A linear map T: ℝ⁶ → ℝ⁴ has rank 3. What is the nullity of T?
A1
B3
C4
D7
The rank-nullity theorem says rank(T) + nullity(T) = dim(domain) = 6. With rank 3, nullity = 6 − 3 = 3. A common error is subtracting from the codomain dimension (4 − 3 = 1), but the theorem accounts for the *input* space, not the output space. The codomain dimension is irrelevant here — what matters is how the 6-dimensional input is divided between what gets erased (kernel) and what survives (image).
Question 2 Multiple Choice
A linear transformation T: ℝ⁴ → ℝ⁶ is injective (one-to-one). What must be true about its rank and nullity?
ARank 6 and nullity 0 — the map fills the codomain
BRank 4 and nullity 0 — the map is injective exactly when the kernel is trivial
CRank 4 and nullity 2 — some dimensions are lost in the larger codomain
DT cannot be injective because the codomain has higher dimension than the domain
Injectivity means no two distinct inputs map to the same output, which is equivalent to the kernel being trivial (nullity = 0). By rank-nullity: rank = 4 − 0 = 4. The image has dimension 4, which sits inside ℝ⁶ — the map can be injective even though it doesn't fill the codomain (that would require surjectivity, which would need rank 6, impossible here). The common confusion is conflating injectivity with surjectivity.
Question 3 True / False
If a linear map T: ℝ⁵ → ℝ³ has rank 3, then T is surjective (onto).
TTrue
FFalse
Answer: True
Surjectivity means the image equals the entire codomain. Here the codomain is ℝ³, dimension 3. If rank(T) = 3 = dim(ℝ³), the image spans the entire codomain — T is surjective. By rank-nullity, nullity = 5 − 3 = 2, meaning a 2-dimensional subspace of ℝ⁵ is collapsed to zero, but the remaining 3 dimensions map onto all of ℝ³.
Question 4 True / False
A linear map T: ℝ³ → ℝ⁵ can be both injective and surjective.
TTrue
FFalse
Answer: False
For T to be bijective (both injective and surjective), it would need rank equal to both dim(domain) = 3 and dim(codomain) = 5 simultaneously — impossible. Surjectivity requires rank = 5, but rank-nullity says rank ≤ dim(domain) = 3. So T can be at most injective (rank 3, nullity 0), but it cannot be surjective. A bijection between finite-dimensional spaces requires equal dimensions.
Question 5 Short Answer
In your own words, explain what the rank-nullity theorem says about where the dimensions of a linear map's input space 'go.'
Think about your answer, then reveal below.
Model answer: The theorem says the input dimension is completely accounted for by two subspaces: the kernel (vectors that get 'erased' — mapped to zero) and the image (vectors that 'survive' — produce distinct nonzero outputs). Their dimensions sum exactly to the input dimension. Nothing is counted twice, and nothing is missed. It is a conservation law for dimension.
The intuition of 'erased vs. survived' is what makes the theorem memorable and useful. Every dimension of the input either gets crushed into the kernel or contributes a dimension to the image. This directly constrains what is possible: a map from a large space to a small one must have a nontrivial kernel (some information is always lost), and a map from a small space to a large one cannot fill the codomain (some output directions are unreachable).