Linear transformations — definition, kernel, image

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1. Definition

WHY these two rules? A vector space is defined by two operations: adding vectors and scaling them. A map "preserves structure" precisely when it commutes with those two operations. Nothing more, nothing less.

HOW to use the definition (Derivation: $T(0)=0$): $T(0) = T(0\cdot v) = 0\cdot T(v) = 0.$ Why this step? I wrote $0$ as the scalar $0$ times any vector, then pulled the scalar out (homogeneity). So every linear map sends the zero vector to the zero vector — a free sanity check.

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2. Kernel (null space)

WHY is it a subspace? If $T(u)=0$ and $T(v)=0$ then $T(cu+dv) = cT(u)+dT(v) = 0$. So sums and scalings of kernel vectors stay in the kernel → it's a subspace of $V$ (and always contains $0$).

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3. Image (range)

WHY is it a subspace of $W$? Any output is $T(v)$; $cT(u)+dT(v) = T(cu+dv)$ is again an output. Closed under the operations → subspace.

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4. Rank–Nullity Theorem (the bridge)

Derivation from scratch. Let $\dim V = n$. Take a basis $\{k_1,\dots,k_p\}$ of $\ker T$ (so nullity $=p$). Extend it to a full basis of $V$: $\{k_1,\dots,k_p, w_1,\dots,w_{q}\}$ with $p+q=n$. Claim: $\{T(w_1),\dots,T(w_q)\}$ is a basis of $\operatorname{im} T$.

• They span the image: any $v=\sum a_i k_i + \sum b_j w_j$, so $T(v)=\sum a_i T(k_i)+\sum b_j T(w_j)=\sum b_j T(w_j)$ (kernel terms vanish). Why? $T(k_i)=0$ by definition of kernel.
• They are independent: if $\sum b_j T(w_j)=0$ then $T(\sum b_j w_j)=0$, so $\sum b_j w_j\in\ker T = \operatorname{span}(k_i)$. Writing it as $\sum c_i k_i$ and using basis independence forces all $b_j=0$. So rank $=q$, hence $n=p+q=$ nullity $+$ rank. $\blacksquare$

Check on our example: $\dim V=3$, nullity $=1$, rank $=2$, and $1+2=3$. ✓

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5. Common mistakes

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6. Active recall

Recall Quick self-test (hide answers)

• State the two defining properties of a linear map. → additivity & homogeneity.
• Why must $T(0)=0$? → $T(0v)=0T(v)=0$.
• $\ker T \subseteq$ ? → $V$. $\operatorname{im} T\subseteq$? → $W$.
• Injective iff? → $\ker T=\{0\}$.
• Rank–nullity statement? → $\dim V=$ nullity $+$ rank.

Recall Feynman: explain to a 12-year-old

Imagine a machine that takes arrows and gives new arrows, with one fair rule: if you add two arrows first then feed them in, you get the same answer as feeding each and adding afterwards (and doubling an arrow doubles the output). The kernel is the pile of arrows that the machine squashes flat to nothing. The image is the collection of every arrow the machine is able to spit out. If only the "do-nothing" arrow gets squashed, the machine never confuses two different arrows. Counting rule: (arrows squashed) + (arrows it can make) = (arrows you started with).

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Flashcards

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Connections

• Matrix multiplication — every $T:\mathbb{R}^n\to\mathbb{R}^m$ is multiplication by its matrix $A=[T(e_1)\cdots T(e_n)]$.
• Column space and null space — image = column space, kernel = null space.
• Rank–Nullity Theorem — the dimension accounting bridge.
• Injective surjective bijective — kernel/image characterizations.
• Change of basis — how the matrix of $T$ transforms.
• Eigenvalues and eigenvectors — $\ker(T-\lambda I)$ is the eigenspace.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, ek linear transformation basically ek "honest machine" hai jo vectors ko vectors mein badalta hai, lekin do rule follow karta hai: agar tum pehle do vectors ko add karke daalo, toh wahi answer aata hai jo alag-alag daal ke add karne pe aata; aur agar vector ko double karo toh output bhi double. In dono rules ka ek seedha consequence — zero hamesha zero pe jaata hai, $T(0)=0$. Isiliye $Ax+b$ wale affine maps linear nahi hote jab $b\neq0$.

Ab do sabse important cheezein. Kernel matlab woh saare input vectors jo machine 0 bana deti hai — yeh domain $V$ ke andar rehta hai, $W$ mein nahi (yeh sabse common galti hai!). Agar kernel mein sirf $0$ hai, toh map injective hai (koi do alag inputs same output nahi dete). Image matlab woh saare outputs jo machine bana sakti hai — yeh $W$ ke andar rehta hai, aur matrix ke case mein yeh column space hota hai.

In dono ko jodne wala magic formula hai Rank–Nullity: $\dim V = \text{nullity} + \text{rank}$, yaani jitne dimensions crush hue (kernel) plus jitne reachable hain (image), total input dimension ke barabar. Humare example $T(x,y,z)=(x+y+z,\,x-z)$ mein nullity $1$, rank $2$, aur $1+2=3$ — perfect. Yaad rakhne ke liye: "KILL goes IN, REACH comes OUT" — kernel input ko maarta hai, image output deta hai. Yeh idea poore linear algebra ki backbone hai, eigenvalues se lekar solving systems tak.