## 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

1. Ji says:

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:

You are correct. Thank you very much for finding this error. I have updated the post with the correct calculation.