## Linear Algebra and Its Applications, Exercise 2.1.3

Exercise 2.1.3. For each of the following two matrices $A = \begin{bmatrix} 1&-1 \\ 0&0 \end{bmatrix} \qquad B = \begin{bmatrix} 0&0&0 \\ 0&0&0 \end{bmatrix}$

describe the matrix’s column space and nullspace.

Answer: The column space for $A$ consists of the linear combinations of its columns: $c_1 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} -1 \\ 0 \end{bmatrix} = \begin{bmatrix} c_1 - c_2 \\ 0 \end{bmatrix}$

The column space $\mathcal{R}(A)$ is therefore the x axis, i.e., the set of vectors of the form $(x_1, 0)$.

The nullspace $\mathcal{N}(A)$ is the set of vectors $x$ for which $Ax = 0$. We then have $Ax = \begin{bmatrix} 1&-1 \\ 0&0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} x_1 - x_2 \\ 0 \end{bmatrix} = 0$

so that we have $x_1 - x_2 = 0$ or $x_1 = x_2$. The nullspace $\mathcal{N}(A)$ is therefore the line $x_2 = x_1$ passing through the origin at a 45 degree angle to the x axis.

For the matrix $B$ the column space $\mathcal{R}(B)$ contains all vectors of the form $c_1 \begin{bmatrix} 0 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 0 \\ 0 \end{bmatrix} + c_3 \begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

Therefore $\mathcal{R}(B)$ consists only of the point $(0, 0)$.

For any vector $x = (x_1, x_2, x_3)$ we have $Bx = \begin{bmatrix} 0&0&0 \\ 0&0&0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

Therefore the nullspace $\mathcal{N}(B) = \mathbf{R}^3$.

UPDATE: Corrected references to $\mathcal{R}(A)$ that should have been to $\mathcal{R}(B)$. Thanks go to nubilaveritas for catching these errors.

UPDATE 2: Corrected the product $Bx$ to be 2 by 1 rather than 3 by 1. Thanks go to James Teow for catching the error.

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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### 4 Responses to Linear Algebra and Its Applications, Exercise 2.1.3

1. nubilaveritas says:

I think it is R(B) at some point

• hecker says:

You are correct; I have fixed these errors. Thank you for finding them!

2. James Teow says:

How does Bx result in a 3 by 1 vector? It’s a 2 x 3 Matrix multiplied by a 3 x 1 vector, shouldn’t it result in a 2 x 1 vector(0, 0)?

• hecker says:

You are correct. I’ve updated the post to correct this. Thanks for reporting this error!