Eigenvalues & Eigenspaces Practice Quiz with a Step-by-Step Interactive Lesson
Use the question set below to practice eigenvalues and eigenspaces: recognizing equations of the form \(Av=\lambda v\), remembering that eigenvectors are nonzero, computing eigenspaces as \(\ker(A-\lambda I)\), solving \(\det(A-\lambda I)=0\), reading diagonal and triangular matrices, using trace and determinant, understanding when \(0\) is an eigenvalue, tracking how powers, shifts, sums on a common eigenvector, scalar multiples, and inverses affect eigenvalues, and knowing that eigenvectors for distinct eigenvalues are linearly independent. If you want a refresher, open the lesson for mentally followable examples and checks.
Answer the question set and review your mistakes at the end.
How this eigenvalues practice works
1. Take the practice set: answer eigenvalue, eigenvector, eigenspace, trace, determinant, and matrix shortcut questions below.
2. Open the lesson: review definitions, characteristic equations, eigenspace computations, and operation rules with worked examples.
3. Retry: return to the question set and translate each question into \(Av=\lambda v\) or \((A-\lambda I)v=0\).
What you will learn in the eigenvalues & eigenspaces lesson
Eigenvalue equation
Eigenpair: \(Av=\lambda v\) with \(v≠0\)
Eigenspace: \(E_\lambda=\ker(A-\lambda I)\), including the zero vector
The zero vector belongs to every eigenspace but is never an eigenvector
Computing eigenvalues
Characteristic equation: \(\det(A-\lambda I)=0\)
Diagonal and triangular matrices have eigenvalues on the diagonal
Trace is the sum and determinant is the product of eigenvalues, counted with algebraic multiplicity
Finding eigenspaces
For each eigenvalue, solve \((A-\lambda I)v=0\)
A one-dimensional eigenspace is a line of eigenvectors plus \(0\)
Repeated eigenvalues require checking eigenspace dimension; eigenvectors for distinct eigenvalues are linearly independent
Structure and traps
\(0\) is an eigenvalue exactly when \(A\) is singular
If \(Av=\lambda v\), then \(A^kv=\lambda^k v\) and \((A-cI)v=(\lambda-c)v\); if also \(Bv=\mu v\), then \((A+B)v=(\lambda+\mu)v\)
Some real matrices, such as a quarter-turn rotation, have no real eigenvalues
Purpose: Learn to see eigenvalue questions as a single equation, \(Av=\lambda v\), and eigenspace questions as a kernel computation, \((A-\lambda I)v=0\). The lesson connects definitions, characteristic equations, trace and determinant shortcuts, triangular matrices, transformations of eigenvalues, common eigenvector sums, linear independence for distinct eigenvalues, and common traps.
Success criteria
Recognize eigenpairs from \(Av=\lambda v\) with \(v≠0\).
Explain why the zero vector is not an eigenvector.
Compute \(E_\lambda=\ker(A-\lambda I)\) for small matrices.
Use \(\det(A-\lambda I)=0\) to find eigenvalues.
Read eigenvalues of diagonal and triangular matrices from the diagonal.
Use trace and determinant as sum and product checks.
Track how powers, shifts, scalar multiples, inverses, and sums on a common eigenvector change eigenvalues.
Recognize that eigenvectors for distinct eigenvalues are linearly independent.
Avoid false converses such as \(A^2v=9v\) forcing \(Av=3v\).
Key vocabulary
Eigenvalue: a scalar \(\lambda\) such that \(Av=\lambda v\) for some nonzero \(v\).
Eigenvector: a nonzero vector whose direction is preserved by \(A\).
Eigenspace: \(E_\lambda=\ker(A-\lambda I)\), the subspace of vectors satisfying \(Av=\lambda v\).
Characteristic polynomial: \(\det(A-\lambda I)\), whose roots are the eigenvalues.
Algebraic multiplicity: multiplicity of a root of the characteristic polynomial.
Geometric multiplicity: dimension of the eigenspace.
Quick pre-check
Pre-check 1: If \(Av=3v\) and \(v≠0\), what is the eigenvalue associated with \(v\)?
Hint: Compare the equation directly with \(Av=\lambda v\).
Pre-check 2: Can the zero vector be an eigenvector?
Hint: The eigenvector definition requires a nonzero vector, even though \(0\) belongs to every eigenspace.
Eigenvectors are directions that only scale
Learning goal: Translate between \(Av=\lambda v\), \((A-\lambda I)v=0\), and the eigenspace \(E_\lambda\).
Key idea
If \(Av=\lambda v\) for a nonzero vector \(v\), then \(A\) sends the line through \(v\) back to itself, scaling by \(\lambda\). Rearranging gives \[(A-\lambda I)v=0,\] so eigenspaces are kernels. The eigenspace includes \(0\), but its nonzero vectors are the eigenvectors for \(\lambda\).
Recognition checklist
First check that the vector is nonzero.
Compute \(Av\) and see whether it is a scalar multiple of \(v\).
If the scalar is \(0\), then \(v\) is a nonzero vector in \(\ker A\).
For a fixed \(\lambda\), solve a homogeneous system: \((A-\lambda I)v=0\).
Worked example
Example: For \(A=\begin{pmatrix}0&1\\1&0\end{pmatrix}\), decide whether \((1,-1)\) is an eigenvector.
The matrix swaps coordinates, so \(A(1,-1)=(-1,1)=-(1,-1)\). The vector is nonzero and is scaled by \(-1\), so it is an eigenvector with eigenvalue \(-1\).
Try it
Try it 1: What is the eigenspace associated with an eigenvalue \(\lambda\)?
Hint: Move all terms in \(Av=\lambda v\) to the left side.
Try it 2: If \(Av=0\) with \(v≠0\), what is an eigenvalue of \(A\)?
Hint: This is the eigenvalue equation with scalar \(0\).
Eigenvalues make \(A-\lambda I\) singular
Learning goal: Find eigenvalues from \(\det(A-\lambda I)=0\) and recognize quick cases.
Key idea
A scalar \(\lambda\) is an eigenvalue exactly when \((A-\lambda I)v=0\) has a nonzero solution. That happens exactly when \(A-\lambda I\) is singular: \[\det(A-\lambda I)=0.\] Some books use \(\det(\lambda I-A)\); the sign convention can change the polynomial by a factor, but the roots are the same.
Recognition checklist
Form \(A-\lambda I\).
Compute the determinant.
Set the determinant equal to \(0\).
Factor or solve for the possible \(\lambda\) values.
After finding a \(\lambda\), solve \((A-\lambda I)v=0\) if an eigenspace is requested.
Worked example
Example: Find the eigenvalues of \(A=\begin{pmatrix}2&1\\0&5\end{pmatrix}\).
The matrix is triangular, so the determinant of \(A-\lambda I\) is \((2-\lambda)(5-\lambda)\). The characteristic equation is \((2-\lambda)(5-\lambda)=0\), so the eigenvalues are \(2\) and \(5\).
Try it
Try it 1: The characteristic equation for eigenvalues is:
Hint: Eigenvalues make \(A-\lambda I\) have a nonzero kernel.
Try it 2: What are the eigenvalues of \(\begin{pmatrix}2&0\\0&5\end{pmatrix}\)?
Hint: For a diagonal matrix, the characteristic equation factors through the diagonal entries.
Use fast invariants before computing
Learning goal: Use diagonal entries, trace, and determinant as fast ways to identify or check eigenvalues.
Key idea
For diagonal and triangular matrices, the eigenvalues are the diagonal entries, counted with multiplicity. For any square matrix over a field where the characteristic polynomial splits, the trace equals the sum of eigenvalues and the determinant equals their product, both counted with algebraic multiplicity.
Worked example
Example: A \(2\times2\) matrix has eigenvalues \(2\) and \(\lambda\), and trace \(5\). Find \(\lambda\).
Trace is the sum of the eigenvalues counted with multiplicity. Thus \(2+\lambda=5\), so \(\lambda=3\). This also gives a quick consistency check on any characteristic polynomial calculation.
Try it
Try it 1: If a \(2\times2\) matrix has trace \(5\) and eigenvalues \(2\) and \(\lambda\), what is \(\lambda\)?
Hint: The trace equals the sum \(2+\lambda\).
Try it 2: If \(A\) has eigenvalues \(-1\) and \(4\), what is \(\det(A)\)?
Hint: The determinant is the product of the eigenvalues.
Solve one kernel for each eigenvalue
Learning goal: Compute and interpret eigenspaces as subspaces, not just as isolated vectors.
Key idea
Once \(\lambda\) is known, the eigenspace is the solution set of a homogeneous system: \[E_\lambda=\ker(A-\lambda I).\] It is a subspace. Every nonzero vector in it is an eigenvector for \(\lambda\). Its dimension is the geometric multiplicity, which is always at least \(1\) for an eigenvalue and no larger than the algebraic multiplicity. Eigenvectors belonging to distinct eigenvalues are linearly independent.
Worked example
Example: For \(A=\begin{pmatrix}0&1\\1&0\end{pmatrix}\), find eigenspaces for \(1\) and \(-1\).
For \(\lambda=1\), the equation \(A(x,y)=(x,y)\) gives \((y,x)=(x,y)\), so \(x=y\), and \(E_1=\operatorname{span}\{(1,1)\}\). For \(\lambda=-1\), \((y,x)=(-x,-y)\), so \(y=-x\), and \(E_{-1}=\operatorname{span}\{(1,-1)\}\).
Try it
Try it 1: For \(A=\begin{pmatrix}0&1\\1&0\end{pmatrix}\), which vector is an eigenvector for eigenvalue \(1\)?
Hint: Eigenvalue \(1\) means the vector is unchanged by swapping coordinates.
Try it 2: If an eigenspace has dimension \(2\), it contains:
Hint: Every nonzero vector in the eigenspace is an eigenvector.
Powers, shifts, scalings, sums, and inverses
Learning goal: Reuse one eigenpair to understand related matrices such as \(A^2\), \(A-I\), \(cA\), \(A+B\) on a common eigenvector, and \(A^{-1}\).
Key idea
If \(Av=\lambda v\), then the same vector \(v\) stays useful for many expressions in \(A\): \[A^kv=\lambda^k v,\qquad (A-cI)v=(\lambda-c)v,\qquad (cA)v=c\lambda v.\] If also \(Bv=\mu v\), then \((A+B)v=(\lambda+\mu)v\). If \(A\) is invertible, then \(\lambda≠0\) and \(A^{-1}v=\lambda^{-1}v\).
Worked example
Example: If \(Av=2v\) and \(A\) is invertible, what is \(A^{-1}v\)?
Apply \(A^{-1}\) to \(Av=2v\): \(v=2A^{-1}v\). Therefore \(A^{-1}v=\frac{1}{2}v\). The eigenvalue of \(A^{-1}\) on the same vector is \(1/2\).
Try it
Try it 1: If \(Av=\lambda v\), what is \(A^2v\)?
Hint: Apply \(A\) to both sides one more time.
Try it 2: If \(A\) has eigenvalue \(\lambda\), then \(A-I\) has eigenvalue:
Hint: \((A-I)v=Av-v\).
Summary
Powers send \(\lambda\) to \(\lambda^k\) on the same eigenvector.
Shifts \(A-cI\) send \(\lambda\) to \(\lambda-c\).
Scalar multiples \(cA\) send \(\lambda\) to \(c\lambda\).
If \(Av=\lambda v\) and \(Bv=\mu v\), then \(A+B\) sends the same \(v\) to \((\lambda+\mu)v\).
For invertible \(A\), inverse eigenvalues are reciprocals.
Zero, nilpotent maps, and real rotations
Learning goal: Recognize common special cases without overgeneralizing.
Key idea
The eigenvalue \(0\) means \(Av=0\) for some nonzero vector, so \(A\) has a nontrivial kernel and is not invertible. If \(A\) is nilpotent, so \(A^k=0\), then any eigenvalue \(\lambda\) must satisfy \(\lambda^k=0\), hence \(\lambda=0\). A real matrix can have no real eigenvalues: the rotation \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\) has characteristic polynomial \(\lambda^2+1\).
Worked example
Example: Why is \(0\) an eigenvalue exactly when \(A\) is not invertible?
\(0\) is an eigenvalue when \(Av=0v=0\) for some nonzero \(v\). That is exactly saying \(\ker A\) contains a nonzero vector, so the matrix is singular and cannot be invertible.
Try it
Try it 1: For \(A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\), are there real eigenvalues?
Hint: The characteristic polynomial is \(\lambda^2+1\), which has no real root.
Try it 2: If \(A\) is nilpotent, what is its only possible eigenvalue?
Hint: If \(A^k=0\), then \(A^kv=\lambda^k v\) for an eigenvector.
Keep eigenvectors, eigenspaces, and implications separate
Learning goal: Avoid common false conclusions and finish with a short final check.
Common traps
The zero vector is not an eigenvector: the definition explicitly excludes it.
An eigenspace does contain \(0\): it is a kernel and therefore a subspace.
\(A^2v=9v\) does not force \(Av=3v\): \(Av=-3v\) also works.
Repeated eigenvalue does not determine eigenspace dimension: solve \((A-\lambda I)v=0\).
Trace and determinant are checks, not full eigenspaces: they do not tell you eigenvectors.
Distinct eigenvalues give independence: nonzero eigenvectors for distinct eigenvalues are linearly independent.
Real versus complex matters: some real matrices have complex eigenvalues but no real eigenvectors.
Worked example
Example: Does \(A^2v=9v\) force \(Av=3v\)?
No. If \(Av=-3v\), then \(A^2v=A(-3v)=-3Av=9v\). Squaring loses the sign, so the conclusion \(Av=3v\) is not forced.
Try it
Try it 1: If \(A\) has eigenvalue \(0\), what can be said about \(A\)?
Hint: \(0\) as an eigenvalue means a nonzero vector maps to \(0\).
Try it 2: If \(A^2v=9v\), does this force \(Av=3v\)?
Hint: Think about a vector with eigenvalue \(-3\).
Final recap
Eigenvectors are nonzero vectors satisfying \(Av=\lambda v\).
Eigenspaces are kernels: \(E_\lambda=\ker(A-\lambda I)\).
Eigenvalues solve \(\det(A-\lambda I)=0\).
Diagonal and triangular matrices reveal eigenvalues on the diagonal.
Trace sums eigenvalues and determinant multiplies them, counted with algebraic multiplicity.
\(0\) is an eigenvalue exactly when \(A\) is singular.
Powers, shifts, scalar multiples, sums on a common eigenvector, and inverses transform eigenvalues in predictable ways.
Eigenvectors for distinct eigenvalues are linearly independent.
Next step: Close this lesson and try the quiz again. For each question, decide first whether it asks for an eigenvalue, an eigenvector, an eigenspace, or a structural consequence like invertibility.
Practice set
Autovalori e autospazi practice questions with instant score
Answer all 10 questions below, then get your final score and a mistake review at the end so you know exactly what to improve.
0/10answered
Question 1Not answered
Se \(Av=3v\) e \(v\ne0\), qual è l'autovalore associato a \(v\)?
Correct answer: C. \(3\)
Explanation: Un autovettore soddisfa \(Av=\lambda v\); qui \(\lambda=3\).
Question 2Not answered
Il vettore nullo può essere un autovettore?
Correct answer: C. No, mai
Explanation: No. Gli autovettori devono essere diversi da zero.
Question 3Not answered
Quali sono gli autovalori di \(\begin{pmatrix}2&0\\0&5\end{pmatrix}\)?
Correct answer: A. \(2\) e \(5\)
Explanation: Per una matrice diagonale, gli autovalori sono gli elementi della diagonale.
Question 4Not answered
Se \(A\) ha autovalore \(0\), cosa si può dire di \(A\)?
Correct answer: D. \(A\) non è invertibile
Explanation: L'autovalore \(0\) significa che qualche vettore non nullo viene mandato in \(0\), quindi \(A\) non è iniettiva.
Question 5Not answered
Per \(A=I_n\), qual è l'unico autovalore?
Correct answer: A. \(1\)
Explanation: L'identità soddisfa \(Iv=v\) per ogni vettore \(v\).
Question 6Not answered
Cos'è l'autospazio associato a un autovalore \(\lambda\)?
Correct answer: A. \(\ker(A-\lambda I)\)
Explanation: È il nucleo di \(A-\lambda I\), incluso il vettore nullo.
Question 7Not answered
Se una matrice \(2\times2\) ha traccia \(5\) e autovalori \(2\) e \(\lambda\), quanto vale \(\lambda\)?
Correct answer: B. \(3\)
Explanation: La traccia è la somma degli autovalori, quindi \(2+\lambda=5\).
Question 8Not answered
Se una matrice \(2\times2\) ha autovalori \(2\) e \(3\), qual è il suo determinante?
Correct answer: D. \(6\)
Explanation: Il determinante è il prodotto degli autovalori.
Question 9Not answered
Per \(A=\begin{pmatrix}0&1\\1&0\end{pmatrix}\), quale vettore è un autovettore per l'autovalore \(1\)?
Correct answer: A. \((1,1)\)
Explanation: La matrice scambia le coordinate, e \((1,1)\) resta invariato.
Question 10Not answered
Per \(A=\begin{pmatrix}0&1\\1&0\end{pmatrix}\), quale vettore è un autovettore per l'autovalore \(-1\)?
Correct answer: D. \((1,-1)\)
Explanation: La matrice scambia le coordinate, mandando \((1,-1)\) in \((-1,1)=-(1,-1)\).
