatrix}$

and then subtract 1 times the second row from the third row:

$\begin{bmatrix} 1&1&1 \\ 0&0&1 \\ 0&0&1 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&1&1 \\ 0&0&1 \\ 0&0&0 \end{bmatrix}$

The resulting echelon matrix has two pivots and thus rank of 2; this is also the rank of $A$.

Solving for the nullspace, we have $x_1$ and $x_3$ as basic variables and $x_2$ as a free variable. Setting $x_2 = 1$, from the second row of the echelon matrix we have $x_3 = 0$ and from the first row we have $x_1 + x_2 +x_3 = x_1 + 1 + 0 = 0$ or $x_1 = -1$.

So the nullspace $\mathcal{N}(A)$ is the 1-dimensional subspace of $\mathbb{R}^3$ with basis vector $(-1, 1, 0)$. (In other words, the nullspace of $A$ is the line passing through the origin and the point $(-1, 1, 0)$.)

Similarly we can use elimination to reduce $B$ to echelon form. We first exchange the first and third rows:

$\begin{bmatrix} 0&0&1&2 \\ 0&0&1&2 \\ 1&1&1&0 \end{bmatrix} \Rightarrow\begin{bmatrix} 1&1&1&0 \\ 0&0&1&2 \\ 0&0&1&2 \end{bmatrix}$

and then subtract 1 times the second row from the third row:

$\begin{bmatrix} 1&1&1&0 \\ 0&0&1&2 \\ 0&0&1&2 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&1&1&0 \\ 0&0&1&2 \\ 0&0&0&0 \end{bmatrix}$

The resulting echelon matrix has two pivots and thus rank of 2; this is also the rank of $B$.

Solving for the nullspace, we have $x_1$ and $x_3$ as basic variables and $x_2$ and $x_4$ as free variables. Setting $x_2 = 1$ and $x_4 = 0$, from the second row of the echelon matrix we have $x_3 + 2 x_4 = x_3 + 2 \cdot 0 = 0$ or $x_3 = 0$, and from the first row we have $x_1 + x_2 +x_3 = x_1 + 1 + 0 = 0$ or $x_1 = -1$.

Setting $x_2 = 0$ and $x_4 = 1$, from the second row of the echelon matrix we have $x_3 + 2 x_4 = x_3 + 2 \cdot 1 = 0$ or $x_3 = -2$, and from the first row we have $x_1 + x_2 +x_3 = x_1 + 0 - 2 = 0$ or $x_1 = 2$.

So the nullspace $\mathcal{N}(B)$ is the 2-dimensional subspace of $\mathbb{R}^4$ with basis vectors $(-1, 1, 0, 0)$ and $(2, 0, -2, 1)$.

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.

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