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Orthogonality, Projections and Linear Systems Notes for GATE DA

Orthogonality, Projections and Linear Systems notes for GATE DA: 22 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice qu

orthogonality projections and linear systems notes

Chapter Roadmap: Orthogonality, Projections and Linear Systems

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The Geometry of Projection Matrices

.geo-svg { width: 100%; max-height: 220px; margin-bottom: 1rem; } .vec-drop { stroke-dasharray: 100; stroke-dashoffset: 100; animation: drop 2s infinite alternate ease-in-out; } @keyframes drop { to { stroke-dashoffset: 0; } } U b p e The Core Idea A projection matrix P maps any vector b \in R^{n} to a vector p in a target subspace U, such that the error vector e = b - p is orthogonal to U . Geometric Meaning p is the "shadow" or closest point in U . e is the "error" or perpendicular drop. P isolates the component of b that lives inside U .

Column Space and Null Space of P

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Question 1

Level 1: Warm-up

A linear system A x = b is consistent if and only if:

Question 2

Level 1: Warm-up

Which of the following pairs of algebraic properties uniquely defines an orthogonal projection matrix P ?

Question 3

Level 1: Warm-up

An orthogonal projection matrix P \in R^{n \times n} has rank k . What are its eigenvalues, and what is its trace?

Question 4

Level 1: Warm-up

A consistent system of linear equations with n unknowns has infinitely many solutions if:

Question 5

Level 1: Warm-up

For the system of linear equations x + y = 2 and 2 x + 2 y = k to have infinitely many solutions, what must be the value of the parameter k ?

Question 6

Level 1: Warm-up

The system of equations x + 2 y = 3 and 2 x + 4 y = k has infinitely many solutions. What is the value of k ?

Question 7

Level 1: Warm-up

Let P be the orthogonal projection matrix onto a subspace U \subseteq R^{n} . Which of the following correctly describes the null space of P ?

Question 8

Level 1: Warm-up

Let a = (21) . What is the orthogonal projection matrix P onto the line spanned by a ?

Question 9

Level 1: Warm-up

Let P be the orthogonal projection matrix onto the line spanned by the vector a = (34) in R^{2} . What is the trace of P ?

Question 10

Level 1: Warm-up

Let A be an m \times n matrix with full column rank. Which of the following expressions gives the orthogonal projection matrix onto the column space of A ?

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Orthogonality, Projections and Linear Systems Notes for GATE DA

Orthogonality, Projections and Linear Systems notes for GATE DA: 22 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Chapter Roadmap: Orthogonality, Projections and Linear Systems

Chapter Journey

Step 1: Projection Matrices, Null Space and Idempotence (Current)

Geometric intuition of orthogonal projections. Algebraic properties: Idempotence and Symmetry. Column space, Null space, and Eigenvalues.

Step 2: Consistency and Solution Sets of Linear Systems

Conditions for a linear system to have a solution. Geometric interpretation of consistency. Structuring the complete solution set.

Goal: Master the transition from projecting vectors to solving and analyzing linear systems.

The Geometry of Projection Matrices

The Core Idea

A projection matrix $P$ maps any vector $b \in R^{n}$ to a vector $p$ in a target subspace $U$ , such that the error vector $e = b - p$ is orthogonal to $U$ .

Geometric Meaning

$p$ is the "shadow" or closest point in $U$ .
$e$ is the "error" or perpendicular drop.
$P$ isolates the component of $b$ that lives inside $U$ .

Column Space and Null Space of P

Space Decomposition

Let $P$ be the projection matrix onto subspace $U$ .

1. Column Space $C (P)$

$C (P) = U$
$P v = v$ for $v \in C (P)$

2. Null Space $N (P)$

$N (P) = U^{⊥}$
$P v = 0$ for $v \in N (P)$

Fundamental Theorem:

R^{n} = C (P) \oplus N (P)

Every vector

b = p + e

, where

p \in C (P)

and

e \in N (P)

Spectral Properties: Eigenvalues and Trace

Eigenvalues

Since $P^{2} = P$ , for any eigenvalue $λ$ :

λ^{2} = λ ⟹ λ \in {0, 1}

$λ = 1$

Eigenvectors form

C (P)

$λ = 0$

Eigenvectors form

N (P)

Trace and Rank

The trace is the sum of eigenvalues. Since they are only 0s and 1s:

Trace (P) = Number of 1 s = Rank (P) = dim (C (P))

Orthogonality, Projections and Linear Systems: Solved Questions with Step-by-Step Explanations (10 Problems)

Question 1 · Linear Algebra MCQ

A linear system $A x = b$ is consistent if and only if:

A.
$Rank (A) = Rank ([A ∣ b])$
B.
$Rank (A) < Rank ([A ∣ b])$
C.
$det (A) \neq = 0$
D.
$b = 0$

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct-recall question about the fundamental condition for consistency.

Step 1: Consistency means there exists at least one solution $x$ .

Step 2: Geometrically, this means $b$ lies in the column space of $A$ .

Step 3: Algebraically, appending $b$ to $A$ to form the augmented matrix $[A ∣ b]$ adds a new column.

Step 4: If $b$ is already in the column space of $A$ , it does not increase the dimension of the column space. Thus, the rank remains unchanged.

Step 5: If $b$ is NOT in the column space, it adds a new independent direction, increasing the rank by 1.

Step 6: Therefore, consistency is equivalent to $Rank (A) = Rank ([A ∣ b])$ .

Answer: A

Question 2 · Linear Algebra MCQ

Which of the following pairs of algebraic properties uniquely defines an orthogonal projection matrix $P$ ?

A.
$P^{2} = I$ and $P^{T} = P$
B.
$P^{2} = P$ and $P^{T} = - P$
C.
$P^{2} = P$ and $P^{T} = P$
D.
$P^{2} = I$ and $P^{T} = - P$

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct recall question about the defining algebraic properties of orthogonal projection matrices.

Step 1: Recall the geometric meaning. An orthogonal projection drops a perpendicular to a target subspace.

Step 2: Algebraically, projecting a vector twice is the same as projecting it once, because the first projection already lands in the subspace. This gives idempotence: $P^{2} = P$ .

Step 3: The projection is "orthogonal", meaning the error vector is perpendicular to the subspace. This geometric requirement translates to the matrix being symmetric: $P^{T} = P$ .

Answer: $P^{2} = P$ and $P^{T} = P$ (Option C).

Question 3 · Linear Algebra MCQ

An orthogonal projection matrix $P \in R^{n \times n}$ has rank $k$ . What are its eigenvalues, and what is its trace?

A.
Eigenvalues are $0$ and $1$ ; Trace is $k$
B.
Eigenvalues are $- 1$ and $1$ ; Trace is $0$
C.
Eigenvalues are $0$ and $1$ ; Trace is $n$
D.
Eigenvalues are $1$ and $k$ ; Trace is $k$

Correct Answer:

Step-by-Step Solution

Key idea: This question tests the spectral properties (eigenvalues and trace) of projection matrices.

Step 1: Since $P^{2} = P$ , any eigenvalue $λ$ must satisfy $λ^{2} = λ$ . Thus, the only possible eigenvalues are $λ \in {0, 1}$ .

Step 2: The rank of $P$ is the dimension of its column space, which equals the number of non-zero eigenvalues. So there are exactly $k$ eigenvalues equal to $1$ , and the remaining $n - k$ eigenvalues are $0$ .

Step 3: The trace is the sum of the eigenvalues: $k \times 1 + (n - k) \times 0 = k$ .

Answer: Eigenvalues are $0$ and $1$ ; Trace is $k$ (Option A).

Question 4 · Linear Algebra MCQ

A consistent system of linear equations with $n$ unknowns has infinitely many solutions if:

A.
$Rank (A) = n$
B.
$Rank (A) < n$
C.
$Rank (A) > n$
D.
$Rank (A) = 0$

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct-recall question linking rank to the number of solutions.

Step 1: For a consistent system, the number of solutions is determined by the number of free variables.

Step 2: Number of free variables $= n - Rank (A)$ .

Step 3: If there are no free variables ( $n - Rank (A) = 0 ⟹ Rank (A) = n$ ), the solution is unique.

Step 4: If there is at least one free variable ( $n - Rank (A) \geq 1 ⟹ Rank (A) < n$ ), the system has infinitely many solutions.

Step 5: $Rank (A) > n$ is impossible since rank cannot exceed the number of columns.

Answer: B

Question 5 · Linear Algebra MCQ

For the system of linear equations $x + y = 2$ and $2 x + 2 y = k$ to have infinitely many solutions, what must be the value of the parameter $k$ ?

A.
4
B.
2
C.
0
D.
Any real number

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct-formula application question involving a parameter that determines consistency.

Step 1: Write the augmented matrix for the system:

$[1212 ∣ ∣ 2 k]$

Step 2: Perform row reduction to echelon form. Apply the operation $R_{2} \to R_{2} - 2 R_{1}$ :

$[1010 ∣ ∣ 2 k - 4]$

Step 3: Analyze the second row. It represents the equation $0 x + 0 y = k - 4$ , or simply $0 = k - 4$ .

Step 4: For the system to have infinitely many solutions, it must first be consistent. This requires the second row to not be a contradiction. Thus, we must have $k - 4 = 0$ , which gives $k = 4$ .

Step 5: If $k = 4$ , the second row becomes $0 = 0$ , leaving one independent equation with two variables, resulting in infinitely many solutions.

Answer: A

Question 6 · Linear Algebra MCQ

The system of equations $x + 2 y = 3$ and $2 x + 4 y = k$ has infinitely many solutions. What is the value of $k$ ?

A.
3
B.
5
C.
6
D.
8

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct-formula application question involving a parameter that determines consistency.

Step 1: Write the augmented matrix for the system:

$[1224 ∣ ∣ 3 k]$

Step 2: Perform row reduction to echelon form. Apply the operation $R_{2} \to R_{2} - 2 R_{1}$ :

$[1020 ∣ ∣ 3 k - 6]$

Step 3: Analyze the second row. It represents the equation $0 x + 0 y = k - 6$ , or simply $0 = k - 6$ .

Step 4: For the system to have infinitely many solutions, it must first be consistent. This requires the second row to not be a contradiction. Thus, we must have $k - 6 = 0$ , which gives $k = 6$ .

Step 5: If $k = 6$ , the second row becomes $0 = 0$ , leaving one independent equation with two variables, resulting in infinitely many solutions.

Answer: C

Question 7 · Linear Algebra MCQ

Let $P$ be the orthogonal projection matrix onto a subspace $U \subseteq R^{n}$ . Which of the following correctly describes the null space of $P$ ?

A.
The subspace $U$ itself
B.
The orthogonal complement $U^{⊥}$
C.
The entire space $R^{n}$
D.
The trivial subspace ${0}$

Correct Answer:

Step-by-Step Solution

Key idea: This question tests the fundamental subspaces of a projection matrix.

Step 1: The column space $C (P)$ is the set of all possible outputs of $P$ . Since $P$ projects onto $U$ , $C (P) = U$ .

Step 2: The null space $N (P)$ is the set of vectors that map to $0$ . Geometrically, these are the vectors that are completely "flattened" or removed by the projection.

Step 3: The vectors removed are exactly those orthogonal to the target subspace $U$ . Thus, $N (P) = U^{⊥}$ .

Answer: The orthogonal complement $U^{⊥}$ (Option B).

Question 8 · Linear Algebra MCQ

Let $a = (21)$ . What is the orthogonal projection matrix $P$ onto the line spanned by $a$ ?

A.
$egin{pmatrix} 1/5 & 2/5 \ 2/5 & 4/5 nd{pmatrix}$
B.
$egin{pmatrix} 4/5 & 2/5 \ 2/5 & 1/5 nd{pmatrix}$
C.
$egin{pmatrix} 4/5 & 1/5 \ 1/5 & 1/5 nd{pmatrix}$
D.
$egin{pmatrix} 2/5 & 4/5 \ 4/5 & 1/5 nd{pmatrix}$

Correct Answer:

Step-by-Step Solution

Key idea: This question requires direct substitution into the formula for the projection matrix onto a 1D subspace (a line).

Step 1: The formula for the projection matrix onto the line spanned by $a$ is $P = \frac{a a ^{T}}{a ^{T} a}$ .

Step 2: Calculate the denominator: $a^{T} a = 2^{2} + 1^{2} = 4 + 1 = 5$ .

Step 3: Calculate the numerator: $a a^{T} = (21) (21) = (4221)$ .

Step 4: Divide by the denominator: $P = \frac{1}{5} (4221) = (4/5 2/5 2/5 1/5)$ .

Answer: Option B.

Question 9 · Linear Algebra NAT

Let $P$ be the orthogonal projection matrix onto the line spanned by the vector $a = (34)$ in $R^{2}$ . What is the trace of $P$ ?

Correct Answer:

Step-by-Step Solution

Key idea: This question tests the spectral property that the trace of a projection matrix equals its rank.

Step 1: The matrix $P$ is a projection onto a line. A line in $R^{2}$ is a 1-dimensional subspace.

Step 2: The rank of a projection matrix is equal to the dimension of its column space, which is the target subspace. Thus, $rank (P) = 1$ .

Step 3: The eigenvalues of any projection matrix are strictly $0$ and $1$ . The number of $1$ s equals the rank. So, $P$ has one eigenvalue of $1$ and one eigenvalue of $0$ .

Step 4: The trace is the sum of the eigenvalues: $1 + 0 = 1$ .

Answer: 1

Question 10 · Linear Algebra MCQ

Let $A$ be an $m \times n$ matrix with full column rank. Which of the following expressions gives the orthogonal projection matrix onto the column space of $A$ ?

A.
$A (A^{T} A)^{- 1} A^{T}$
B.
$(A^{T} A)^{- 1} A^{T}$
C.
$A^{T} (A A^{T})^{- 1} A$
D.
$A (A^{T} A) A^{T}$

Correct Answer:

Step-by-Step Solution

Key idea: This question tests the standard formula for constructing a projection matrix onto a general subspace defined by the columns of a matrix $A$ .

Step 1: The projection of a vector $b$ onto the column space of $A$ is given by $p = A x$ , where $x$ solves the normal equations $A^{T} A x = A^{T} b$ .

Step 2: Since $A$ has full column rank, $A^{T} A$ is invertible. Thus, $x = (A^{T} A)^{- 1} A^{T} b$ .

Step 3: Substitute $x$ back into $p$ :

$p = A ((A^{T} A)^{- 1} A^{T} b) = (A (A^{T} A)^{- 1} A^{T}) b$

Step 4: The matrix that maps $b$ to $p$ is the projection matrix $P = A (A^{T} A)^{- 1} A^{T}$ .

Answer: Option A.

orthogonality projections and linear systems notes

Chapter Roadmap: Orthogonality, Projections and Linear Systems

Chapter Journey

The Geometry of Projection Matrices

The Core Idea

Column Space and Null Space of P

Space Decomposition

Try a question

Login to view the complete 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

Orthogonality, Projections and Linear Systems Notes for GATE DA

Chapter Roadmap: Orthogonality, Projections and Linear Systems

Chapter Journey

The Geometry of Projection Matrices

The Core Idea

Column Space and Null Space of P

Space Decomposition

Spectral Properties: Eigenvalues and Trace

Eigenvalues

Trace and Rank

Orthogonality, Projections and Linear Systems: 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

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