Intuition The one-line soul of the theoremA linear map T : V → W T:V\to W T : V → W n n n two jobs : some dimensions get crushed to zero (the kernel), and the rest get faithfully shipped out (the image). No dimension is lost or created — it just gets sorted. So
dim ( ker T ) ⏟ crushed + dim ( im T ) ⏟ shipped = dim V ⏟ total input . \underbrace{\dim(\ker T)}_{\text{crushed}} + \underbrace{\dim(\operatorname{im} T)}_{\text{shipped}} = \underbrace{\dim V}_{\text{total input}}. crushed dim ( ker T ) + shipped dim ( im T ) = total input dim V .
Definition Rank–Nullity TheoremLet V V V finite-dimensional vector space over a field F F F T : V → W T:V\to W T : V → W rank ( T ) + nullity ( T ) = dim V \operatorname{rank}(T) + \operatorname{nullity}(T) = \dim V rank ( T ) + nullity ( T ) = dim V
Rank = dim ( im T ) =\dim(\operatorname{im} T) = dim ( im T ) image (range) of T T T Nullity = dim ( ker T ) =\dim(\ker T) = dim ( ker T ) kernel (null space) of T T T { v ∈ V : T ( v ) = 0 } \{v\in V : T(v)=0\} { v ∈ V : T ( v ) = 0 }
ker T \ker T ker T V V V input side).im T \operatorname{im} T im T W W W output side).Note both numbers add up to dim V \dim V dim V the dimension of the domain , not the codomain W W W
Intuition Conservation of dimensionsPicture the basis of V V V n n n T T T
Some combinations collapse to 0 0 0 ker T \ker T ker T k k k
The remaining n − k n-k n − k independently in the output (if any survivor were dependent on the others, it would secretly belong to the kernel). Those n − k n-k n − k span and are independent in im T \operatorname{im} T im T
So image dimension = n − k = n-k = n − k = k =k = k n n n
We build it carefully. The key trick is the Basis Extension Theorem : any basis of a subspace can be extended to a basis of the whole space.
Setup. Let dim V = n \dim V = n dim V = n nullity ( T ) = k \operatorname{nullity}(T)=k nullity ( T ) = k { u 1 , … , u k } a basis of ker T . \{u_1,\dots,u_k\} \quad\text{a basis of } \ker T. { u 1 , … , u k } a basis of ker T .
Step 1 — Extend. Since ker T ⊆ V \ker T \subseteq V ker T ⊆ V V V V { u 1 , … , u k , v 1 , … , v n − k } a basis of V . \{u_1,\dots,u_k,\; v_1,\dots,v_{n-k}\} \quad\text{a basis of } V. { u 1 , … , u k , v 1 , … , v n − k } a basis of V . Why this step? The basis-extension theorem guarantees we can add n − k n-k n − k V V V
Claim: B = { T ( v 1 ) , … , T ( v n − k ) } \;B=\{T(v_1),\dots,T(v_{n-k})\} B = { T ( v 1 ) , … , T ( v n − k )} im T \operatorname{im} T im T dim ( im T ) = n − k \dim(\operatorname{im} T)=n-k dim ( im T ) = n − k rank + nullity = ( n − k ) + k = n = dim V . ■ \operatorname{rank}+\operatorname{nullity}=(n-k)+k=n=\dim V. \quad\blacksquare rank + nullity = ( n − k ) + k = n = dim V . ■
So we must show B B B (a) spans im T \operatorname{im} T im T (b) is linearly independent .
Step 2 — B B B im T \operatorname{im} T im T
Take any w ∈ im T w\in \operatorname{im} T w ∈ im T w = T ( v ) w=T(v) w = T ( v ) v ∈ V v\in V v ∈ V v v v v = ∑ i = 1 k a i u i + ∑ j = 1 n − k b j v j . v=\sum_{i=1}^{k}a_i u_i+\sum_{j=1}^{n-k}b_j v_j. v = ∑ i = 1 k a i u i + ∑ j = 1 n − k b j v j . T T T T ( u i ) = 0 T(u_i)=0 T ( u i ) = 0 u i ∈ ker T u_i\in\ker T u i ∈ ker T w = T ( v ) = ∑ a i T ( u i ) ⏟ = 0 + ∑ b j T ( v j ) = ∑ j = 1 n − k b j T ( v j ) . w=T(v)=\sum a_i\underbrace{T(u_i)}_{=0}+\sum b_j T(v_j)=\sum_{j=1}^{n-k} b_j\,T(v_j). w = T ( v ) = ∑ a i = 0 T ( u i ) + ∑ b j T ( v j ) = ∑ j = 1 n − k b j T ( v j ) . Why this step? Every output is built only from the survivors T ( v j ) T(v_j) T ( v j ) B B B
Step 3 — B B B
Suppose ∑ j = 1 n − k c j T ( v j ) = 0 \sum_{j=1}^{n-k} c_j\,T(v_j)=0 ∑ j = 1 n − k c j T ( v j ) = 0 T ( ∑ j c j v j ) = 0 ⟹ ∑ j c j v j ∈ ker T . T\!\left(\sum_j c_j v_j\right)=0 \;\Longrightarrow\; \sum_j c_j v_j \in \ker T. T ( ∑ j c j v j ) = 0 ⟹ ∑ j c j v j ∈ ker T . Why this step? If a combination of the survivor-outputs is zero, then the matching combination of inputs lives in the kernel — pulling us back to the u i u_i u i
Since { u 1 , … , u k } \{u_1,\dots,u_k\} { u 1 , … , u k } ker T \ker T ker T ∑ j c j v j = ∑ i d i u i ⟹ ∑ j c j v j − ∑ i d i u i = 0. \sum_j c_j v_j=\sum_i d_i u_i \;\Longrightarrow\; \sum_j c_j v_j-\sum_i d_i u_i=0. ∑ j c j v j = ∑ i d i u i ⟹ ∑ j c j v j − ∑ i d i u i = 0. { u 1 , … , u k , v 1 , … , v n − k } \{u_1,\dots,u_k,v_1,\dots,v_{n-k}\} { u 1 , … , u k , v 1 , … , v n − k } V V V linearly independent , so all coefficients vanish:
c 1 = ⋯ = c n − k = 0 ( and d i = 0 ) . c_1=\dots=c_{n-k}=0\;(\text{and }d_i=0). c 1 = ⋯ = c n − k = 0 ( and d i = 0 ) . Why this step? The independence of the whole basis of V V V c j c_j c j
Both (a) and (b) hold, so B B B im T \operatorname{im} T im T n − k n-k n − k
Worked example Example 1 — a projectionR 3 → R 3 \mathbb R^3\to\mathbb R^3 R 3 → R 3 T ( x , y , z ) = ( x , y , 0 ) T(x,y,z)=(x,y,0) T ( x , y , z ) = ( x , y , 0 ) z z z
Kernel: T ( v ) = 0 T(v)=0 T ( v ) = 0 x = y = 0 x=y=0 x = y = 0 z z z ⇒ ker T = { ( 0 , 0 , z ) } \Rightarrow \ker T=\{(0,0,z)\} ⇒ ker T = {( 0 , 0 , z )} nullity = 1 \operatorname{nullity}=1 nullity = 1 Why? Only the z z z Image: outputs are all ( x , y , 0 ) (x,y,0) ( x , y , 0 ) x y xy x y ⇒ rank = 2 \Rightarrow \operatorname{rank}=2 ⇒ rank = 2 Why? Two independent survivors T ( e 1 ) , T ( e 2 ) T(e_1),T(e_2) T ( e 1 ) , T ( e 2 ) Check: 2 + 1 = 3 = dim R 3 2+1=3=\dim\mathbb R^3 2 + 1 = 3 = dim R 3
Worked example Example 2 — a2 × 3 2\times 3 2 × 3 A = ( 1 2 3 2 4 6 ) A=\begin{pmatrix}1&2&3\\2&4&6\end{pmatrix} A = ( 1 2 2 4 3 6 ) T : R 3 → R 2 T:\mathbb R^3\to\mathbb R^2 T : R 3 → R 2
Row reduce: row 2 = 2 × = 2\times = 2 × 1 pivot ⇒ rank = 1 \Rightarrow \operatorname{rank}=1 ⇒ rank = 1 Why? Rank = number of independent rows/pivots = dimension of column space (the image).
Rank–nullity: nullity = dim V − rank = 3 − 1 = 2 \operatorname{nullity}=\dim V-\operatorname{rank}=3-1=2 nullity = dim V − rank = 3 − 1 = 2 Why? We get nullity for free without solving the system — that is the theorem's power.
Verify: A x = 0 ⇒ x 1 + 2 x 2 + 3 x 3 = 0 Ax=0\Rightarrow x_1+2x_2+3x_3=0 A x = 0 ⇒ x 1 + 2 x 2 + 3 x 3 = 0 3 3 3 ⇒ \Rightarrow ⇒ ⇒ \Rightarrow ⇒ 2 2 2
Worked example Example 3 — derivative on polynomialsV = P 3 V=P_3 V = P 3 ≤ 3 \le 3 ≤ 3 dim = 4 \dim=4 dim = 4 T = d d x : P 3 → P 3 T=\dfrac{d}{dx}:P_3\to P_3 T = d x d : P 3 → P 3
Kernel: p ′ = 0 p'=0 p ′ = 0 p p p ⇒ ker T = { constants } \Rightarrow\ker T=\{\text{constants}\} ⇒ ker T = { constants } = 1 =1 = 1 Image: derivatives of { 1 , x , x 2 , x 3 } \{1,x,x^2,x^3\} { 1 , x , x 2 , x 3 } { 0 , 1 , 2 x , 3 x 2 } \{0,1,2x,3x^2\} { 0 , 1 , 2 x , 3 x 2 } P 2 P_2 P 2 ⇒ \Rightarrow ⇒ = 3 =3 = 3 Check: 3 + 1 = 4 = dim P 3 3+1=4=\dim P_3 3 + 1 = 4 = dim P 3 Why this matters: the theorem applies to abstract spaces, not just R n \mathbb R^n R n
Common mistake "It should equaldim W \dim W dim W Why it feels right: the image lives inside W W W W W W The fix: rank + nullity counts how the input V V V W W W W W W = dim V =\dim V = dim V domain ).
Common mistake "Just take the survivorsv j v_j v j V V V T ( v j ) T(v_j) T ( v j ) Why it feels right: linear maps of independent vectors sometimes stay independent.
The fix: Not automatic! A linear map can collapse independent vectors. We needed Step 3's argument (going into the kernel and using the full basis) precisely because independence is not preserved in general — it works here only because we excluded the kernel directions.
Common mistake "ForgettingV V V Why it feels right: the formula looks dimension-free.
The fix: the basis-extension step and finite counting need dim V < ∞ \dim V<\infty dim V < ∞
Recall Feynman: explain it to a 12-year-oldImagine a machine with 3 input levers (3 dimensions of input). When you push the levers, some pushes do nothing — the machine ignores them (those are the "kernel" levers). The pushes that do something make the output move. The big idea: every lever is either ignored or makes a unique new output . So (ignored levers) + (useful levers) = (total levers). That's the whole theorem!
Mnemonic Remember the formula"RaN = Domain" → R ank a nd N ullity sum to the Domain 's dimension.
Also: "Kernel kills, image keeps; together they count the input."
#flashcards/maths
What does the rank–nullity theorem state? For a linear map
T : V → W T:V\to W T : V → W with
V V V finite-dimensional,
rank ( T ) + nullity ( T ) = dim V \operatorname{rank}(T)+\operatorname{nullity}(T)=\dim V rank ( T ) + nullity ( T ) = dim V .
Rank is the dimension of which space? The image (range) of
T T T ,
dim ( im T ) \dim(\operatorname{im}T) dim ( im T ) .
Nullity is the dimension of which space? The kernel (null space) of
T T T ,
dim ( ker T ) \dim(\ker T) dim ( ker T ) .
The sum equals the dimension of the domain or the codomain? What theorem lets us extend a kernel basis to a basis of V V V The basis-extension theorem.
In the proof, what set is shown to be a basis of the image? { T ( v 1 ) , … , T ( v n − k ) } \{T(v_1),\dots,T(v_{n-k})\} { T ( v 1 ) , … , T ( v n − k )} , images of the basis vectors outside the kernel.
Why do the T ( v j ) T(v_j) T ( v j ) Any
v v v splits into kernel part (mapped to 0) plus the
v j v_j v j part, so every output is a combination of the
T ( v j ) T(v_j) T ( v j ) .
Why are the T ( v j ) T(v_j) T ( v j ) If
∑ c j T ( v j ) = 0 \sum c_j T(v_j)=0 ∑ c j T ( v j ) = 0 then
∑ c j v j ∈ ker T \sum c_j v_j\in\ker T ∑ c j v j ∈ ker T ; writing it via the
u i u_i u i and using full-basis independence forces all
c j = 0 c_j=0 c j = 0 .
For a 2 × 3 2\times3 2 × 3 Is independence of vectors always preserved by a linear map? No; a map can collapse independent vectors — that's why the proof argues via the kernel.
Kernel of T, subspace of V
Image of T, subspace of W
Intuition Hinglish mein samjhoDekho, rank–nullity theorem ka idea bahut simple hai. Maan lo tumhare paas ek linear map T : V → W T:V\to W T : V → W V V V n n n n n n T T T zero ban jaati hai (yeh kernel hai), ya phir wo output me zinda bachti hai (yeh image banati hai). Theorem kehta hai: kernel ka dimension (nullity) plus image ka dimension (rank) hamesha barabar hota hai dim V \dim V dim V
Proof ka asli trick basis extension hai. Pehle kernel ka basis lo: u 1 , … , u k u_1,\dots,u_k u 1 , … , u k V V V v 1 , … , v n − k v_1,\dots,v_{n-k} v 1 , … , v n − k T ( v 1 ) , … , T ( v n − k ) T(v_1),\dots,T(v_{n-k}) T ( v 1 ) , … , T ( v n − k ) T ( u i ) = 0 T(u_i)=0 T ( u i ) = 0 T ( v j ) T(v_j) T ( v j ) ∑ c j T ( v j ) = 0 \sum c_j T(v_j)=0 ∑ c j T ( v j ) = 0 ∑ c j v j \sum c_j v_j ∑ c j v j c j c_j c j
Yeh kyun important hai? Kyunki bina system solve kiye tum nullity nikaal sakte ho: nullity = n − rank \text{nullity}=n-\text{rank} nullity = n − rank
Yaad rakhna: sum hamesha domain V V V W W W T T T
