Eigenvalues and Eigenvectors Short Notes for GATE CS
Eigenvalues and Eigenvectors short notes for GATE CS: 7 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.
eigenvalues and eigenvectors short notes
Topic Summary: Determinants and Multiplicity
Topic Summary: Determinants and Multiplicity
Core Formulas
Product:det(A)=∏λi
Sum:Tr(A)=∑λi
Geometric Multiplicity:GM=n−rank(A−λI)
The Golden Inequality
1≤GM≤AM
Key Exam Takeaways
Maximum algebraic multiplicity in an n×n matrix is n.
Never calculate eigenvalues just to find their product or sum; use determinant and trace.
A matrix is defective (not diagonalizable) if GM<AM for any eigenvalue.
If rows/columns are linearly dependent, det(A)=0, meaning at least one eigenvalue is zero.
Verification Quick Checklist
Non-zero rule:v=0 is mandatory.
The Test: Compute Av.
The Check: Does Av=cv for some scalar c?
Yes → Eigenvector (λ=c).
No → Not an eigenvector.
Equivalence:v and kv (k=0) represent the same eigenvector direction and share the same λ.
Formula Matrix
Rule
Formula & Condition
Trace
∑λi=tr(A)=0(Simple graph)
Regular
λmax=k(Connected k-regular)
Bipartite
λ∈Spec⟺−λ∈Spec(Bipartite)
Complete Kn
{n−1,−1(n−1)}(Complete graph)
Complete Bip. Km,n
{mn,−mn,0(m+n−2)}(Complete bipartite)
Disjoint Union
Spec(G1∪G2)=Spec(G1)∪Spec(G2)
Sum of Squares
∑λi2=tr(A2)=2∣E∣(Simple graph)
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Question 1
Level 1: Warm-up
The eigenvalues of a 3×3 matrix are 2, −3, and 4. What is the determinant of the matrix?
Question 2
Level 1: Warm-up
Let A=[4007] and v=[10]. If Av=λv, what is the value of λ?
Question 3
Level 1: Warm-up
For an n×n matrix, the characteristic equation ∣A−λI∣=0 yields a polynomial in λ. What is the maximum possible degree of this polynomial for a 5×5 matrix?
Question 4
Level 1: Warm-up
The trace of a 3×3 matrix is 12. If two of its eigenvalues are 4 and 5, consider the following statement: "The third eigenvalue is 3." Is this statement true or false?
Question 5
Level 1: Warm-up
Let A be a 3×3 matrix where the third row is exactly the sum of the first two rows. Which of the following must be an eigenvalue of A?
Question 6
Level 1: Warm-up
A linear transformation scales the vector v by a factor of −5 and reverses its direction. What is the eigenvalue associated with v?
Question 7
Level 1: Warm-up
Consider the characteristic equation of a 4×4 matrix: (λ−3)2(λ+2)2=0.
Assertion (A): The algebraic multiplicity of λ=3 is 2.
Reason (R): The geometric multiplicity of λ=3 is 4.
Question 8
Level 1: Warm-up
The characteristic equation of a matrix has roots λ1=−3 and λ2=2. The algebraic multiplicity of λ1 is 3, and for λ2 it is 1. What is the sum of the maximum possible geometric multiplicities for all eigenvalues?
Question 9
Level 1: Warm-up
Consider the following statements regarding eigenvector verification for a large 4×4 matrix:
Assertion (A): To verify if a given vector is an eigenvector, one should first compute the characteristic equation det(A−λI)=0.
Reason (R): Matrix-vector multiplication is computationally cheaper than expanding a 4×4 determinant.
Question 10
Level 1: Warm-up
By definition, an eigenvector of a square matrix must satisfy a strict non-zero condition. What is the maximum number of zero vectors that can be included in a valid set of eigenvectors for any matrix?
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Eigenvalues and Eigenvectors Short Notes for GATE CS
Eigenvalues and Eigenvectors short notes for GATE CS: 7 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.
Topic Summary: Determinants and Multiplicity
Topic Summary: Determinants and Multiplicity
Core Formulas
Product:det(A)=∏λi
Sum:Tr(A)=∑λi
Geometric Multiplicity:GM=n−rank(A−λI)
The Golden Inequality
1≤GM≤AM
Key Exam Takeaways
Maximum algebraic multiplicity in an n×n matrix is n.
Never calculate eigenvalues just to find their product or sum; use determinant and trace.
A matrix is defective (not diagonalizable) if GM<AM for any eigenvalue.
If rows/columns are linearly dependent, det(A)=0, meaning at least one eigenvalue is zero.
Verification Quick Checklist
Non-zero rule:v=0 is mandatory.
The Test: Compute Av.
The Check: Does Av=cv for some scalar c?
Yes → Eigenvector (λ=c).
No → Not an eigenvector.
Equivalence:v and kv (k=0) represent the same eigenvector direction and share the same λ.
Formula Matrix
Rule
Formula & Condition
Trace
∑λi=tr(A)=0(Simple graph)
Regular
λmax=k(Connected k-regular)
Bipartite
λ∈Spec⟺−λ∈Spec(Bipartite)
Complete Kn
{n−1,−1(n−1)}(Complete graph)
Complete Bip. Km,n
{mn,−mn,0(m+n−2)}(Complete bipartite)
Disjoint Union
Spec(G1∪G2)=Spec(G1)∪Spec(G2)
Sum of Squares
∑λi2=tr(A2)=2∣E∣(Simple graph)
If You See This, Do This
Stem Trigger
First Move
"sum of eigenvalues"
Check trace. Simple graph →0.
"sum of squares"
Calculate 2×∣E∣.
"k-regular" + "largest"
Answer is k.
"bipartite" + "λ"
Write −λ immediately.
"complete Kn"
Write n−1 and (n−1) copies of −1.
"complete bip. Km,n"
Write ±mn and (m+n−2) zeros.
"disjoint union"
Concatenate individual spectra.
Eigenvalues and Eigenvectors: Solved Questions with Step-by-Step Explanations (10 Problems)
Question 1 · Engineering MathematicsMCQ
The eigenvalues of a 3×3 matrix are 2, −3, and 4. What is the determinant of the matrix?
A.
-24
B.
24
C.
-1
D.
3
Correct Answer:
A
Step-by-Step Solution
Key idea: The determinant of a matrix is exactly equal to the product of all its eigenvalues.
Step 1: Identify the eigenvalues: λ1=2, λ2=−3, λ3=4.
Step 2: Multiply them together: det(A)=2×(−3)×4.
Step 3: Calculate the product: −6×4=−24.
Answer: -24
Question 2 · Engineering MathematicsMCQ
Let A=[4007] and v=[10]. If Av=λv, what is the value of λ?
A.
4
B.
7
C.
11
D.
28
Correct Answer:
A
Step-by-Step Solution
Key idea: This is a direct substitution question for the eigenvalue equation Av=λv.
Step 1: Multiply matrix A by vector v.
Av=[4007][10]=[40]
Step 2: Set this equal to λv.
[40]=λ[10]=[λ0]
Step 3: Compare the components to find λ=4.
Answer: 4
Question 3 · Engineering MathematicsMCQ
For an n×n matrix, the characteristic equation ∣A−λI∣=0 yields a polynomial in λ. What is the maximum possible degree of this polynomial for a 5×5 matrix?
A.
5
B.
10
C.
25
D.
125
Correct Answer:
A
Step-by-Step Solution
Key idea: The degree of the characteristic polynomial is always exactly equal to the size of the square matrix.
Step 1: The matrix is of size 5×5, so n=5.
Step 2: The determinant ∣A−λI∣ expands to a polynomial of degree n.
Step 3: Therefore, the degree is 5.
Answer: 5
Question 4 · Engineering MathematicsMCQ
The trace of a 3×3 matrix is 12. If two of its eigenvalues are 4 and 5, consider the following statement: "The third eigenvalue is 3." Is this statement true or false?
A.
True, because the sum of eigenvalues equals the trace
B.
True, because the product of eigenvalues equals the trace
C.
False, because the third eigenvalue is 21
D.
False, because the third eigenvalue is 60
Correct Answer:
A
Step-by-Step Solution
Key idea: The trace of a matrix is equal to the sum of its eigenvalues.
Step 1: Let the third eigenvalue be λ3. The trace is the sum of eigenvalues: 4+5+λ3=12.
Step 2: Solve for λ3: 9+λ3=12⟹λ3=3.
Step 3: The statement is true, and the reason is that the sum of eigenvalues equals the trace.
Answer: True, because the sum of eigenvalues equals the trace.
Question 5 · Engineering MathematicsMCQ
Let A be a 3×3 matrix where the third row is exactly the sum of the first two rows. Which of the following must be an eigenvalue of A?
A.
0
B.
1
C.
3
D.
-1
Correct Answer:
A
Step-by-Step Solution
Key idea: This is a singular matrix recognition question. Linearly dependent rows imply a zero determinant, which forces at least one eigenvalue to be zero.
Step 1: The third row is the sum of the first two rows, meaning the rows of A are linearly dependent.
Step 2: A matrix with linearly dependent rows is singular, which means its determinant is exactly zero (det(A)=0).
Step 3: The product of all eigenvalues equals the determinant. Since det(A)=0, at least one eigenvalue must be 0.
Answer: 0
Question 6 · Engineering MathematicsMCQ
A linear transformation scales the vector v by a factor of −5 and reverses its direction. What is the eigenvalue associated with v?
A.
5
B.
-5
C.
0.2
D.
-0.2
Correct Answer:
B
Step-by-Step Solution
Key idea: The eigenvalue λ represents the pure scaling factor, including any sign change for direction reversal.
Step 1: The problem states the scaling factor is −5 and the direction is reversed.
Step 2: Reversing direction means the scalar is negative. The magnitude of scaling is 5.
Step 3: Therefore, the eigenvalue is λ=−5.
Answer: -5
Question 7 · Engineering MathematicsMCQ
Consider the characteristic equation of a 4×4 matrix: (λ−3)2(λ+2)2=0.
Assertion (A): The algebraic multiplicity of λ=3 is 2.
Reason (R): The geometric multiplicity of λ=3 is 4.
A.
Both A and R are true and R is the correct explanation of A
B.
Both A and R are true but R is NOT the correct explanation of A
C.
A is true but R is false
D.
A is false but R is true
Correct Answer:
C
Step-by-Step Solution
Key idea: Algebraic multiplicity is the power of the factor in the characteristic equation, while geometric multiplicity is the number of independent eigenvectors (bounded by algebraic multiplicity).
Step 1: The factor (λ−3) is squared, so the algebraic multiplicity of λ=3 is 2. Assertion (A) is true.
Step 2: The geometric multiplicity must be between 1 and the algebraic multiplicity (2). It cannot be 4. Reason (R) is false.
Step 3: Therefore, A is true but R is false.
Answer: A is true but R is false.
Question 8 · Engineering MathematicsMCQ
The characteristic equation of a matrix has roots λ1=−3 and λ2=2. The algebraic multiplicity of λ1 is 3, and for λ2 it is 1. What is the sum of the maximum possible geometric multiplicities for all eigenvalues?
A.
3
B.
4
C.
5
D.
6
Correct Answer:
B
Step-by-Step Solution
Key idea: The geometric multiplicity (GM) of an eigenvalue is bounded by its algebraic multiplicity (AM). The maximum possible GM is exactly equal to the AM.
Step 1: For λ1=−3, the algebraic multiplicity is 3. The maximum possible geometric multiplicity is therefore 3.
Step 2: For λ2=2, the algebraic multiplicity is 1. The maximum possible geometric multiplicity is 1.
Step 3: The sum of the maximum possible geometric multiplicities is 3+1=4.
Answer: 4
Question 9 · Engineering MathematicsMCQ
Consider the following statements regarding eigenvector verification for a large 4×4 matrix:
Assertion (A): To verify if a given vector is an eigenvector, one should first compute the characteristic equation det(A−λI)=0.
Reason (R): Matrix-vector multiplication is computationally cheaper than expanding a 4×4 determinant.
A.
Both A and R are true and R is the correct explanation of A
B.
Both A and R are true but R is NOT the correct explanation of A
C.
A is false but R is true
D.
A is true but R is false
Correct Answer:
C
Step-by-Step Solution
Key idea: When candidate vectors are provided, direct verification via multiplication is vastly superior to solving the characteristic equation.
Step 1: Evaluate Assertion (A): Computing the characteristic equation for a 4×4 matrix is computationally heavy and unnecessary if you only need to verify specific candidate vectors. Thus, A is false.
Step 2: Evaluate Reason (R): Matrix-vector multiplication requires O(n2) operations, while expanding a 4×4 determinant is much more complex and prone to error. Thus, R is true.
Step 3: Since A is false and R is true, the correct option is the third one.
Answer: A is false but R is true.
Question 10 · Engineering MathematicsMCQ
By definition, an eigenvector of a square matrix must satisfy a strict non-zero condition. What is the maximum number of zero vectors that can be included in a valid set of eigenvectors for any matrix?
A.
0
B.
1
C.
2
D.
Depends on the matrix size
Correct Answer:
A
Step-by-Step Solution
Key idea: The definition of an eigenvector explicitly excludes the zero vector.
Step 1: An eigenvector v of a matrix A must satisfy Av=λv.
Step 2: While the zero vector trivially satisfies A0=λ0 for any λ, it provides no information about directional scaling.
Step 3: Therefore, the strict definition requires v=0. The zero vector can never be an eigenvector.