Linear Maps, Kernel & Image Practice Questions, Quiz, and Step-by-Step Lesson - improve your math skills with focused questions and clear explanations.
Linear Maps, Kernel & Image Practice Quiz with a Step-by-Step Interactive Lesson
Use the quiz at the top of the page to practice linear maps, kernel, and image: checking whether a map is linear, using \(T(0)=0\), finding \(\ker T=\{v\in V:T(v)=0\}\), describing \(\operatorname{Im}T=\{T(v):v\in V\}\), linking injectivity to \(\ker T=\{0\}\), linking surjectivity to \(\operatorname{Im}T=W\), reading matrix maps through column space and null space, using rank-nullity, and handling composition facts like \(S\circ T\) injective implies \(T\) injective. If you want a refresher, open the lesson for mentally followable examples and checks.
How this linear maps practice works
1. Take the quiz: answer the linear map, kernel, image, injectivity, and surjectivity questions at the top of the page.
2. Open the lesson: review definitions, matrix-map shortcuts, rank-nullity, and composition facts with worked examples.
3. Retry: return to the quiz and use the kernel/image language immediately.
What you will learn in the linear maps, kernel & image lesson
Recognize linear maps
Linearity test: \(T(u+v)=T(u)+T(v)\) and \(T(cv)=cT(v)\)
Zero check: every linear map sends \(0_V\) to \(0_W\)
Spot affine and nonlinear traps such as \((x,y)\mapsto(x+1,y)\) or \((x,y)\mapsto(x^2,y)\)
Kernel and injectivity
Kernel: \(\ker T=\{v\in V:T(v)=0\}\)
\(\ker T\) is a subspace of the domain
Injective: \(T\) is one-to-one exactly when \(\ker T=\{0\}\)
Image and surjectivity
Image: all outputs \(T(v)\), always a subspace of the codomain
For \(x\mapsto Ax\), the image is the column space of \(A\)
When you are ready, return to the quiz at the top of the page and keep practicing linear maps, kernels, and images.
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Advanced Linear Algebra
Linear Maps, Kernel & Image Lesson
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Lesson overview
Purpose: Build a clear picture of a linear map \(T:V\to W\), then use its two most important subspaces: the kernel in the domain and the image in the codomain. You will connect these ideas to injectivity, surjectivity, matrix maps, rank-nullity, and composition.
Success criteria
Check linearity using additivity and scalar multiplication.
Use \(T(0_V)=0_W\) to reject shifted maps quickly.
Find \(\ker T\) by solving \(T(v)=0\), and know that it is a subspace of \(V\).
Find \(\operatorname{Im}T\) by describing all outputs, often as a span.
Recognize injective maps by \(\ker T=\{0\}\) and surjective maps by \(\operatorname{Im}T=W\).
For \(x\mapsto Ax\), read the image as column space and the kernel as the solution set of \(Ax=0\).
Use \(\dim V=\dim\ker T+\dim\operatorname{Im}T\) to count dimensions.
Use basic composition facts without confusing the roles of \(S\) and \(T\).
Key vocabulary
Linear map: a function \(T:V\to W\) with \(T(u+v)=T(u)+T(v)\) and \(T(cv)=cT(v)\).
Kernel: \(\ker T=\{v\in V:T(v)=0_W\}\), a subspace of the domain.
Image: \(\operatorname{Im}T=\{T(v):v\in V\}\), a subspace of the codomain.
Rank: \(\dim\operatorname{Im}T\).
Nullity: \(\dim\ker T\).
Isomorphism: a bijective linear map.
Quick pre-check
Pre-check 1: For a linear map \(T:V\to W\), what is \(T(0_V)\)?
Hint: Use \(T(0)=T(0+0)=T(0)+T(0)\), then subtract \(T(0)\) in \(W\).
Pre-check 2: What is the kernel of a linear map \(T:V\to W\)?
Hint: The kernel lives in the domain and consists of the inputs sent to the zero vector.
Linearity is about preserving structure
Learning goal: Decide quickly whether a formula defines a linear map, and use \(T(0)=0\) as a fast first filter.
Key idea
A map \(T:V\to W\) is linear when it respects linear combinations: \[T(cu+dv)=cT(u)+dT(v)\] for all vectors \(u,v\) and scalars \(c,d\). This single formula contains both additivity and scalar multiplication.
Recognition checklist
Zero test: if \(T(0)≠ 0\), the map is not linear.
No shifts: constants like \(x+1\) usually make an affine map, not a linear one.
No nonlinear operations: squares, absolute values, products of coordinates, and case changes usually break linearity.
Coordinate formulas: sums like \(x-y\) and scalar multiples like \(2x\) are compatible with linearity.
Scalar consequences: \(T(-v)=-T(v)\), \(T(u-v)=T(u)-T(v)\), and \(T(v/2)=T(v)/2\).
Worked example
Example: Is \(T:\mathbb{R}^2\to\mathbb{R}^2\), \(T(x,y)=(x+1,y)\), linear?
No. The zero vector maps to \(T(0,0)=(1,0)\), not \((0,0)\). A linear map must send the zero vector of the domain to the zero vector of the codomain.
Try it
Try it 1: Which map is linear on \(\mathbb{R}^2\)?
Hint: Look for a formula made only from coordinate sums and scalar multiples, with no shift and no nonlinear operation.
Try it 2: If \(T(v)=w\), what is \(T(3v)\)?
Hint: Scalar multiplication passes through a linear map: \(T(3v)=3T(v)\).
The kernel measures collapse to zero
Learning goal: Find kernels from simple formulas and use \(\ker T=\{0\}\) as the injectivity test.
Key idea
The kernel is the set of inputs that disappear: \[\ker T=\{v\in V:T(v)=0_W\}.\] It is always a subspace of \(V\). For injectivity, the only input allowed to disappear is the zero vector: \(T\) is injective exactly when \(\ker T=\{0\}\).
Worked example
Example: For \(T:\mathbb{R}^3\to\mathbb{R}^2\), \(T(x,y,z)=(x,z)\), find \(\ker T\).
Solve \(T(x,y,z)=(0,0)\). This gives \(x=0\) and \(z=0\), while \(y\) is free. So \[\ker T=\{(0,y,0):y\in\mathbb{R}\}=\operatorname{span}\{(0,1,0)\}.\]
Try it
Try it 1: For \(T:\mathbb{R}^2\to\mathbb{R}^2\), \(T(x,y)=(x,0)\), what is \(\ker T\)?
Hint: Set \((x,0)=(0,0)\). The first coordinate is forced, but the second coordinate is free.
Try it 2: If \(T(v)=T(w)\), which vector is in \(\ker T\)?
Hint: Subtract the two equal images using linearity.
Summary
To find a kernel, solve \(T(v)=0\).
Kernel vectors live in the domain.
\(\ker T=\{0\}\) is the clean test for injectivity.
The image is the set of reachable outputs
Learning goal: Describe images as spans or coordinate conditions, and use \(\operatorname{Im}T=W\) as the surjectivity test.
Key idea
The image is the set of all outputs: \[\operatorname{Im}T=\{T(v):v\in V\}.\] It is always a subspace of the codomain. A map \(T:V\to W\) is surjective exactly when every vector of \(W\) is reached, so \(\operatorname{Im}T=W\).
Worked example
Example: For \(T:\mathbb{R}^2\to\mathbb{R}^2\), \(T(x,y)=(x+y,x+y)\), describe the image.
Every output has equal coordinates: \((x+y,x+y)=(t,t)=t(1,1)\). Conversely, every \(t(1,1)\) is reached by taking \(x=t,y=0\). Thus \(\operatorname{Im}T=\operatorname{span}\{(1,1)\}\), a one-dimensional line.
Try it
Try it 1: For \(T:\mathbb{R}^3\to\mathbb{R}^2\), \(T(x,y,z)=(x,z)\), what is \(\operatorname{Im}T\)?
Hint: Given any pair \((a,b)\), choose \(x=a\) and \(z=b\).
Try it 2: If \(\operatorname{Im}T=W\), what is the usual name for \(T\)?
Hint: Every vector in the codomain is reached.
Column space and null space
Learning goal: Translate kernel and image questions for \(x\mapsto Ax\) into familiar matrix language.
Key idea
For a matrix map \(T(x)=Ax\), the image is the span of the columns of \(A\), also called the column space. The kernel is the solution set of \(Ax=0\), also called the null space. Pivot columns tell you the image dimension; free variables tell you the kernel dimension.
Worked example
Example: Let \(A=\begin{pmatrix}1&0&2\\0&1&-1\end{pmatrix}\). Describe the kernel of \(x\mapsto Ax\).
Solve \(Ax=0\): \[x_1+2x_3=0,\qquad x_2-x_3=0.\] Let \(x_3=t\). Then \(x_1=-2t\), \(x_2=t\), so \[x=t(-2,1,1).\] Therefore \(\ker A=\operatorname{span}\{(-2,1,1)\}\).
Try it
Try it 1: For a matrix map \(x\mapsto Ax\), what is the image of the map?
Hint: Outputs of \(Ax\) are linear combinations of the columns of \(A\).
Try it 2: If the columns of a \(3\times2\) matrix are independent, what is its kernel?
Hint: Independent columns mean the only solution to \(Ax=0\) is the zero vector in the domain.
Counting dimensions with rank-nullity
Learning goal: Use rank-nullity to find a missing dimension without solving every vector equation.
Key idea
For a linear map \(T:V\to W\) with finite-dimensional domain, \[\dim V=\dim\ker T+\dim\operatorname{Im}T.\] The image dimension is the rank; the kernel dimension is the nullity. Notice that the domain dimension appears on the left.
Worked example
Example: If \(T:\mathbb{R}^4\to\mathbb{R}^3\) is surjective, what is \(\dim\ker T\)?
Surjective means \(\operatorname{Im}T=\mathbb{R}^3\), so rank \(=3\). Rank-nullity gives \(4=\dim\ker T+3\). Therefore \(\dim\ker T=1\).
Try it
Try it 1: If \(T:\mathbb{R}^3\to\mathbb{R}^2\) has rank \(2\), what is \(\dim\ker T\)?
Hint: Use \(\dim V=3=\operatorname{rank}T+\operatorname{nullity}T\).
Try it 2: If \(T:\mathbb{R}^2\to\mathbb{R}^3\) is injective, what is \(\dim\operatorname{Im}T\)?
Hint: Injective means nullity \(0\), so the rank equals the domain dimension.
How kernels and images behave in compositions
Learning goal: Use basic composition facts without reversing the roles of the inner and outer maps.
Key idea
The composition of linear maps is linear. If \(S\circ T\) is injective, then \(T\) must be injective: equal \(T\)-outputs would become equal \((S\circ T)\)-outputs. If \(S\circ T\) is surjective, then \(S\) must be surjective: every final target value is reached by applying \(S\) to something in the image of \(T\).
Worked example
Example: Why does \((S\circ T)\) injective force \(T\) injective?
Assume \(T(v)=T(w)\). Applying \(S\) gives \(S(T(v))=S(T(w))\). Since \(S\circ T\) is injective, this implies \(v=w\). Therefore \(T\) is injective.
Try it
Try it 1: If \(S\circ T\) is injective, what must be true about \(T\)?
Hint: Start with \(T(v)=T(w)\), then apply \(S\).
Try it 2: If a linear map is both injective and surjective, what is it called?
Hint: A bijective linear map gives a structure-preserving correspondence between vector spaces.
Keep domain, codomain, kernel, and image separate
Learning goal: Avoid common mistakes and finish with a quick final check.
Common traps
Kernel lives in the domain: it is made of inputs.
Image lives in the codomain: it is made of outputs.
The image is never empty: a linear map always outputs at least \(0_W\).
\(\ker T=\{0\}\) means injective, not surjective.
\(\operatorname{Im}T=W\) means surjective, not automatically injective.
Extreme maps: the zero map has \(\ker T=V\) and \(\operatorname{Im}T=\{0\}\); the identity map has \(\ker T=\{0\}\) and \(\operatorname{Im}T=V\).
Do not accept shifted formulas: \(T(0)≠0\) immediately breaks linearity.
Worked example
Example: For \(T(x,y)=(x,0)\), what are the kernel and image?
The kernel condition \((x,0)=(0,0)\) gives \(x=0\), so \(\ker T=\{(0,y):y\in\mathbb{R}\}\). The outputs have form \((x,0)\), so \(\operatorname{Im}T=\{(x,0):x\in\mathbb{R}\}\). Kernel is a vertical line in the domain; image is a horizontal line in the codomain.
Try it
Try it 1: If \(T:\mathbb{R}^2\to\mathbb{R}^2\) is the zero map, what is \(\operatorname{Im}T\)?
Hint: The zero map has exactly one output.
Try it 2: For \(T(x,y)=(x+y,x+y)\), which vector lies in \(\ker T\)?
Hint: The kernel condition is \(x+y=0\).
Final recap
Linearity means preserving linear combinations.
\(\ker T=\{v\in V:T(v)=0_W\}\), and \(\ker T=\{0\}\) exactly when \(T\) is injective.
\(\operatorname{Im}T=\{T(v):v\in V\}\), and \(\operatorname{Im}T=W\) exactly when \(T\) is surjective.
For \(x\mapsto Ax\), image means column space and kernel means the solution set of \(Ax=0\).
The zero map has kernel \(V\) and image \(\{0\}\); the identity map has kernel \(\{0\}\) and image \(V\).
Rank-nullity counts dimensions from the domain: \(\dim V=\dim\ker T+\dim\operatorname{Im}T\).
Composition facts track the inner map for injectivity and the outer map for surjectivity.
Next step: Close this lesson and try the quiz again. When a question mentions \(T(v)=0\), think kernel; when it asks what outputs are possible, think image.
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