GATE DA

Special matrices

Projection, orthogonal, idempotent and partitioned matrices, and quadratic forms

Orthogonal, idempotent and projection matrices, partitioned (block) matrices, and quadratic forms with their definiteness tests. The key facts are the eigenvalues of each special type, rank = trace for idempotent matrices, the block determinant, transpose and inverse rules, and the eigenvalue and minor tests for definiteness. With these, many long computations become one-line checks.

7 previous-year questions · 7 solved · 6 practice questions

Conventions and the family of special matrices

Throughout, matrices are real, vectors are column vectors written in bold ( $x$ ), $A^{T}$ is the transpose, $I$ is the identity of the size that makes the product defined, $λ$ denotes an eigenvalue and $∥ x ∥ = x^{T} x$ is the Euclidean norm. $A$ is $n \times n$ unless a size is stated.

Type	Defining condition	Eigenvalues	Determinant
Symmetric	$A^{T} = A$	all real; eigenvectors for distinct eigenvalues are orthogonal	no restriction
Skew-symmetric	$A^{T} = - A$	$0$ or purely imaginary, in $\pm bi$ pairs	$0$ if $n$ is odd; $\geq 0$ if $n$ is even
Orthogonal	$Q^{T} Q = I$ ( $Q$ square)	modulus $1$	$+ 1$ or $- 1$
Idempotent	$A^{2} = A$	only $0$ or $1$	$1$ if $A = I$ , otherwise $0$
Orthogonal projection	$P^{2} = P$ and $P^{T} = P$	only $0$ or $1$	$1$ if $P = I$ , otherwise $0$
Triangular or diagonal	all entries below the diagonal are zero (upper triangular) or all above are zero (lower triangular); both for diagonal	the diagonal entries	product of the diagonal entries

SYMMETRIC PLUS SKEW DECOMPOSITION

A = symmetric \frac{1}{2} (A + A^{T}) + skew-symmetric \frac{1}{2} (A - A^{T})

Holds for every square real matrix, and the split is unique.

SKEW-SYMMETRIC FORM VANISHES

x^{T} S x = 0 for every x when S^{T} = - S

$S$ is $n \times n$ and $x$ is any real vector in $R^{n}$ . Because $x^{T} S x$ is a scalar, it equals its own transpose, which is $- x^{T} S x$ .

Orthogonal matrices

A square matrix $Q$ is orthogonal when its columns are orthonormal, that is $Q^{T} Q = I$ . For a square matrix this forces $Q^{- 1} = Q^{T}$ and hence also $Q Q^{T} = I$ , so the rows are orthonormal too. Geometrically $Q$ keeps lengths and angles: it is a rotation when $det Q = + 1$ , and involves a reflection when $det Q = - 1$ .

ORTHOGONAL MATRIX

Q^{T} Q = Q Q^{T} = I, Q^{- 1} = Q^{T}

Square real $Q$ . Columns (and rows) form an orthonormal set.

LENGTH AND ANGLE PRESERVATION

∥ Q x ∥ = ∥ x ∥, (Q x)^{T} (Q y) = x^{T} y

Holds for every $x, y$ when $Q$ is orthogonal.

DETERMINANT AND EIGENVALUES OF AN ORTHOGONAL MATRIX

det Q = \pm 1, ∣ λ ∣ = 1 for every eigenvalue λ

A real eigenvalue can only be $+ 1$ or $- 1$ . Complex eigenvalues come in conjugate pairs of modulus 1.

EVERY 2×2 ORTHOGONAL MATRIX

R (θ) = [cos θ sin θ - sin θ cos θ], F (θ) = [cos θ sin θ sin θ - cos θ]

Every real $2 \times 2$ orthogonal matrix is one of these two forms. $R (θ)$ is a rotation by $θ$ : $det = + 1$ , eigenvalues $cos θ \pm i sin θ$ . $F (θ)$ is a reflection: symmetric, $det = - 1$ , eigenvalues $1$ and $- 1$ .

If $Q$ is orthogonal, so are $Q^{T} = Q^{- 1}$ , every power $Q^{k}$ and $- Q$ . The product of two orthogonal matrices of the same size is orthogonal.
A symmetric orthogonal matrix satisfies $Q^{2} = Q^{T} Q = I$ , so its eigenvalues are $\pm 1$ only.
If $det Q = - 1$ then $- 1$ is an eigenvalue of $Q$ , for any $n$ . If $n$ is odd and $det Q = + 1$ then $1$ is an eigenvalue. For $n = 3$ the eigenvector for $1$ is the rotation axis.
Permutation matrices are orthogonal.

EXAM TIP

$det Q = \pm 1$ is necessary, not sufficient. $diag (2, \frac{1}{2})$ has determinant $1$ and is not orthogonal. To test orthogonality, check that the columns have unit length and are mutually perpendicular.

Idempotent matrices

A square matrix $A$ is idempotent when $A^{2} = A$ . If $A v = λ v$ with $v \neq = 0$ , then $A^{2} v = λ^{2} v = λ v$ , so $λ^{2} = λ$ and $λ \in {0, 1}$ . An idempotent matrix is always diagonalisable (its minimal polynomial divides $x^{2} - x$ , which has distinct roots), so the number of eigenvalues equal to $1$ is the rank. If $A$ is idempotent, so is $A^{T}$ .

IDEMPOTENT: EIGENVALUES AND RANK

A^{2} = A \Rightarrow λ \in {0, 1}, rank (A) = trace (A)

Square matrix with $A^{2} = A$ . The converse is false: $diag (3, - 1)$ has rank $2 =$ trace $2$ and is not idempotent. For a general matrix, rank and trace are unrelated.

COMPLEMENT OF AN IDEMPOTENT

(I - A)^{2} = I - A, A (I - A) = (I - A) A = 0

If $A$ ( $n \times n$ ) is idempotent, then so is $I - A$ , and $rank (A) + rank (I - A) = n$ .

POWERS AND A QUICK INVERSE

A^{k} = A (k \geq 1), (I + A)^{- 1} = I - \frac{1}{2} A

For idempotent $A$ . Check the inverse by multiplying out: $(I + A) (I - \frac{1}{2} A) = I + A - \frac{1}{2} A - \frac{1}{2} A^{2} = I$ .

EXAM TIP

An idempotent matrix need not be symmetric. $[1010]$ squares to itself, has eigenvalues $1, 0$ and is not symmetric. It projects onto the $x_{1}$ -axis along the direction $(1, - 1)^{T}$ , which is not perpendicular to that axis. Only symmetric idempotents are orthogonal projections. Also, the only invertible idempotent is $I$ .

Projection matrices

Let $A$ be an $n \times k$ matrix with linearly independent columns. The orthogonal projection of $b \in R^{n}$ onto the column space of $A$ is the point $A \hat{x}$ closest to $b$ . The residual $b - A \hat{x}$ is perpendicular to every column of $A$ , which gives the normal equations $A^{T} A \hat{x} = A^{T} b$ .

PROJECTION MATRIX ONTO THE COLUMN SPACE OF A

P = A (A^{T} A)^{- 1} A^{T} (P is n \times n)

Needs the $k$ columns of $A$ linearly independent, so that the $k \times k$ matrix $A^{T} A$ is invertible. $P b$ is the point of the column space nearest to $b$ .

SPECIAL CASES OF THE PROJECTION MATRIX

P = \frac{a a ^{T}}{a ^{T} a}, P = Q Q^{T} (Q has orthonormal columns)

First: projection onto the line through $a \neq = 0$ . Second: $Q$ is $n \times k$ with $Q^{T} Q = I_{k}$ , and need not be square.

PROPERTIES OF AN ORTHOGONAL PROJECTION

P^{2} = P, P^{T} = P, trace (P) = rank (P) = k, λ \in {0, 1}

For $P = A (A^{T} A)^{- 1} A^{T}$ with $A$ of size $n \times k$ and rank $k$ . $P$ has eigenvalue $1$ on the column space (multiplicity $k$ ) and $0$ on its orthogonal complement (multiplicity $n - k$ ). $P$ depends only on the column space, not on which basis forms the columns of $A$ .

PYTHAGORAS FOR A PROJECTION

P b ⊥ (b - P b), ∥ b ∥^{2} = ∥ P b ∥^{2} + ∥(I - P) b ∥^{2}

$I - P$ projects onto the orthogonal complement of the column space. Use it to find the distance from $b$ to the subspace.

REFLECTION MATRIX

H = I - 2 \frac{a a ^{T}}{a ^{T} a}, H^{T} = H, H^{T} H = H^{2} = I, det H = - 1

$a \neq = 0$ in $R^{n}$ , and $H$ is $n \times n$ . $H = I - 2 P$ , with $P$ the projection onto the line through $a$ . $H$ reflects across the hyperplane perpendicular to $a$ . Its eigenvalues are $- 1$ once (along $a$ ) and $1$ with multiplicity $n - 1$ . More generally, $I - 2 P$ is symmetric and orthogonal for every orthogonal projection $P$ .

Check that the columns of $A$ are independent. If they are not, replace $A$ by a matrix whose columns are a basis of its column space.
Form $A^{T} A$ ( $k \times k$ ) and $A^{T} b$ .
Solve $A^{T} A \hat{x} = A^{T} b$ .
The projection is $A \hat{x}$ , and the distance is $∥ b - A \hat{x} ∥$ .
Verify that $A^{T} (b - A \hat{x}) = 0$ .

WORKED EXAMPLE

Let $A = 101011$ ( $3 \times 2$ ) and $b = 300$ . Find the projection of $b$ onto the column space of $A$ , and $trace (P)$ .

Solution

$A^{T} A = [2112]$ , with determinant $3$ , so $(A^{T} A)^{- 1} = \frac{1}{3} [2 - 1 - 1 2]$ .
$A^{T} b = [30]$ , so $\hat{x} = \frac{1}{3} [6 - 3] = [2 - 1]$ .
$P b = A \hat{x} = 2101 - 011 = 2 - 1 1$ .
Check: the residual $11 - 1$ is perpendicular to both columns (both dot products are $0$ ).
$trace (P) = rank (P) = 2$ .

Answer: $P b = (2, - 1, 1)^{T}$ and $trace (P) = 2$ .

Partitioned (block) matrices

A partitioned matrix is a matrix cut by horizontal and vertical lines into sub-matrices (blocks). Blocks add and multiply by the same rules as numbers, provided the block sizes are compatible and the order of every product is kept, because matrix products do not commute.

BLOCK MULTIPLICATION

[A C B D] [E G F H] = [A E + B G C E + D G A F + B H C F + D H]

The column partition of the left factor must match the row partition of the right factor, so that every product such as $A E$ and $B G$ is defined.

TRANSPOSE OF A PARTITIONED MATRIX

[A C B D]^{T} = [A^{T} B^{T} C^{T} D^{T}]

Transpose each block and also swap the two off-diagonal blocks.

DETERMINANT OF A BLOCK TRIANGULAR MATRIX

det [A 0 B D] = det A \cdot det D

$A$ and $D$ square. The same holds for block lower triangular and block diagonal matrices. The eigenvalues of the whole matrix are those of $A$ together with those of $D$ , with multiplicities.

INVERSES OF BLOCK DIAGONAL AND BLOCK TRIANGULAR MATRICES

[A 0 0 D]^{- 1} = [A^{- 1} 0 0 D^{- 1}], [A 0 B D]^{- 1} = [A^{- 1} 0 - A^{- 1} B D^{- 1} D^{- 1}]

$A$ and $D$ square. The matrix is invertible exactly when $A$ and $D$ are both invertible.

SCHUR COMPLEMENT DETERMINANT

det [A C B D] = det A \cdot det (D - C A^{- 1} B)

Requires $A$ square and invertible, and $D$ square. $D - C A^{- 1} B$ is the Schur complement of $A$ . If instead $D$ is invertible, $det [A C B D] = det D \cdot det (A - B D^{- 1} C)$ .

RANK AND POWERS OF A BLOCK DIAGONAL MATRIX

rank [A 0 0 D] = rank A + rank D, [A 0 0 D]^{k} = [A^{k} 0 0 D^{k}]

The rank statement allows diagonal blocks of any size. The power statement needs square blocks.

EXAM TIP

Do not copy the $2 \times 2$ rule: in general $det [A C B D] \neq = det A det D - det B det C$ . The formula $det (A D - B C)$ is valid only when all four blocks are square, of equal size, and $C$ and $D$ commute. Otherwise use the Schur complement or a block triangular form.

Quadratic forms and definiteness

A quadratic form in $x = (x_{1}, \dots, x_{n})^{T}$ is $q (x) = x^{T} M x$ for an $n \times n$ matrix $M$ . The skew part of $M$ contributes nothing, so only $\frac{1}{2} (M + M^{T})$ matters. We therefore always work with the unique symmetric matrix $A$ : its diagonal entries are the coefficients of $x_{i}^{2}$ , and each off-diagonal pair $a_{ij} = a_{j i}$ is half the coefficient of $x_{i} x_{j}$ .

QUADRATIC FORM AND ITS SYMMETRIC MATRIX

q (x) = x^{T} A x = i \sum a_{ii} x_{i}^{2} + 2 i < j \sum a_{ij} x_{i} x_{j}, A = \frac{1}{2} (M + M^{T})

$A$ is real symmetric. Example: $x_{1}^{2} + 6 x_{1} x_{2} + 4 x_{2}^{2}$ has $A = [1334]$ .

PRINCIPAL AXES (ORTHOGONAL DIAGONALISATION)

A = Q Λ Q^{T}, q (x) = i = 1 \sum n λ_{i} y_{i}^{2} with y = Q^{T} x

A real symmetric $A$ always has real eigenvalues $λ_{i}$ and an orthogonal $n \times n$ matrix $Q$ whose columns are eigenvectors. $Λ = diag (λ_{1}, \dots, λ_{n})$ . The signs of the $λ_{i}$ decide the definiteness.

Class	Condition on $q (x)$	Eigenvalue test	Minor test
Positive definite	$> 0$ for all $x \neq = 0$	all $λ_{i} > 0$	all leading principal minors $> 0$
Positive semi-definite	$\geq 0$ for all $x$	all $λ_{i} \geq 0$	all principal minors (every one, not only the leading ones) $\geq 0$
Negative definite	$< 0$ for all $x \neq = 0$	all $λ_{i} < 0$	leading minors alternate in sign, starting negative: $Δ_{1} < 0, Δ_{2} > 0, Δ_{3} < 0, \dots$ (that is, $(- 1)^{k} Δ_{k} > 0$ )
Negative semi-definite	$\leq 0$ for all $x$	all $λ_{i} \leq 0$	every principal minor of order $k$ is $\leq 0$ for odd $k$ and $\geq 0$ for even $k$ (that is, $- A$ is positive semi-definite)
Indefinite	takes both positive and negative values	at least one positive and one negative eigenvalue	none of the four patterns above; for example, any principal minor of even order is negative, or two diagonal entries have opposite signs

LEADING PRINCIPAL MINOR TEST

Δ_{k} = det (A_{1 : k, 1 : k}) > 0 for k = 1, \dots, n ⟺ A positive definite

$A$ real symmetric. $Δ_{k}$ is the determinant of the top-left $k \times k$ block. For $n = 2$ : $[a b b c]$ is positive definite exactly when $a > 0$ and $a c - b^{2} > 0$ .

EXTREME VALUES OF A QUADRATIC FORM ON THE UNIT SPHERE

λ_{m i n} x^{T} x \leq x^{T} A x \leq λ_{m a x} x^{T} x, ∥ x ∥ = 1 max x^{T} A x = λ_{m a x}, ∥ x ∥ = 1 min x^{T} A x = λ_{m i n}

$A$ real symmetric. Equality holds when $x$ is an eigenvector for $λ_{m a x}$ (or for $λ_{m i n}$ ). It follows from $q = \sum λ_{i} y_{i}^{2}$ with $∥ y ∥ = ∥ x ∥$ .

CHANGE OF VARIABLES AND THE LAW OF INERTIA

x = T y (T invertible) \Rightarrow q = y^{T} (T^{T} A T) y

$T$ is an invertible $n \times n$ matrix. $T^{T} A T$ is symmetric and has the same numbers of positive, negative and zero eigenvalues as $A$ , although the eigenvalues themselves may differ. So the definiteness class, the rank, and the number of positive terms in any diagonal form of $q$ do not change.

For every real $m \times n$ matrix $B$ , the $n \times n$ matrix $B^{T} B$ is positive semi-definite. It is positive definite exactly when the columns of $B$ are independent.
A positive definite matrix has positive diagonal entries and a positive determinant. It is invertible, and $A^{- 1}$ is positive definite (eigenvalues $1/ λ_{i}$ ).
The sum of a positive definite and a positive semi-definite matrix of the same size is positive definite.
$trace (A) = \sum λ_{i}$ and $det A = \prod λ_{i}$ (eigenvalues counted with multiplicity). Use them to read off signs quickly.
An orthogonal projection matrix $P$ is positive semi-definite, because $x^{T} P x = x^{T} P^{T} P x = ∥ P x ∥^{2} \geq 0$ .

WORKED EXAMPLE

Classify $q (x) = 2 x_{1}^{2} + 2 x_{2}^{2} + 2 x_{3}^{2} + 2 x_{1} x_{2} + 2 x_{2} x_{3}$ .

Solution

Each cross-term coefficient $2$ gives $a_{12} = a_{23} = 1$ . There is no $x_{1} x_{3}$ term, so $a_{13} = 0$ . Hence $A = 210121012$ .
$Δ_{1} = 2 > 0$ .
$Δ_{2} = det [2112] = 3 > 0$ .
$Δ_{3} = det A = 2 (4 - 1) - 1 (2 - 0) = 4 > 0$ .
All leading minors are positive, so $A$ is positive definite. Its eigenvalues are $2$ and $2 \pm 2$ , all positive: check $trace = 6$ and $det = 2 \cdot (4 - 2) = 4$ .

Answer: Positive definite.

EXAM TIP

The leading-minor test with minors $\geq 0$ does NOT prove semi-definiteness. $A = [00 0 - 1]$ has leading minors $0, 0$ , yet it is negative semi-definite. For semi-definiteness you need all principal minors, or simply the eigenvalues. Also, always symmetrise first: the matrix of $x_{1}^{2} + 4 x_{1} x_{2} + x_{2}^{2}$ is $[1221]$ (indefinite, $det = - 3$ ), not $[1041]$ .

PREVIOUS-YEAR QUESTIONS

7 questions from the DA papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE DA 2026 · GATE DA 2025 · GATE DA 2024

GATE 2026 DA Q.21

GATE 2026 · DAQ.21 · 1 mark · MCQ

Let $M = (cos θ sin θ - sin θ cos θ)$ be a $2 \times 2$ matrix, where $θ = \frac{2 π}{5}$ , and $I_{2} = (1001)$ . Which of the following options is equal to $M^{2026}$ ?

(A)
$M^{2}$
(B)
$M$
(C)
$M^{- 1}$
(D)
$I_{2}$

Answer and solution

GATE 2026 DA Q.21: answer and solution

See the question in the paper

Tap to open full size

GATE 2026 DA Q.21: answer and solution

GATE 2026 DA Q.52

GATE 2026 · DAQ.52 · 2 marks · MSQ

Let $M = (I_{n} - \frac{1}{n} 1 1^{T})$ be a matrix, where $1 = (1, 1, 1, \dots, 1)^{T} \in R^{n}$ and $I_{n}$ is the identity matrix of order $n$ .

Which of the following options is/are correct?

(A)
$M^{T} = M$
(B)
$M^{2} = I_{n}$
(C)
$Trace (M) = n$
(D)
$M$ is a projection matrix

Answer and solution

GATE 2026 DA Q.52: answer and solution

See the question in the paper

Tap to open full size

GATE 2026 DA Q.52: answer and solution

GATE 2026 DA Q.65

GATE 2026 · DAQ.65 · 2 marks · NAT

Let $A = (I_{n} - \frac{1}{n} 1 1^{T})$ be a matrix, where $1 = (1, 1, 1, \dots, 1)^{T} \in R^{n}$ and $I_{n}$ is the identity matrix of order $n$ .

The value of $max_{S} x^{T} A x$ , where $S = {x \in R^{n} ∣ x^{T} x = 1}$ , is ________. (Answer in integer)

Answer and solution

GATE 2026 DA Q.65: answer and solution

See the question in the paper

Tap to open full size

GATE 2026 DA Q.65: answer and solution

GATE 2025 DA Q.38

GATE 2025 · DAQ.38 · 2 marks · MCQ

Let ${x_{1}, x_{2}, \dots, x_{n}}$ be a set of linearly independent vectors in $R^{n}$ . Let the $(i, j)$ -th element of matrix $A \in R^{n \times n}$ be given by $A_{ij} = x_{i}^{⊤} x_{j}$ , $1 \leq i, j \leq n$ . Which one of the following statements is correct?

(A)
$A$ is invertible
(B)
$0$ is a singular value of $A$
(C)
Determinant of $A$ is $0$
(D)
$z^{⊤} A z = 0$ for some non-zero $z \in R^{n}$

Answer and solution

GATE 2025 DA Q.38: answer and solution

See the question in the paper

Tap to open full size

GATE 2025 DA Q.38: answer and solution

GATE 2025 DA Q.50

GATE 2025 · DAQ.50 · 2 marks · MSQ

Let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be a system of orthonormal vectors in $R^{10}$ . Consider the matrix $A = x_{1} x_{1}^{⊤} + \dots + x_{5} x_{5}^{⊤}$ . Which of the following statements is/are correct?

(A)
Singular values of $A$ are also its eigenvalues
(B)
Singular values of $A$ are either 0 or 1
(C)
Determinant of $A$ is 1
(D)
$A$ is invertible

Answer and solution

GATE 2025 DA Q.50: answer and solution

See the question in the paper

Tap to open full size

GATE 2025 DA Q.50: answer and solution

GATE 2025 DA Q.52

GATE 2025 · DAQ.52 · 2 marks · MSQ

An $n \times n$ matrix $A$ with real entries satisfies the property: $∥ A x ∥^{2} = ∥ x ∥^{2}$ , for all $x \in R^{n}$ , where $∥ \cdot ∥$ denotes the Euclidean norm. Which of the following statements is/are ALWAYS correct?

(A)
$A$ must be orthogonal
(B)
$A = I$ , where $I$ denotes the identity matrix, is the only solution
(C)
The eigenvalues of $A$ are either $+ 1$ or $- 1$
(D)
$A$ has full rank

Answer and solution

GATE 2025 DA Q.52: answer and solution

See the question in the paper

Tap to open full size

GATE 2025 DA Q.52: answer and solution

GATE 2024 DA Q.49

GATE 2024 · DAQ.49 · 2 marks · MSQ

Let $R$ be the set of real numbers, $U$ be a subspace of $R^{3}$ and $M \in R^{3 \times 3}$ be the matrix corresponding to the projection on to the subspace $U$ .

Which of the following statements is/are TRUE?

(A)
If $U$ is a 1-dimensional subspace of $R^{3}$ , then the null space of $M$ is a 1-dimensional subspace.
(B)
If $U$ is a 2-dimensional subspace of $R^{3}$ , then the null space of $M$ is a 1-dimensional subspace.
(C)
$M^{2} = M$
(D)
$M^{3} = M$

Answer and solution

GATE 2024 DA Q.49: answer and solution

See the question in the paper

Tap to open full size

GATE 2024 DA Q.49: answer and solution

PRACTICE

6 questions to test yourself

PRACTICE 11 mark · NAT

A real $4 \times 4$ matrix $A$ satisfies $A^{2} = A$ and $trace (A) = 3$ . Find $det (A + I)$ , where $I$ is the $4 \times 4$ identity.

Answer and solution

Answer: $8$

An idempotent matrix is diagonalisable with eigenvalues in ${0, 1}$ , and $rank = trace = 3$ .
So the eigenvalues are $1, 1, 1, 0$ .
The eigenvalues of $A + I$ are $2, 2, 2, 1$ .
$det (A + I) = 2 \cdot 2 \cdot 2 \cdot 1 = 8$ .

PRACTICE 22 marks · NAT

Let $a = (1, 2, 2)^{T}$ and $b = (3, 0, 0)^{T}$ . Let $P$ be the $3 \times 3$ orthogonal projection matrix onto the line spanned by $a$ . Find the squared Euclidean distance $∥ b - P b ∥^{2}$ from $b$ to that line.

Answer and solution

Answer: $8$

$P = a a^{T} / (a^{T} a)$ , with $a^{T} a = 9$ .
$a^{T} b = 3$ , so $P b = \frac{3}{9} a = (\frac{1}{3}, \frac{2}{3}, \frac{2}{3})^{T}$ and $∥ P b ∥^{2} = \frac{1}{9} \cdot 9 = 1$ .
$P b$ and $b - P b$ are perpendicular, so $∥ b - P b ∥^{2} = ∥ b ∥^{2} - ∥ P b ∥^{2} = 9 - 1 = 8$ .
Direct check: $b - P b = (\frac{8}{3}, - \frac{2}{3}, - \frac{2}{3})^{T}$ , whose squared length is $\frac{64 + 4 + 4}{9} = 8$ .

PRACTICE 32 marks · MSQ

Let $Q = \frac{1}{3} 122 21 - 2 2 - 2 1$ . Which of the following statements are true?

(A)
$Q^{- 1} = Q^{T}$
(B)
$det Q = + 1$
(C)
$Q^{2} = I$
(D)
$trace (Q) = 1$

Answer and solution

Answer: (A), (C), (D)

Each row has squared length $(1 + 4 + 4) /9 = 1$ . The pairwise dot products are $2 + 2 - 4 = 0$ , $2 - 4 + 2 = 0$ and $4 - 2 - 2 = 0$ (all divided by $9$ ). So $Q$ is orthogonal, and (A) is true.
$Q$ is also symmetric, so $Q^{2} = Q^{T} Q = I$ , and (C) is true.
The trace is $\frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1$ , so (D) is true.
From $Q^{2} = I$ the eigenvalues are $\pm 1$ , and they sum to $1$ , so they are $1, 1, - 1$ and $det Q = - 1$ . Directly: $det 122 21 - 2 2 - 2 1 = - 3 - 12 - 12 = - 27$ , and $- 27/27 = - 1$ . Hence (B) is false.

PRACTICE 41 mark · MCQ

The quadratic form $q (x_{1}, x_{2}) = x_{1}^{2} + 4 x_{1} x_{2} + k x_{2}^{2}$ is positive definite for which values of the real constant $k$ ?

(A)
$k > 4$
(B)
$k > 2$
(C)
$k > 0$
(D)
$k < 4$

Answer and solution

Answer: (A)

The symmetric matrix is $A = [12 2 k]$ .
Leading minors: $Δ_{1} = 1 > 0$ always, and $Δ_{2} = k - 4$ .
Positive definiteness needs $Δ_{2} > 0$ , that is $k > 4$ .

PRACTICE 52 marks · NAT

Let $M = [A 0 B C]$ be a $5 \times 5$ matrix, where $A = [2112]$ , $C = diag (1, 2, - 1)$ and $B$ is an arbitrary $2 \times 3$ real matrix. Find $det (2 M)$ .

Answer and solution

Answer: $- 192$

$M$ is block upper triangular, so $det M = det A \cdot det C$ , whatever $B$ is.
$det A = 4 - 1 = 3$ and $det C = 1 \cdot 2 \cdot (- 1) = - 2$ , so $det M = - 6$ .
For a $5 \times 5$ matrix, $det (2 M) = 2^{5} det M = 32 \cdot (- 6) = - 192$ .

PRACTICE 61 mark · NAT

Find the maximum value of $q (x_{1}, x_{2}) = 3 x_{1}^{2} + 2 x_{1} x_{2} + 3 x_{2}^{2}$ subject to $x_{1}^{2} + x_{2}^{2} = 1$ .

Answer and solution

Answer: $4$

The symmetric matrix is $A = [3113]$ .
Its eigenvalues are $3 \pm 1$ , that is $4$ and $2$ (trace $6$ , determinant $8$ ).
On the unit circle the maximum of $x^{T} A x$ is $λ_{m a x} = 4$ .
Check: at $x = \frac{1}{2} (1, 1)^{T}$ , $q = \frac{3}{2} + 1 + \frac{3}{2} = 4$ .

OFFICIAL GATE 2027 SYLLABUS WORDING

Copied from the GATE 2027 syllabus; see all 2027 syllabi.

GATE DA

matrices, projection matrix, orthogonal matrix, idempotent matrix, partition matrix and their properties, quadratic forms

Preparing in Kolkata? IAE Academy's course for GATE DA.