LINEAR-ALGEBRA · Interactive Practice | Unit 29 · Video 1

The Two Rules That Define Linearity

IKey Formulas

Formula	Name	Description
$T(v + w) = T(v) + T(w)$	Additivity	Transforming a sum equals summing the transforms
$T(cv) = c\,T(v)$	Homogeneity	Scaling commutes with the transformation
$T(cv + dw) = c\,T(v) + d\,T(w)$	Unified linearity rule	Both rules in one equation
$T(\mathbf{0}) = \mathbf{0}$	Zero diagnostic	Required for linearity (fastest check)

IIVisualization 1 — The Linearity Test

Linearity demands $T(cv + dw) = c\,T(v) + d\,T(w)$ for every choice — do both sides land on the same vector?

Transform T Scalar c 1.5 Scalar d 1

INPUT — combine first

OUTPUT — T(cv+dw) vs cT(v)+dT(w)

💡 Every $c$ and $d$ must agree — a single disagreeing pair is enough to prove $T$ is not linear.

IIIVisualization 2 — The Zero-Vector Diagnostic

A linear map must fix the origin; a shift drags it to $(1, 2)$ and fails on sight.

Transform T t 1.00

💡 Passing the diagnostic can't prove linearity — but failing it disproves linearity instantly.

IVVisualization 3 — Every Linear Map Is a Matrix

Every linear map is a matrix: its columns are just where the basis vectors $\hat{\imath}$ and $\hat{\jmath}$ land.

💡 $\det A$ is the signed area of the transformed unit square — a negative value means the map flips orientation, and $0$ collapses the plane onto a line.

VQuiz Questions

Question 1

Which single equation captures both the additivity and homogeneity rules of linearity?

✅ Correct! This unified equation encodes both rules at once.

❌ Not quite. Look for the form that scales each vector by its scalar before transforming, then keeps those scalars on the right side.

Show solution

Solution:

Additivity says $T(v + w) = T(v) + T(w)$ and homogeneity says $T(cv) = c\,T(v)$ . Combining them:

$T(cv + dw) = T(cv) + T(dw) = c\,T(v) + d\,T(w)$

This single equation, required to hold for every choice of vectors $v, w$ and scalars $c, d$ , is the complete definition of linearity.

Question 2

Consider the shift transformation $T(v) = v + v_0$ where $v_0 = (1, 2)$ is a fixed nonzero vector. Which statement is true?

✅ Correct! The fastest diagnostic — origin must map to origin — rules shifts out immediately.

❌ Not quite. Apply the zero-vector diagnostic: what is $T(\mathbf{0})$ ?

Show solution

Solution:

Apply the zero-vector diagnostic:

$T(\mathbf{0}) = \mathbf{0} + v_0 = (1, 2) \neq \mathbf{0}$

Since linear transformations must send $\mathbf{0}$ to $\mathbf{0}$ , $T$ cannot be linear.

You can also see the failure of additivity directly: $T(v + w) = v + w + v_0 \quad \text{but} \quad T(v) + T(w) = v + w + 2v_0$

An extra $v_0$ appears on the right. Shifts are called affine, not linear.

Question 3

Let $T(v) = \|v\|$ (the norm/length of $v$ ). Suppose $v$ is a vector with $\|v\| = 3$ . Computing $T(-2v)$ and $-2 \cdot T(v)$ gives:

✅ Correct! Norm is non-negative, so it can't track the sign flip — homogeneity fails.

❌ Not quite. Remember that $\|-2v\| = 2\|v\|$ (always non-negative), but $-2 \cdot \|v\|$ keeps the negative sign.

Show solution

Solution:

Norm is always non-negative, so: $T(-2v) = \|-2v\| = 2\|v\| = 2 \cdot 3 = 6$

But scaling the output by $-2$ : $-2 \cdot T(v) = -2 \cdot \|v\| = -2 \cdot 3 = -6$

Since $6 \neq -6$ , homogeneity fails for the negative scalar $c = -2$ . Therefore norm is not a linear transformation, even though $T(\mathbf{0}) = 0$ passes the diagnostic. Negative scalars expose the failure.

Question 4

True or False: If $T(\mathbf{0}) = \mathbf{0}$ , then $T$ must be linear.

✅ Correct! The diagnostic is necessary but not sufficient — norm is a clean counterexample.

❌ Not quite. Can you think of a transformation that fixes the origin but still violates one of the two linearity rules? (Hint: think about norm.)

Show solution

Solution:

False. The condition $T(\mathbf{0}) = \mathbf{0}$ is necessary but not sufficient for linearity.

Counterexample: the norm $T(v) = \|v\|$ satisfies $T(\mathbf{0}) = 0$ , but as we saw, it fails homogeneity for negative scalars. So $T$ is not linear.

The zero diagnostic is a fast way to rule out linearity (if $T(\mathbf{0}) \neq \mathbf{0}$ , definitely not linear), but to confirm linearity you must verify $T(cv + dw) = c\,T(v) + d\,T(w)$ for all vectors and scalars.

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