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Vector Spaces Practice Questions for GATE DA

GATE DA Vector Spaces: 5 chapters, 5 previous year questions (25% of Linear Algebra), 266 practice questions and one solved question from each chapter.

A question from this chapter

Question 1

Level 3: Exam Standard

Apply Gram-Schmidt to the ordered list w_{1} = (1, 1, 0), w_{2} = (1, - 1, 1), w_{3} = (2, 0, 1) in R^{3} . How many of the intermediate unnormalized residuals z_{k} (before normalizing) turn out to be the zero vector?

Question 2

Level 3: Exam Standard

Let P be the orthogonal projection matrix onto a subspace U of R^{4} . Let Q be the orthogonal projection matrix onto the orthogonal complement U^{⊥} . Consider the matrix M = P - 3 Q . Let S be the set of all possible values of det (M) as U varies over all possible subspaces of R^{4} . Find the sum of all elements in S .

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Vector Spaces Practice Questions for GATE DA

GATE DA Vector Spaces: 5 chapters, 5 previous year questions (25% of Linear Algebra), 266 practice questions and one solved question from each chapter.

About Vector Spaces Practice Questions

266 practice questions for Vector Spaces in GATE DA, sorted chapter by chapter and graded from basic to exam level, each with a full solution.

Vector Spaces Weightage in GATE DA

Vector Spaces accounts for 5 of 20 Linear Algebra previous year questions in our bank (25%), about 1.7 per paper across 3 papers.

Vector Spaces Chapter Matrix

Chapter	Topics	PYQs	Share of unit PYQs	Practice questions
Chapter 1 — Vector Spaces		0	0%	0
Vector Spaces, Subspaces and Bases	Subspaces, Linear Independence and Orthonormal Bases, Geometry of Subspaces and Norm Balls	3	60%	154
Chapter 2 — Vector Spaces		0	0%	0
Orthogonality, Projections and Linear Systems	Projection Matrices, Null Space and Idempotence, Consistency and Solution Sets of Linear Systems	2	40%	112
Chapter 3 — Vector Spaces		0	0%	0

One Solved Question from Each Vector Spaces Chapter

Question 1 · Vector Spaces, Subspaces and Bases NAT

Apply Gram-Schmidt to the ordered list $w_{1} = (1, 1, 0)$ , $w_{2} = (1, - 1, 1)$ , $w_{3} = (2, 0, 1)$ in $R^{3}$ . How many of the intermediate unnormalized residuals $z_{k}$ (before normalizing) turn out to be the zero vector?

Correct Answer:

0.00

Step-by-Step Solution

Key idea: estimation/counting over algorithm outputs; the overcounting trap is declaring a residual zero merely because dot products look small, rather than exact cancellation.

Step 1: z1 = w1 = (1,1,0). Nonzero. u1=z1/√2.

Step 2: z2 = w2 - (w2·u1)u1. w2·u1 = ((1)(1)+(-1)(1)+(1)(0))/√2 = 0/√2 =0. So z2=w2=(1,-1,1). Nonzero.

Step 3: z3 = w3 - (w3·u1)u1 - (w3·u2')u2' where u2'=z2 normalized. w3·u1 = (2+0+0)/√2=2/√2=√2. Projection onto u1 gives √2*u1=(1,1,0). w3·z2 = (2)(1)+(0)(-1)+(1)(1)=3. ||z2||^2=1+1+1=3. Component along z2 direction: (3/3)z2 = z2=(1,-1,1). Subtract both: w3-(1,1,0)-(1,-1,1)=(2-1-1, 0-1+1, 1-0-1)=(0,0,0). Zero!

Correction: z3 IS zero ⇒ exactly one zero residual. Answer should be 1.00.

Recompute cleanly: After step1 u1=(1/√2,1/√2,0). Step2 z2=(1,-1,1), u2=(1,-1,1)/√3. Step3: proj_u1 w3 = (w3·u1)u1 = √2 (1/√2,1/√2,0)=(1,1,0). proj_u2 w3=(w3·u2)u2; w3·u2=(2-0+1)/√3=3/√3=√3; times u2 = √3(1,-1,1)/√3=(1,-1,1). Sum projections=(2,0,1). w3 minus sum = (0,0,0). Yes z3=0. Count zeros among z1,z2,z3 = 1.

NAT answer: "1.00".

Question 2 · Orthogonality, Projections and Linear Systems MCQ

Let $P$ be the orthogonal projection matrix onto a subspace $U$ of $R^{4}$ . Let $Q$ be the orthogonal projection matrix onto the orthogonal complement $U^{⊥}$ . Consider the matrix $M = P - 3 Q$ . Let $S$ be the set of all possible values of $det (M)$ as $U$ varies over all possible subspaces of $R^{4}$ . Find the sum of all elements in $S$ .

A.
121
B.
61
C.
81
D.
54

Correct Answer:

Step-by-Step Solution

Key idea: This is a casework problem on the dimension of the subspace $U$ . The eigenvalues of an orthogonal projection matrix are strictly $0$ and $1$ .

Step 1: Let $d = dim (U)$ . Since $P$ projects onto $U$ , its eigenvalues are $1$ (with multiplicity $d$ ) and $0$ (with multiplicity $4 - d$ ).

Step 2: Since $Q$ projects onto $U^{⊥}$ , its eigenvalues are $0$ (with multiplicity $d$ ) and $1$ (with multiplicity $4 - d$ ).

Step 3: The matrix $M = P - 3 Q$ shares the same eigenvectors as $P$ and $Q$ because $U$ and $U^{⊥}$ are orthogonal complements.

For $v \in U$ , $M v = P v - 3 Q v = v - 0 = v$ . So $1$ is an eigenvalue of $M$ with multiplicity $d$ .

For $v \in U^{⊥}$ , $M v = P v - 3 Q v = 0 - 3 v = - 3 v$ . So $- 3$ is an eigenvalue of $M$ with multiplicity $4 - d$ .

Step 4: The determinant of $M$ is the product of its eigenvalues:

$det (M) = (1)^{d} (- 3)^{4 - d} = (- 3)^{4 - d}$ .

Step 5: Casework on $d \in {0, 1, 2, 3, 4}$ :

$d = 0 ⟹ det (M) = (- 3)^{4} = 81$
$d = 1 ⟹ det (M) = (- 3)^{3} = - 27$
$d = 2 ⟹ det (M) = (- 3)^{2} = 9$
$d = 3 ⟹ det (M) = (- 3)^{1} = - 3$
$d = 4 ⟹ det (M) = (- 3)^{0} = 1$

Step 6: The set of possible values is $S = {81, - 27, 9, - 3, 1}$ .

The sum of all elements in $S$ is $81 - 27 + 9 - 3 + 1 = 61$ .

Answer: 61

A question from this chapter

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Vector Spaces Practice Questions for GATE DA

About Vector Spaces Practice Questions

Vector Spaces Weightage in GATE DA

Vector Spaces Chapter Matrix

More from Linear Algebra

One Solved Question from Each Vector Spaces Chapter

Step-by-Step Solution

Step-by-Step Solution