Projections onto coordinate axes | Year 11

01 / LEARN

Understand the idea.

Keep one component, discard the other

Orthogonal projection onto the $x$ -axis keeps the horizontal coordinate and replaces the vertical coordinate with zero: $(x, y) \mapsto (x, 0)$ . Onto the $y$ -axis it gives $(0, y)$ . The removed displacement is perpendicular to the target axis.

Projection and reflection behave differently. Projection discards the perpendicular component; reflection reverses it. Points on the target axis are fixed by both maps, but points off that axis distinguish the two.

Projection is idempotent

The coordinate projection matrices keep one component and remove the other. Applying the same projection twice changes nothing further: $P_{x}^{2} = P_{x}$ and $P_{y}^{2} = P_{y}$ . This property is called idempotence.

A coordinate projection is singular because many inputs share the same retained component. The diagram shows the triangle collapsing onto the $x$ -axis; the output alone cannot recover the original vertical coordinates, so no inverse exists.

An equal-scale unit grid shows a gold triangle with vertices zero zero, two zero and one two. Its sage x-axis projection is the segment from zero zero to two zero, with the middle projected vertex at one zero.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Two coordinate projections

Project $(3, - 4)$ onto the x-axis and $(- 2, 5)$ onto the y-axis.

Worked solution

1
Write the two projection matrices.
$P_{x} = (\begin{array}{cccccccccc} 1 & 0 \\ 0 & 0 \end{array}), P_{y} = (\begin{array}{cccccccccc} 0 & 0 \\ 0 & 1 \end{array})$
The diagonal 1 keeps a coordinate and the zero row removes the other.
2
Project the first point onto the horizontal axis.
$\begin{aligned} P_{x} (\begin{array}{cccccccccc} 3 \\ - 4 \end{array}) & = (\begin{array}{cccccccccc} 1 (3) + 0 (- 4) \\ 0 (3) + 0 (- 4) \end{array}) \\ = (\begin{array}{cccccccccc} 3 \\ 0 \end{array}) \end{aligned}$
The horizontal coordinate stays 3 while the vertical component is discarded.
3
Project the second point onto the vertical axis.
$\begin{aligned} P_{y} (\begin{array}{cccccccccc} - 2 \\ 5 \end{array}) & = (\begin{array}{cccccccccc} 0 (- 2) + 0 (5) \\ 0 (- 2) + 1 (5) \end{array}) \\ = (\begin{array}{cccccccccc} 0 \\ 5 \end{array}) \end{aligned}$
The vertical coordinate stays 5 while the horizontal component is discarded.

Answer

$(3, 0)$ and $(0, 5)$ .

EXAMPLE 02

Combine projections

Find $P_{x} + P_{y}$ and $P_{x} P_{y}$ .

Worked solution

1
Add corresponding entries.
$P_{x} + P_{y} = (\begin{array}{cccccccccc} 1 + 0 & 0 + 0 \\ 0 + 0 & 0 + 1 \end{array}) = (\begin{array}{cccccccccc} 1 & 0 \\ 0 & 1 \end{array}) = I$
The two separate projections supply the horizontal and vertical components respectively.
2
Multiply in the stated order.
$\begin{aligned} P_{x} P_{y} & = (\begin{array}{cccccccccc} 1 & 0 \\ 0 & 0 \end{array}) (\begin{array}{cccccccccc} 0 & 0 \\ 0 & 1 \end{array}) \\ = (\begin{array}{cccccccccc} 0 & 0 \\ 0 & 0 \end{array}) \end{aligned}$
Every row-column dot product is zero.
3
Interpret the product on any point.
$(x, y) → P_{y} (0, y) → P_{x} (0, 0)$
Keeping only the vertical component first leaves no horizontal component for the second projection to retain.

Answer

$P_{x} + P_{y} = I$ and $P_{x} P_{y} = 0$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Find $P_{y} {(7, - 2)}^{T}$ .

Show answer & reasoning

${(0, - 2)}^{T}$ .

$P_{y}$ removes the horizontal component and keeps the vertical one. Therefore $P_{y} {(7, - 2)}^{T} = {(0, - 2)}^{T}$ ; the negative sign on the retained coordinate remains unchanged.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Give the matrix for projection onto the $y$ -axis.
2 marks
Find $P_{x} {(- 5, 6)}^{T}$ .
2 marks
Find all points whose $x$ -axis projection is $(2, 0)$ .
2 marks
Calculate $P_{y}^{2}$ .
2 marks
Explain why a coordinate projection has no inverse.
2 marks
Compare the $x$ -axis projection and $x$ -axis reflection of $(3, 4)$ .
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Give the matrix for projection onto the $y$ -axis.
2 marks
The unit-vector images are $(0, 0)$ and $(0, 1)$ .
$(\begin{array}{cccccccccc} 0 & 0 \\ 0 & 1 \end{array})$ .
Find $P_{x} {(- 5, 6)}^{T}$ .
2 marks
Keep $x$ and remove $y$ .
${(- 5, 0)}^{T}$ .
Find all points whose $x$ -axis projection is $(2, 0)$ .
2 marks
The $x$ -coordinate must be 2; the $y$ -coordinate is unrestricted.
$(2, t)$ for any real $t$ .
Calculate $P_{y}^{2}$ .
2 marks
Multiply the two diagonal projection matrices.
The surviving diagonal entry remains 1.
$P_{y}$ .
Explain why a coordinate projection has no inverse.
2 marks
Many inputs have the same image, and the determinant is zero.
It loses one coordinate.
Compare the $x$ -axis projection and $x$ -axis reflection of $(3, 4)$ .
2 marks
Projection discards $y$ ; reflection reverses $y$ .
Projection $(3, 0)$ ; reflection $(3, - 4)$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: matrices and transformations; original lessons follow the approved school-sized chapter structure.

Official curriculum source ↗

In this lesson

Understand the idea.

Keep one component, discard the other

Projection is idempotent

See the method in action.

Two coordinate projections

Write the two projection matrices.

Project the first point onto the horizontal axis.

Project the second point onto the vertical axis.

Combine projections

Add corresponding entries.

Multiply in the stated order.

Interpret the product on any point.

Try it yourself.

Check your understanding

Keep practising