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Vector Spaces, Subspaces and Bases Short Notes for GATE DA

Vector Spaces, Subspaces and Bases short notes for GATE DA: 2 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice question

vector spaces subspaces and bases short notes

Quick Recap: Basis & Orthonormality

Quick Recap: Basis & Orthonormality Basis Linearly Independent + Spans V . Dimension Number of vectors in any basis. Orthonormal u_{i} \cdot u_{j} = δ_{ij} (Kronecker delta). Coordinates If B is orthonormal, v = \sum (v \cdot u_{i}) u_{i} . Gram-Schmidt Sequentially remove projections to orthogonalize. Subspace Test 0 \in W, Closed under +, Closed under \cdot .

Quick Recap: Geometry of Subspaces and Norms

Final Revision Points L_{2} Norm ∥ x ∥_{2} = x^{T} x . Standard Euclidean distance. Solid L_{2} Ball x^{T} x \leq R^{2} . Filled geometric region. Subspace k -dimensional flat passing through the origin. Intersection (Solid n -D ball, radius R) \cap (k -D subspace) = Solid k -D ball, radius R . 2D Area If k = 2, area is π R^{2} . Watch out x^{T} x = R^{2} (hollow) vs x^{T} x \leq R^{2} (solid).

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Question 1

Level 1: Warm-up

Assertion (A): In the Gram-Schmidt process, the first step to find u_{1} is to normalize the initial vector v_{1} . Reason (R): Normalization ensures that u_{1} is orthogonal to all other vectors in the basis.

Question 2

Level 1: Warm-up

Assertion (A): In the Gram-Schmidt process, the vector w_{2} is constructed by subtracting the projection of v_{2} onto u_{1} from v_{2} . Reason (R): This subtraction step ensures that the resulting vector w_{2} has a unit length.

Question 3

Level 1: Warm-up

Assertion (A): In the Gram-Schmidt process, the final step to find u_{2} is to divide the orthogonalized vector w_{2} by its norm ∥ w_{2} ∥ . Reason (R): This division step ensures that the resulting vector u_{2} is orthogonal to u_{1} .

Question 4

Level 1: Warm-up

Assertion (A): In the Gram-Schmidt process, the projection of v_{2} onto u_{1} is calculated as (v_{2} \cdot u_{1}) u_{1} . Reason (R): This projection vector is orthogonal to u_{1} .

Question 5

Level 1: Warm-up

Let {u_{1}, u_{2}} be an orthonormal basis for R^{2}, where u_{1} = (0, 1) and u_{2} = (- 1, 0) . If a vector v = (3, 4) is expressed as v = c_{1} u_{1} + c_{2} u_{2}, what is the value of the coefficient c_{2} ?

Question 6

Level 1: Warm-up

What is the minimum number of vectors in R^{4} that guarantees the set is linearly dependent, regardless of the specific vectors chosen?

Question 7

Level 1: Warm-up

Let {u_{1}, u_{2}} be an orthonormal basis for R^{2} with u_{1} = (1, 0) and u_{2} = (0, - 1) . If a vector v = (4, - 5) is expressed as v = c_{1} u_{1} + c_{2} u_{2}, what is the value of c_{1} - c_{2} ?

Question 8

Level 1: Warm-up

If a matrix A is formed by placing 4 vectors from R^{6} as its columns, what is the minimum rank A must have for these vectors to be linearly independent?

Question 9

Level 1: Warm-up

Let {u_{1}, u_{2}} be an orthonormal basis for R^{2}, where u_{1} = (\frac{3}{5}, \frac{4}{5}) and u_{2} = (- \frac{4}{5}, \frac{3}{5}) . If a vector v = (10, 5) is expressed as v = c_{1} u_{1} + c_{2} u_{2}, what is the value of c_{1} + c_{2} ?

Question 10

Level 1: Warm-up

If a matrix A is formed by placing 3 vectors from R^{5} as its columns, what is the minimum rank A must have for these vectors to be linearly independent?

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Vector Spaces, Subspaces and Bases Short Notes for GATE DA

Vector Spaces, Subspaces and Bases short notes for GATE DA: 2 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Quick Recap: Basis & Orthonormality

Basis

Linearly Independent + Spans

V

Dimension

Number of vectors in any basis.

Orthonormal

u_{i} \cdot u_{j} = δ_{ij}

(Kronecker delta).

Coordinates

B

is orthonormal,

v = \sum (v \cdot u_{i}) u_{i}

Gram-Schmidt

Sequentially remove projections to orthogonalize.

Subspace Test

0 \in W

, Closed under

+

, Closed under

\cdot

Quick Recap: Geometry of Subspaces and Norms

Final Revision Points

L_{2}

Norm

∥ x ∥_{2} = x^{T} x

. Standard Euclidean distance.

Solid

L_{2}

Ball

x^{T} x \leq R^{2}

. Filled geometric region.

Subspace

k

-dimensional flat passing through the origin.

Intersection

(Solid

n

-D ball, radius

R

)

\cap

(

k

-D subspace) = Solid $k$ -D ball, radius $R$ .

2D Area

k = 2

, area is

π R^{2}

Watch out

x^{T} x = R^{2}

(hollow) vs

x^{T} x \leq R^{2}

(solid).

Vector Spaces, Subspaces and Bases: Solved Questions with Step-by-Step Explanations (10 Problems)

Question 1 · Linear Algebra MCQ

Assertion (A): In the Gram-Schmidt process, the first step to find $u_{1}$ is to normalize the initial vector $v_{1}$ .

Reason (R): Normalization ensures that $u_{1}$ is orthogonal to all other vectors in the basis.

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:

Step-by-Step Solution

Key idea: Distinguish between normalization (unit length) and orthogonalization (perpendicularity).

Step 1: Evaluate Assertion (A). The Gram-Schmidt process starts by taking $v_{1}$ and dividing by its norm to get $u_{1}$ . This is normalization. So, A is true.

Step 2: Evaluate Reason (R). Normalization only ensures the vector has a length of 1 ( $∥ u_{1} ∥ = 1$ ). It does not make the vector orthogonal to anything. Orthogonality is achieved in subsequent steps by subtracting projections. So, R is false.

Step 3: Match with options. A is true, R is false.

Answer: A is true, but R is false.

Question 2 · Linear Algebra MCQ

Assertion (A): In the Gram-Schmidt process, the vector $w_{2}$ is constructed by subtracting the projection of $v_{2}$ onto $u_{1}$ from $v_{2}$ .

Reason (R): This subtraction step ensures that the resulting vector $w_{2}$ has a unit length.

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:

Step-by-Step Solution

Key idea: Distinguish between orthogonalization (subtraction) and normalization (division by norm). Step 1: Evaluate Assertion (A). The formula for

w_{2}

is indeed

w_{2} = v_{2} - (v_{2} \cdot u_{1}) u_{1}

. This is correct. Step 2: Evaluate Reason (R). The subtraction step removes the component of

v_{2}

that is parallel to

u_{1}

, making

w_{2}

orthogonal to

u_{1}

. It does not ensure unit length. Normalization (

u_{2} = w_{2} /∥ w_{2} ∥

) is required for unit length. Step 3: Conclude that A is true, but R is false. Answer: A is true, but R is false.

Question 3 · Linear Algebra MCQ

Assertion (A): In the Gram-Schmidt process, the final step to find $u_{2}$ is to divide the orthogonalized vector $w_{2}$ by its norm $∥ w_{2} ∥$ .

Reason (R): This division step ensures that the resulting vector $u_{2}$ is orthogonal to $u_{1}$ .

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:

Step-by-Step Solution

Key idea: Distinguish between orthogonalization (subtraction) and normalization (division by norm). Step 1: Evaluate Assertion (A). The formula for

u_{2}

is indeed

u_{2} = w_{2} /∥ w_{2} ∥

. This is correct. Step 2: Evaluate Reason (R). The division step scales

w_{2}

to have a length of 1. It does not ensure orthogonality. Orthogonality was already achieved in the previous step when

w_{2}

was constructed by subtracting the projection of

v_{2}

onto

u_{1}

. Step 3: Conclude that A is true, but R is false. Answer: A is true, but R is false.

Question 4 · Linear Algebra MCQ

Assertion (A): In the Gram-Schmidt process, the projection of $v_{2}$ onto $u_{1}$ is calculated as $(v_{2} \cdot u_{1}) u_{1}$ .

Reason (R): This projection vector is orthogonal to $u_{1}$ .

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:

Step-by-Step Solution

Key idea: Distinguish between the projection vector and the orthogonalized remainder vector. Step 1: Evaluate Assertion (A). The formula for the projection of

v_{2}

onto the unit vector

u_{1}

is indeed

(v_{2} \cdot u_{1}) u_{1}

. This is correct. Step 2: Evaluate Reason (R). The projection vector

(v_{2} \cdot u_{1}) u_{1}

is a scalar multiple of

u_{1}

, meaning it is parallel to

u_{1}

, not orthogonal to it. Orthogonality is achieved by the remainder vector

w_{2} = v_{2} - (v_{2} \cdot u_{1}) u_{1}

. Step 3: Conclude that A is true, but R is false. Answer: A is true, but R is false.

Question 5 · Linear Algebra MCQ

Let ${u_{1}, u_{2}}$ be an orthonormal basis for $R^{2}$ , where $u_{1} = (0, 1)$ and $u_{2} = (- 1, 0)$ . If a vector $v = (3, 4)$ is expressed as $v = c_{1} u_{1} + c_{2} u_{2}$ , what is the value of the coefficient $c_{2}$ ?

A.
-3
B.
3
C.
4
D.
-4

Correct Answer:

Step-by-Step Solution

Key idea: In an orthonormal basis, coefficients are found using the dot product.

Step 1: Recall the formula for coordinates in an orthonormal basis: $c_{i} = v \cdot u_{i}$ .

Step 2: We need $c_{2}$ , so we compute the dot product of $v$ and $u_{2}$ .

Step 3: $c_{2} = v \cdot u_{2} = (3, 4) \cdot (- 1, 0) = (3) (- 1) + (4) (0) = - 3 + 0 = - 3$ .

Answer: -3

Question 6 · Linear Algebra MCQ

What is the minimum number of vectors in $R^{4}$ that guarantees the set is linearly dependent, regardless of the specific vectors chosen?

A.
3
B.
4
C.
5
D.
6

Correct Answer:

Step-by-Step Solution

Key idea: In $R^{n}$ , any set of more than $n$ vectors is linearly dependent.

Step 1: Identify the dimension of the space. Here, the space is $R^{4}$ , so $n = 4$ .

Step 2: Apply the theorem: Any set of $k$ vectors in $R^{n}$ is linearly dependent if $k > n$ .

Step 3: We need the minimum $k$ such that $k > 4$ . The smallest integer greater than 4 is 5.

Answer: 5

Question 7 · Linear Algebra MCQ

Let ${u_{1}, u_{2}}$ be an orthonormal basis for $R^{2}$ with $u_{1} = (1, 0)$ and $u_{2} = (0, - 1)$ . If a vector $v = (4, - 5)$ is expressed as $v = c_{1} u_{1} + c_{2} u_{2}$ , what is the value of $c_{1} - c_{2}$ ?

A.
-1
B.
9
C.
-9
D.
1

Correct Answer:

Step-by-Step Solution

Key idea: In an orthonormal basis, coordinates are found via dot products.

Step 1: Use the formula $c_{i} = v \cdot u_{i}$ .

Step 2: Calculate $c_{1} = (4, - 5) \cdot (1, 0) = 4 (1) + (- 5) (0) = 4$ .

Step 3: Calculate $c_{2} = (4, - 5) \cdot (0, - 1) = 4 (0) + (- 5) (- 1) = 5$ .

Step 4: Compute the requested value: $c_{1} - c_{2} = 4 - 5 = - 1$ .

Answer: -1

Question 8 · Linear Algebra MCQ

If a matrix $A$ is formed by placing 4 vectors from $R^{6}$ as its columns, what is the minimum rank $A$ must have for these vectors to be linearly independent?

A.
2
B.
4
C.
6
D.
10

Correct Answer:

Step-by-Step Solution

Key idea: The rank method states that vectors are independent if the rank equals the number of vectors.

Step 1: Identify the number of vectors. Here, there are $k = 4$ vectors.

Step 2: Apply the decision rule from the rank method: The vectors are linearly independent if and only if $Rank (A) = k$ .

Step 3: Therefore, the minimum (and exact) rank required is 4. The dimension of the space ( $n = 6$ ) is irrelevant as long as $k \leq n$ .

Answer: 4

Question 9 · Linear Algebra MCQ

Let ${u_{1}, u_{2}}$ be an orthonormal basis for $R^{2}$ , where $u_{1} = (\frac{3}{5}, \frac{4}{5})$ and $u_{2} = (- \frac{4}{5}, \frac{3}{5})$ . If a vector $v = (10, 5)$ is expressed as $v = c_{1} u_{1} + c_{2} u_{2}$ , what is the value of $c_{1} + c_{2}$ ?

A.
5
B.
21
C.
-5
D.
15

Correct Answer:

Step-by-Step Solution

Key idea: In an orthonormal basis, coordinates are found via dot products.

Step 1: Use the formula $c_{i} = v \cdot u_{i}$ .

Step 2: Calculate $c_{1} = (10, 5) \cdot (\frac{3}{5}, \frac{4}{5}) = 10 (\frac{3}{5}) + 5 (\frac{4}{5}) = 6 + 4 = 10$ .

Step 3: Calculate $c_{2} = (10, 5) \cdot (- \frac{4}{5}, \frac{3}{5}) = 10 (- \frac{4}{5}) + 5 (\frac{3}{5}) = - 8 + 3 = - 5$ .

Step 4: Compute the requested sum: $c_{1} + c_{2} = 10 + (- 5) = 5$ .

Answer: 5

Question 10 · Linear Algebra MCQ

If a matrix $A$ is formed by placing 3 vectors from $R^{5}$ as its columns, what is the minimum rank $A$ must have for these vectors to be linearly independent?

A.
2
B.
3
C.
5
D.
8

Correct Answer:

Step-by-Step Solution

Key idea: The rank method states that vectors are independent if the rank equals the number of vectors.

Step 1: Identify the number of vectors. Here, there are $k = 3$ vectors.

Step 2: Apply the decision rule from the rank method: The vectors are linearly independent if and only if $Rank (A) = k$ .

Step 3: Therefore, the minimum (and exact) rank required is 3. The dimension of the space ( $n = 5$ ) is irrelevant as long as $k \leq n$ .

Answer: 3

vector spaces subspaces and bases short notes

Quick Recap: Basis & Orthonormality

Quick Recap: Basis & Orthonormality

Quick Recap: Geometry of Subspaces and Norms

Final Revision Points

Try a question

Login to view the complete short notes

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course

Vector Spaces, Subspaces and Bases Short Notes for GATE DA

Quick Recap: Basis & Orthonormality

Quick Recap: Basis & Orthonormality

Quick Recap: Geometry of Subspaces and Norms

Final Revision Points

Vector Spaces, Subspaces and Bases: Solved Questions with Step-by-Step Explanations (10 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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