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

nubilaveritas says:

January 19, 2014 at 10:33 am

I think it is R(B) at some point

- hecker says:
  
  January 21, 2014 at 7:00 pm
  
  You are correct; I have fixed these errors. Thank you for finding them!
  
James Teow says:

October 12, 2017 at 10:37 pm

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:
  
  October 14, 2017 at 6:41 pm
  
  You are correct. I’ve updated the post to correct this. Thanks for reporting this error!