Linear Algebra and Its Applications, Exercise 2.6.7

Exercise 2.6.7. Describe the matrices representing the following transformations:

i) projecting all vectors onto the $x$ – $y$ plane

ii) reflecting all vectors through the $x$ – $y$ plane

iii) rotating all vectors in the $x$ – $y$ plane by 90 degrees, leaving the $z$ axis unchanged

iv) rotating the $x$ – $y$ plane by 90 degrees, followed by rotating the $x$ – $z$ plane by 90 degrees, and finally rotating the $y$ – $z$ plane by 90 degrees

v) again carrying out the three rotations in succession, but through 180 degrees each time instead

Answer: i) The following matrix projects all vectors onto the $x$ – $y$ plane:

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

Note that this has the effect of multiplying the $z$ components of all vectors by zero, and multiplying the $x$ and $y$ components by the identity matrix:

$\begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&0 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \begin{bmatrix} v_1 \\ v_2 \\ 0 \end{bmatrix}$

ii) The following matrix reflects all vectors through the $x$ – $y$ plane:

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

Note that this has the effect of negating the $z$ components of all vectors, and multiplying the $x$ and $y$ components by the identity matrix:

$\begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&-1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \begin{bmatrix} v_1 \\ v_2 \\ -v_3 \end{bmatrix}$

iii) In order to carry out this transformation, we start with the 2 by 2 matrix for rotation in the $x$ – $y$ plane by 90 degrees:

$\begin{bmatrix} \cos 90^{\circ}&-\sin 90^{\circ} \\ \sin 90^{\circ}&\cos 90^{\circ} \end{bmatrix} = \begin{bmatrix} 0&-1 \\ 1&0 \end{bmatrix}$

Moving to three dimensions, the following matrix rotates all vectors in the $x$ – $y$ plane by 90 degrees, leaving the $z$ components alone:

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

Note that this has the effect of preserving the $z$ components of all vectors, and multiplying the $x$ and $y$ components by the 2 by 2 matrix for a 90 degree rotation:

$\begin{bmatrix} 0&-1&0 \\ 1&0&0 \\ 0&0&1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \begin{bmatrix} -v_2 \\ v_1 \\ v_3 \end{bmatrix}$

iv) In order to carry out this transformation, we first use the matrix from the previous exercise that rotates all vectors in the $x$ – $y$ plane by 90 degrees, leaving the $z$ components alone:

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

We then use the matrix that rotates all vectors in the $x$ – $z$ plane by 90 degrees, leaving the $y$ components alone:

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

Note that this matrix is constucted by taking the 2 by 2 matrix for rotation by 90 degrees in the $x$ – $z$ plane and extending it to three dimensions to leave the $y$ components as is.

Finally we use the matrix that rotates all vectors in the $y$ – $z$ plane by 90 degrees, leaving the $x$ components alone:

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

We perform all three rotations by multiplying the matrices in reverse order:

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

$= \begin{bmatrix} 0&0&-1 \\ -1&0&0 \\ 0&1&0 \end{bmatrix} \begin{bmatrix} 0&-1&0 \\ 1&0&0 \\ 0&0&1 \end{bmatrix} = \begin{bmatrix} 0&0&-1 \\ 0&1&0 \\ 1&0&0 \end{bmatrix}$

Note that this transformation is equivalent to rotating all vectors in the $x$ – $z$ plane by 90 degrees, leaving the $y$ component unchanged.

v) In order to carry out this transformation, we start with the 2 by 2 matrix for rotation in the $x$ – $y$ plane by 180 degrees:

$\begin{bmatrix} \cos 180^{\circ}&-\sin 180^{\circ} \\ \sin 180^{\circ}&\cos 180^{\circ} \end{bmatrix} = \begin{bmatrix} -1&0 \\ 0&-1 \end{bmatrix}$

Moving to three dimensions, we first use the matrix that rotates all vectors in the $x$ – $y$ plane by 180 degrees, leaving the $z$ components alone:

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

We then use the matrix that rotates all vectors in the $x$ – $z$ plane by 180 degrees, leaving the $y$ components alone:

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

Finally we use the matrix that rotates all vectors in the $y$ – $z$ plane by 180 degrees, leaving the $x$ components alone:

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

We perform all three rotations by multiplying the matrices in reverse order:

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

$= \begin{bmatrix} -1&0&0 \\ 0&-1&0 \\ 0&0&1 \end{bmatrix} \begin{bmatrix} -1&0&0 \\ 0&-1&0 \\ 0&0&1 \end{bmatrix} = \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix}$

Note that this transformation is equivalent to multiplying a vector by the identity matrix, leaving it unchanged.

UPDATE: Corrected the calculation of the cumulative effect of the three rotations in the answer to (iv). Thanks to Ji for pointing out my 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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2 Responses to Linear Algebra and Its Applications, Exercise 2.6.7

Ji says:

November 8, 2015 at 9:46 am

In solution 4, there is calculation mistake.
{{1, 0, 0} ,{0,0,-1}, {0,1,0}} {{0, 0, -1}, {0,1,-0}, {1,0,0}} = {{0, 0, -1} ,{-1,0,0}, {0,1,0}}

- hecker says:
  
  November 8, 2015 at 6:28 pm
  
  You are correct. Thank you very much for finding this error. I have updated the post with the correct calculation.