HSN.VM.C.12Common CoreMathNumber and QuantityGrades 9-12
HSN.VM.C.12: 2 × 2 Matrices as Transformations of the Plane and the Determinant as Area
In plain English: HSN.VM.C.12 is an advanced (+) Common Core number and quantity standard that asks students to use 2 × 2 matrices to transform the whole plane, for example by rotating, reflecting, stretching or shearing figures, and to interpret the absolute value of the determinant as the factor by which the matrix multiplies area. It is usually taught in Precalculus.
(+) Work with 2 × 2 matrices as transformations of the plane, and interpret the absolute value of the determinant in terms of area.
Common Core State Standards for Mathematics · Domain: Vector and Matrix Quantities (VM) · Cluster: Perform operations on matrices and use matrices in applications. Also written as HSN-VM.C.12 or N-VM.12 · Official standard
Students learn to see a 2 × 2 matrix as a transformation of the whole plane. Every point (x, y) is written as the column vector [x; y], and the matrix sends it to a new point. The origin never moves, lines go to lines, and the unit square with corners (0, 0), (1, 0), (1, 1) and (0, 1) goes to the parallelogram built on the two columns of the matrix. Students transform squares, rectangles and triangles, and they name the transformations they see: rotations, reflections, dilations, stretches and shears.
The second idea is area. For A = [a b; c d], the determinant is det A = ad - bc. Students find the area of the image of the unit square with the box method and discover that it equals |ad - bc|. Because every figure can be filled with small squares, the matrix multiplies every area by the same factor |det A|. A negative determinant also means the figure is flipped (its orientation is reversed), and a determinant of 0 means the plane is squashed onto a line or a point. Matrices are written row by row with semicolons between rows: [2 1; 1 3] has first row 2, 1 and second row 1, 3, and the point (4, -1) is the column vector [4; -1], with 4 on top.
Learning Objectives
By the end of this lesson, students will be able to:
Find the image of a point, a segment or a polygon under a 2 × 2 matrix and sketch it
Recognize and write the matrices of rotations, reflections, dilations, stretches and shears, using the images of (1, 0) and (0, 1)
Compute the determinant of a 2 × 2 matrix and explain why its absolute value is the area of the image of the unit square
Use |det A| as an area scale factor for any figure, and interpret a negative or zero determinant
Prior Knowledge Required
Students should already be comfortable with:
Multiplying a vector by a matrix and matrices as transformations of vectors HSN.VM.C.11
Describing dilations, rotations and reflections with coordinates 8.G.A.3
Finding the areas of triangles and rectangles from coordinates HSG.GPE.B.7
Give each student grid paper and a coordinate rule to apply to a small triangle.
Warm-Up Prompt
"Plot the triangle with vertices (0, 0), (2, 0) and (0, 1). Apply the rule (x, y) → (3x, 2y) to each vertex and plot the new triangle. What was the area before? What is it after? By what number was the area multiplied?"
The new vertices are (0, 0), (6, 0) and (0, 2), so the area goes from 1 to 6, and 6 = 3 × 2. Write the rule as the matrix [3 0; 0 2]. Ask students to predict the area factor for [5 0; 0 4] (20) and then ask the question the lesson answers: what is the area factor when the matrix has no zeros, such as [2 1; 1 3]?
Direct Instruction20-25 minutes
Part 1: Transforming the plane. Multiply the matrix by each vertex of a figure and connect the images in the same order. Stress two facts. First, A[1; 0] is the first column of A and A[0; 1] is the second column, so the unit square goes to the parallelogram with sides along the two columns. Second, the image of a straight segment is a straight segment, so transforming the vertices is enough.
Image of the unit square
Find the image of the unit square under A = [2 1; 1 3] and its area (Diagram 1).
Equation: Corners go to (0, 0), (2, 1), (3, 4) and (1, 3); box method: 12 - 7 = 5, and det A = 2(3) - 1(1) = 5
A shear keeps area
Apply S = [1 2; 0 1] to the rectangle with vertices (0, 0), (3, 0), (3, 1) and (0, 1).
Equation: Image (0, 0), (3, 0), (5, 1), (2, 1): a parallelogram with base 3 and height 1, area 3; det S = 1
Negative determinant
Apply M = [1 0; 0 -2] to the triangle with vertices (0, 0), (3, 0) and (0, 2), which has area 3.
Equation: Image (0, 0), (3, 0), (0, -4), area 6; det M = -2, so area doubles and the triangle is flipped across the x-axis
Zero determinant
Describe what P = [1 2; 1 2] does to the plane.
Equation: P[x; y] = [x + 2y; x + 2y], so every point lands on the line y = x; det P = 2 - 2 = 0 and every area becomes 0
Area of any figure
The triangle with vertices (1, 1), (4, 1) and (1, 3) has area 3. Find the area of its image under [3 1; -1 2].
Equation: det = 6 - (-1) = 7, so the image has area 7 × 3 = 21 (check: the images (4, 1), (13, -2) and (6, 5) give area 21)
Part 2: The determinant as area. After the first example, walk through the box method in Diagram 1: the image parallelogram sits in a 3 × 4 box, and the six pieces outside it have total area 7, so the parallelogram has area 5. Then do the same with letters for A = [a b; c d] with positive entries: the box is (a + b) × (c + d), the pieces outside have area ac + bd + 2bc, and what is left is ad - bc. Explain the absolute value: when ad - bc is negative, as in the third example, the area is still |ad - bc|, and the sign only tells you that the figure was flipped. Use Diagram 2 to compare a shear (det 1), a reflection combined with a stretch (det -2) and a collapse (det 0). Close Part 2 with the key sentence: a 2 × 2 matrix multiplies every area in the plane by |det A|.
Guided Practice15-20 minutes
Pairs work through four tasks on grid paper and compare with another pair after each one.
(a) Find and sketch the image of the unit square under [3 -1; 1 2]. Find its area with the box method and with the determinant. (Corners (0, 0), (3, 1), (2, 3), (-1, 2); box 4 × 3 = 12 minus four triangles of total area 5 leaves 7; det = 6 + 1 = 7.)
(b) Describe [-3 0; 0 -3] and give its area factor. (It rotates the plane 180° and dilates it by 3; det 9, so areas are multiplied by 9 while lengths are multiplied by 3.)
(c) A region has area 4. Find the area of its image under [2 -3; 1 1]. (det 2 + 3 = 5, so 20.)
(d) Show that [3 6; 1 2] sends (1, 0) and (0, 1) to points on the same line through the origin. What is its determinant? ((3, 1) and (6, 2), both on y = x/3; det 6 - 6 = 0.)
Listen for these errors: computing ad + bc, forgetting the absolute value, multiplying area by the scale factor 3 instead of 9 in (b), and connecting image vertices in a different order than the original.
Independent Practice15 minutes
Students complete four problems on their own.
Transform the square with vertices (0, 0), (2, 0), (2, 2) and (0, 2) by [1 3; 0 2]. (Image (0, 0), (2, 0), (8, 4), (6, 4); det 2, so the area goes from 4 to 8.)
Find the area factor of [4 -2; 3 1]. (det 4 + 6 = 10.)
Decide whether [-1 2; 3 -6] can be undone. (det 6 - 6 = 0: it squashes the plane onto a line, so no matrix can undo it.)
Find every k for which [k 2; 3 4] triples areas. (|4k - 6| = 3, so k = 9/4 or k = 3/4.)
Closure5 minutes
Exit ticket: (1) Find det [5 2; 3 1] and say what the matrix does to areas and to orientation. (det = -1: areas stay the same and figures are flipped.) (2) A matrix has determinant -4. A circle of area 3π is transformed. What is the area of its image? (12π.) (3) In one sentence, what does a determinant of 0 mean for the picture?
Differentiation Strategies
For Struggling Students
Always start with the unit square: mark the images of (1, 0) and (0, 1) first, since they are just the columns of the matrix
Give a box-method template with the six outside pieces already outlined, so students only fill in the lengths
Use a free geometry app to drag a figure and watch its image and area change as the matrix entries change
For Advanced Students
Ask students to carry out the box-method derivation of ad - bc with letters, and to explain what changes when b is negative
Ask why applying A and then B multiplies areas by |det A| × |det B|, and test it on two matrices from the lesson (this previews det(BA) = det(B)det(A))
Ask for a matrix that doubles every area but is not a dilation, and one that keeps every area but changes every length
Assessment Guidance
What to Look For
Check that students transform figures by multiplying each vertex and connect the images in order, and that they can write the matrix of a described transformation from the images of (1, 0) and (0, 1). For area, look for the absolute value in the final answer, a correct determinant (ad - bc, not ad + bc), and an explanation of the area factor that refers to the unit square. Ask students who only compute to say what a negative or zero determinant looks like in a sketch.
02
Classroom Activities
3 Activities
1
Parallelogram Area Detective
20 minPairs
Pairs draw the image of the unit square under four matrices, find each area by boxing and subtracting, and look for a rule that gives the area from the matrix entries.
Cards
[3 1; 1 2], [4 2; 1 3], [2 3; 1 4] and [1 4; 2 1]
Expected areas: 5, 10, 5 and 7
Procedure
For each card, plot the images of (1, 0), (0, 1) and (1, 1), draw the parallelogram and the smallest box around it
Subtract the pieces outside the parallelogram from the box to get its area
Record a, b, c, d and the area in a table and look for a rule; test your rule on the last card, where ad - bc = 1 - 8 = -7
Discussion Questions
Why does the last card give a negative number, and how does its parallelogram differ from the others?
[3 1; 1 2] and [2 3; 1 4] have different entries but the same area. Are their parallelograms congruent?
Modification for Distance Learning
Students use a free geometry app to plot the parallelograms and its polygon tool to measure the areas, then share one screenshot per card.
2
Transformation Matching
15 minGroups of 3-4
Groups receive 8 cards: 4 matrices and 4 descriptions. They match each matrix to its description, then find the area of the image of an L-shaped figure made of 3 unit squares.
Description cards: reflection across the y-axis; dilation by one half; vertical shear; horizontal stretch by 3
Procedure
Test each matrix on (1, 0) and (0, 1) before choosing a description
Compute each determinant: -1, 0.25, 1 and 3
Use the determinants to find the area of the L-shape's image under each matrix: 3, 0.75, 3 and 9
Draw the image under one matrix and count squares to confirm
Discussion Questions
The dilation halves every length. Why does it multiply area by one quarter?
Which card changes the shape of the L but not its area?
3
Logo Designer
20 minGroups of 3
A design team has a triangular logo with vertices (0, 0), (4, 0) and (1, 3), with area 6. Each group must find a matrix that meets a client's request and prove that it works.
Requests
Request 1: a mirror image of the logo with exactly three times the area (one answer: [-3 0; 0 1], with det -3, giving the image (0, 0), (-12, 0), (-3, 3) and area 18)
Request 2: a slanted version with the same area and no flip (one answer: a shear such as [1 1; 0 1])
Request 3: a version with area 24 that is not a dilation
Procedure
Propose a matrix, compute its determinant and predict the image's area and orientation
Transform the three vertices, draw the image and check the area with coordinates
Present one request to the class, explaining how the determinant guided the choice
Challenge Variation
Going further: ask groups to find a matrix for Request 1 that also rotates the logo, and to explain why many different matrices meet the same area request.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: The Unit Square, Its Image and the Box Method
A = [2 1; 1 3] sends the unit square (gray) to the parallelogram with corners (0, 0), (2, 1), (3, 4) and (1, 3), drawn to scale. The dashed pieces between the parallelogram and the 3 × 4 box have total area 7, so the parallelogram has area 12 - 7 = 5, which equals det A.
Diagram 2: Three Matrices and What Their Determinants Say
Images of the unit square, drawn to scale. The shear keeps area. The matrix [-2 0; 0 1] doubles area and flips the square: P' is now to the left of Q'. The matrix [1 2; 1 2] sends the whole plane onto the line y = x, so every area becomes 0.
04
Homework Assignment
~30 min
HSN.VM.C.12 Homework: Matrix Transformations and Area
Directions: Use grid paper. For every transformation, list the image of each vertex and sketch the figure and its image. Write matrices row by row with semicolons between rows. For every area, show the determinant and the absolute value.
Part 1: Transformations of the Plane (Problems 1-3)
Find the image of the triangle with vertices (1, 0), (3, 1) and (2, 4) under [2 0; 0 -1]. Sketch both triangles and describe the transformation in words.
(a) Find the matrix that sends (1, 0) to (1, 2) and (0, 1) to (3, 1), and list the images of the four corners of the unit square. (b) Write the matrix of a rotation of 90° clockwise about the origin and explain how you found it.
Apply the vertical shear [1 0; 3 1] to the rectangle with vertices (0, 0), (2, 0), (2, 1) and (0, 1). Sketch the image and explain, using base and height, why its area equals the area of the rectangle. Confirm with the determinant.
Part 2: The Determinant and Area (Problems 4-6)
For each matrix, find the determinant and the area of the image of the unit square: (a) [5 2; 2 3] (b) [1 -3; 2 4] (c) [2 -4; -1 2]. Describe what happens to the plane in (c).
A flowerbed on a design grid has area 12 square meters. The designer can enlarge it with [3 1; 0 2] or with [1 4; 2 3]. (a) Find the area of each enlarged flowerbed. (b) Which matrix also flips the design, and how do you know?
The triangle T has vertices (0, 0), (3, 1) and (1, 2). (a) Find its area. (b) Find the image of T under [2 1; -1 1]. (c) Find the area of the image in two ways: from its coordinates and from the determinant.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Images of Figures
Every vertex image correct and sketched in order
One or two vertex errors
Images missing or incorrect
Matrices of Transformations
Matrices found from the images of (1, 0) and (0, 1) and described correctly
Matrix correct without reasoning, or description incomplete
Missing or incorrect
Determinant and Area
Determinants correct and absolute value used for every area
Correct determinants, one area error or missing absolute value
Determinants incorrect
Interpretation
Negative and zero determinants explained with the sketch
Partly explained
No interpretation
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again. Matrices are written row by row, as in the lesson plan, and the point (x, y) is the vector [x; y].
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
What is the determinant of [4 3; 2 5]?
Answer: B
det = ad - bc = 4(5) - 3(2) = 20 - 6 = 14. Choice A adds the two products instead of subtracting. Choice C subtracts in the wrong order, bc - ad.
Question 2 of 20 · Multiple Choice
Where does the matrix [3 1; 0 2] send the point (2, -1)?
Answer: D
[3 1; 0 2][2; -1] = [3(2) + 1(-1); 0(2) + 2(-1)] = [5; -2], so the image is (5, -2). Choice A multiplies only the diagonal entries. Choice C loses the sign in the second row.
Question 3 of 20 · Multiple Choice
The unit square is transformed by [4 1; 2 3]. What is the area of its image?
Answer: A
The image is the parallelogram on the columns (4, 2) and (1, 3), and its area is |4(3) - 1(2)| = 10. Choice B uses ad + bc. Choice C is ad alone, the area of a box that is too big.
Question 4 of 20 · Multiple Choice
Which matrix reflects the plane across the x-axis?
Answer: C
A reflection across the x-axis keeps (1, 0) and sends (0, 1) to (0, -1), so its columns are [1; 0] and [0; -1]. Choice A reflects across the y-axis, choice B across the line y = x, and choice D rotates by 180°.
Question 5 of 20 · Multiple Choice
A region with area 5 is transformed by a matrix whose determinant is -3. What is the area of the image?
Answer: B
The area is multiplied by |det| = 3, so the image has area 15. An area cannot be negative (choice A): the negative sign only means the region is flipped. Choice C divides instead of multiplying, and choice D gives the area factor, not the area.
Question 6 of 20 · Multiple Choice
Which matrix sends the whole plane onto a line?
Answer: D
[2 -6; -1 3] has determinant 2(3) - (-6)(-1) = 6 - 6 = 0, and its second column is -3 times its first, so every image point is a multiple of [2; -1]. Choice B looks similar, but its determinant is 6 + 6 = 12, so it only stretches the plane. Choices A and C have determinants 2 and 14.
Question 7 of 20 · Multiple Choice
A matrix has determinant 1. What must be true about the transformation?
Answer: A
|det| = 1 means areas are unchanged. It does not force lengths or angles to stay the same: the shear [1 2; 0 1] has determinant 1 but changes both, which rules out choices B, C and D.
Question 8 of 20 · Multiple Choice
The shear [1 5; 0 1] is applied to a rectangle with area 8. What is the area of the image?
Answer: C
det = 1(1) - 5(0) = 1, so the area stays 8. The shear slides each horizontal line sideways, which keeps the base and height. Choice A multiplies by the entry 5, and choice B adds 5.
Question 9 of 20 · Multiple Choice
Under [2 -1; 3 1], what is the image of the corner (1, 1) of the unit square?
Answer: A
[2 -1; 3 1][1; 1] = [2 - 1; 3 + 1] = [1; 4]. This corner goes to the sum of the two columns. Choice B is the image of (1, 0), the first column, and choice D is the image of (0, 1).
Question 10 of 20 · Multiple Choice
What is the determinant of the rotation matrix [0 -1; 1 0], and what does it say?
Answer: C
det = 0(0) - (-1)(1) = 1, so a rotation keeps every area, and the positive sign shows it does not flip figures. Choice A comes from computing 0 - 1 and dropping the sign of -1.
Question 11 of 20 · Multiple Choice
For which value of k does [k 2; 3 6] send the whole plane onto a line?
Answer: B
The plane collapses when det = 6k - 6 = 0, so k = 1. Then the columns [1; 3] and [2; 6] point the same way. Choice A comes from 6k - 6 = 18, and choice C from multiplying 3 by 3 instead of solving.
Question 12 of 20 · Multiple Choice
A triangle has area 5. What is the area of its image under [4 0; 0 4]?
Answer: D
det = 16, so the area becomes 16 × 5 = 80. The matrix multiplies every length by 4, and area grows by the square of that factor. Choice A multiplies the area by 4, the length factor, instead of 16. Choice B multiplies by 4 + 4 = 8, and choice C adds 4 to the area.
Question 13 of 20 · Multiple Choice
A matrix has determinant -8. Which statement is true?
Answer: A
|det| = 8 multiplies areas by 8, and the negative sign means figures are flipped, as in a reflection. Choice C ignores the sign, and choice D treats the sign as part of the area.
Question 14 of 20 · Multiple Choice
Which description fits [0 3; 3 0]?
Answer: C
[0 3; 3 0] = 3[0 1; 1 0]: swap the coordinates (reflect across y = x) and then triple them. det = 0 - 9 = -9, so areas are multiplied by 9 and orientation flips. Choice A misses the flip and uses the length factor for area.
Question 15 of 20 · Short Answer
Find the image of the triangle with vertices (0, 0), (2, 0) and (0, 3) under [1 2; -1 1]. Find the area of the triangle and of its image.
Images: (0, 0), (2, -2) and (6, 3). The original area is (1/2)(2)(3) = 3. det = 1 + 2 = 3, so the image has area 9; from coordinates, (1/2)|2(3) - (-2)(6)| = (1/2)(18) = 9.
Question 16 of 20 · Short Answer
A matrix M sends (1, 0) to (4, -1) and (0, 1) to (2, 3). Write M and find the area of the image of the unit square.
The images are the columns, so M = [4 2; -1 3]. det M = 12 - (-2) = 14, so the image parallelogram has area 14.
Question 17 of 20 · Short Answer
Show that [2 -1; -4 2] sends (1, 0), (0, 1) and (1, 1) to points on one line through the origin. Find the line and the determinant.
Images: (2, -4), (-1, 2) and (1, -2). Each has y = -2x, so all three lie on the line y = -2x. det = 4 - 4 = 0, which is why the plane collapses onto a line.
Question 18 of 20 · Short Answer
Use the box method to find the area of the parallelogram with vertices (0, 0), (4, 1), (5, 4) and (1, 3). Then confirm with a determinant.
The box is 5 × 4 = 20. Outside pieces: two triangles of area (1/2)(4)(1) = 2, two of area (1/2)(1)(3) = 1.5 and two 1 × 1 rectangles, total 9. Area = 20 - 9 = 11. The sides are the columns of [4 1; 1 3], and det = 12 - 1 = 11.
Question 19 of 20 · Short Answer
Find every value of k for which [k 1; 2 3] multiplies areas by 7.
We need |3k - 2| = 7. Either 3k - 2 = 7, so k = 3, or 3k - 2 = -7, so k = -5/3. With k = -5/3 the determinant is -7, so figures are also flipped.
Question 20 of 20 · Short Answer
The matrix [-2 1; 0 3] transforms a circle of radius 1. The image is an ellipse. Find its area, and say whether the matrix flips figures.
det = -6 - 0 = -6. The circle's area π is multiplied by |-6| = 6, so the ellipse has area 6π (about 18.85). The determinant is negative, so the matrix flips figures.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSN.VM.C.12 mean?
HSN.VM.C.12 asks students to treat a 2 × 2 matrix as a transformation of the whole plane and to connect its determinant to area. Each point (x, y) is the vector [x; y], and the matrix sends it to a new point. For A = [a b; c d], the number |ad - bc| is the area of the image of the unit square, and the matrix multiplies the area of every figure by that same number.
Is HSN.VM.C.12 taught in Precalculus?
Yes, it is usually taught in Precalculus. The (+) marks it as additional mathematics for students who take advanced courses. It follows naturally from multiplying vectors by matrices (HSN.VM.C.11) and connects to geometry, where students first studied rotations, reflections and dilations with coordinates.
How does a 2 × 2 matrix transform the plane?
It sends every point [x; y] to A[x; y]. The origin stays fixed, straight lines go to straight lines, and parallel lines stay parallel. The images of (1, 0) and (0, 1) are the two columns of the matrix, so the grid of unit squares becomes a grid of parallelograms built on those columns. This is why transforming the vertices of a polygon is enough to draw its image.
What is the determinant of a 2 × 2 matrix?
For A = [a b; c d], det A = ad - bc: the product of the main diagonal minus the product of the other diagonal. For example, det [2 1; 1 3] = 6 - 1 = 5. Geometrically, |det A| is the area of the parallelogram built on the columns of A, and the sign tells you whether the matrix keeps or reverses orientation.
Why is the area of the image of the unit square equal to |ad - bc|?
Draw the image parallelogram inside the smallest rectangle around it. With positive entries, the rectangle is (a + b) by (c + d). The pieces between the parallelogram and the rectangle are two triangles of area ac/2, two of area bd/2 and two rectangles of area bc. Subtracting gives (a + b)(c + d) - ac - bd - 2bc = ad - bc. Other sign cases work the same way and give the absolute value.
Why does the standard say the absolute value of the determinant?
Because area is never negative, while a determinant can be. A negative determinant, such as -2 for [-2 0; 0 1], means the matrix flips figures, the way a reflection does: vertices that went counterclockwise now go clockwise. The area factor is still |-2| = 2.
What does a determinant of 0 mean?
The matrix squashes the whole plane onto a line through the origin (or onto the origin itself, for the zero matrix). Its columns point along the same line, so the unit square becomes a segment with area 0. Such a matrix cannot be undone, which is why a matrix has an inverse exactly when its determinant is not 0 (HSN.VM.C.10).
Does the area rule work for circles and other curved shapes?
Yes. Any region can be filled, as closely as you like, with small squares, and the matrix multiplies the area of each small square by |det A|. So the whole region's area is multiplied by the same factor. For example, [2 0; 0 5] turns a circle of area π into an ellipse of area 10π.
What are common mistakes with matrix transformations and area?
A common error is computing ad + bc or bc - ad instead of ad - bc. Others are reporting a negative area, writing the images of (1, 0) and (0, 1) as rows instead of columns, and assuming that a matrix that triples lengths triples area (it multiplies area by 9). Students also sometimes connect image vertices in a different order from the original figure.
How does HSN.VM.C.12 connect to other standards?
It extends HSN.VM.C.11 from single vectors to the whole plane, and it ties matrices to the transformations of geometry (8.G.A.3 and HSG.CO.A.2). The determinant it interprets is the same number that decides whether a matrix has an inverse (HSN.VM.C.10) and therefore whether a system can be solved with an inverse matrix (HSA.REI.C.9).
07
Related Standards
5 standards
These standards connect to HSN.VM.C.12: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSN.VM.C.11Prerequisite
Multiply a vector by a matrix; work with matrices as transformations of vectors
