## 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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