GATE DA Linear Algebra and Calculus Formula Sheet with Traps

Q: What is the relation between eigenvalues, trace and determinant?

The sum of the eigenvalues of a square matrix equals its trace, and their product equals its determinant, counting repeated eigenvalues. For a 2 × 2 matrix this gives the characteristic equation λ² - (tr A)λ + det A = 0. A zero eigenvalue therefore means a zero determinant, so the matrix is singular.

Q: What is the formula for a projection matrix?

The projection onto the column space of A, whose columns are independent, is P = A(A^(top) A)⁻¹A^(top). Onto a single unit vector u it is uu^(top). Every projection matrix is symmetric and idempotent, so P² = P, its eigenvalues are only 0 and 1, and its rank equals its trace.

Q: How do you check whether a matrix is positive definite?

A symmetric matrix is positive definite exactly when every eigenvalue is positive, or equivalently when every leading principal minor is positive. For positive semidefinite you need every eigenvalue to be non-negative. Checking only that the leading minors are non-negative is not enough for semidefiniteness, which is a classic trap.

Q: When does gradient descent converge on a quadratic?

For f(x) = ax² with a > 0, the update x ← x - η f'(x) multiplies x by 1 - 2aη at each step. It converges exactly when 0 < η < 1/a. At η = 1/2a it reaches the minimum in one step, and beyond 1/a the iterates grow without bound.

Q: Does continuity imply differentiability?

No. Differentiability implies continuity, but not the other way round. The functions | x | and ReLU, max(x, 0), are continuous at 0 but have a corner there, so they are not differentiable at 0. For a piecewise function, match the values at the join for continuity, and also match the one-sided derivatives for differentiability.

By MD ANISH AHAMADUpdated 4 Oct 20267 min read

GATE DA Linear Algebra and Calculus Formula Sheet with Traps

GATE DA linear algebra rests on a few results about rank, eigenvalues, projections and the singular value decomposition. Calculus and optimisation rest on limits, Taylor series and the tests for maxima and minima. This sheet gives each formula with its condition, a one-line example of my own and the trap beside it, all checked against the GATE DA 2027 book.

In this guide

Key takeaways
The terms this sheet uses
Rank, systems and determinants
Eigenvalues and special matrices
Quadratic forms and the SVD
Limits, continuity and derivatives
Taylor series and approximation
Maxima, minima and optimisation
Using the sheet in the exam
Quick revision

Key takeaways

Rank plus nullity equals the number of columns, and rank alone decides how many solutions $Ax = b$ has.
Eigenvalues sum to the trace and multiply to the determinant; a polynomial in $A$ has the same polynomial in $\lambda$ .
Projection and idempotent matrices have eigenvalues 0 and 1 only, so their rank equals their trace.
The squared singular values of $A$ are the eigenvalues of $A^\top A$ .
Differentiable implies continuous, never the reverse.
On a closed interval, the extremes are at the end points or the interior critical points: compare the values.

The terms this sheet uses

The rank of a matrix is the number of independent columns, which equals the number of independent rows. The nullity is the dimension of the set of solutions of $Ax = 0$ . An eigenvalue $\lambda$ of $A$ is a number with $Av = \lambda v$ for some non-zero vector $v$ , its eigenvector.

The trace, $\operatorname{tr} A$ , is the sum of the diagonal. $A^\top$ is the transpose. A stationary point of $f$ is where $f'(x) = 0$ . $\ln$ is the natural logarithm, and $\eta$ is the step size, or learning rate, of gradient descent.

Rank, systems and determinants

Formula	Watch out for
$\operatorname{rank} A + \operatorname{nullity} A = \text{number of columns}$	Columns, not rows
$\operatorname{rank} A = \operatorname{rank} A^\top = \operatorname{rank} A^\top A$	$\operatorname{rank}(AB) \le \min(\operatorname{rank} A, \operatorname{rank} B)$
Unique solution iff $\operatorname{rank} A = \operatorname{rank}[A \,\vert\, b] = n$	$n$ is the number of unknowns
Infinitely many iff equal ranks $< n$ ; none iff $\operatorname{rank} A < \operatorname{rank}[A \,\vert\, b]$	Compare with the augmented matrix
$Ax = 0$ has a non-trivial solution iff $\operatorname{rank} A < n$	Square $A$ : iff $\det A = 0$
More than $n$ vectors in $\mathbb{R}^n$ are dependent	$n$ vectors are independent iff their $\det \ne 0$
$\det(AB) = \det A \det B$ ; $\det(kA) = k^n \det A$	Not $k \det A$
$\displaystyle \det A^{-1} = \frac{1}{\det A}$ ; a row swap flips the sign	Triangular: product of the diagonal
$(AB)^{-1} = B^{-1}A^{-1}$ ; $(AB)^\top = B^\top A^\top$	The order reverses

The $2 \times 2$ inverse swaps the diagonal, negates the other two entries and divides by the determinant:

\begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} = \frac{1}{ad - bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}, \qquad ad - bc \ne 0

Examples: a $3 \times 5$ matrix of rank 2 has nullity 3. A $3 \times 3$ matrix with $\det A = 5$ gives $\det(2A) = 8 \times 5 = 40$ .

LU decomposition writes $A = LU$ , with $L$ unit lower triangular holding the elimination multipliers. You solve $Ly = b$ forward, then $Ux = y$ backward. It exists without row exchanges when every leading principal minor is non-zero; otherwise $PA = LU$ .

Trap: $\det(kA)$ multiplies every one of the $n$ rows by $k$ , so the factor is $k^n$ . Writing $k \det A$ is the most common determinant slip.

Eigenvalues and special matrices

Formula	Watch out for
$\det(A - \lambda I) = 0$ ; $\sum \lambda_i = \operatorname{tr} A$ ; $\prod \lambda_i = \det A$	Count repeated roots
$2 \times 2$ : $\lambda^2 - (\operatorname{tr} A)\lambda + \det A = 0$	Fastest route for small matrices
$A^k$ has $\lambda^k$ ; $A^{-1}$ has $\displaystyle \frac{1}{\lambda}$ ; $A + cI$ has $\lambda + c$	Same eigenvectors
Triangular: eigenvalues on the diagonal	Even if the matrix is not symmetric
Symmetric: real eigenvalues, orthogonal eigenvectors, $A = Q\Lambda Q^\top$	Needs a symmetric $A$
$uv^\top$ has rank 1, one eigenvalue $v^\top u$ , the rest 0	$v^\top u$ is a number, $uv^\top$ a matrix
Orthogonal: $Q^\top Q = I$ , $\det Q = \pm 1$ , every $\lvert \lambda \rvert = 1$	Preserves lengths
Projection: $P = A(A^\top A)^{-1}A^\top$ , $P^\top = P$ , $P^2 = P$	$I - P$ projects onto the complement
Idempotent: eigenvalues 0 or 1, $\operatorname{rank} = \operatorname{tr}$	$I - A$ is idempotent too
Centring $\displaystyle C = I - \frac{1}{n}\mathbf{1}\mathbf{1}^\top$ : rank $n - 1$	Eigenvalue 0 on the vector of ones

Example: $A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}$ has trace 7 and determinant 10, so $\lambda^2 - 7\lambda + 10 = 0$ gives 5 and 2. Then $A^2$ has 25 and 4, and $A + 3I$ has 8 and 5. For $u = (1, 2)$ and $v = (3, 1)$ , $uv^\top$ has non-zero eigenvalue $v^\top u = 5$ .

Cayley–Hamilton: every square matrix satisfies its own characteristic equation.

Quadratic forms and the SVD

Formula	Watch out for
Positive definite iff all $\lambda > 0$ iff all leading principal minors $> 0$	Semidefinite needs all $\lambda \ge 0$ ; minors $\ge 0$ are not enough
Over unit $x$ : $\max x^\top A x = \lambda_{\max}$ , $\min x^\top A x = \lambda_{\min}$	$A$ symmetric
$A^\top A$ is positive semidefinite; definite iff the columns of $A$ are independent	Only then is $A^\top A$ invertible
$A = U\Sigma V^\top$ , $\sigma_i \ge 0$ , $\sigma_i^2$ eigenvalues of $A^\top A$	Rank is the count of non-zero $\sigma_i$
$\lVert A \rVert_2 = \sigma_1$ ; $\lVert A \rVert_F^2 = \sum \sigma_i^2$	Symmetric $A$ : $\sigma_i = \lvert \lambda_i \rvert$
Keep the $k$ largest $\sigma_i$ for the best rank- $k$ approximation	Drop the smallest $\sigma_i$

Examples: $\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$ has leading minors 2 and 3, so it is positive definite, with eigenvalues 3 and 1. The diagonal matrix with entries 3 and $-4$ has singular values 4 and 3, so $\lVert A \rVert_2 = 4$ and $\lVert A \rVert_F^2 = 25$ .

The book's last-minute sheet covers all seven technical sections this way, with every result's condition printed beside it. It is part of the GATE DA 2027 book, with 907 questions with worked solutions and 10 full mock tests.

Limits, continuity and derivatives

Formula	Watch out for
As $x \to 0$ : $\displaystyle \frac{\sin x}{x} \to 1$ , $\displaystyle \frac{1 - \cos x}{x^2} \to \frac{1}{2}$ , $\displaystyle \frac{e^x - 1}{x} \to 1$ , $\displaystyle \frac{\ln(1 + x)}{x} \to 1$	Rescale the argument first
$\displaystyle \left(1 + \frac{1}{n}\right)^n \to e$	A $1^\infty$ form, not 1
L'Hôpital: $\displaystyle \lim \frac{f}{g} = \lim \frac{f'}{g'}$	Only for $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$
Continuity at $a$ : left limit $=$ right limit $= f(a)$	All three must exist
$(f \circ g)' = f'(g)\,g'$ ; $\displaystyle \left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$	Keep the minus sign in order
$\sigma'(x) = \sigma(x)\big(1 - \sigma(x)\big)$ , largest $\displaystyle \frac{1}{4}$ at $0$ ; $\tanh' = 1 - \tanh^2$	$\sigma'$ is never above $\displaystyle \frac{1}{4}$
$(a^x)' = a^x \ln a$ ; $\displaystyle (\ln x)' = \frac{1}{x}$	The factor is $\ln a$ , not $\log_{10} a$

Example: $\displaystyle \lim_{x \to 0} \frac{1 - \cos 2x}{x^2} = 4 \times \frac{1}{2} = 2$ , after writing $u = 2x$ .

Taylor series and approximation

Formula	Watch out for
$\displaystyle f(x) = \sum_{n \ge 0} \frac{f^{(n)}(a)}{n!}(x - a)^n$	About $a = 0$ it is the Maclaurin series
$\displaystyle e^x = \sum \frac{x^n}{n!}$ ; $\displaystyle \sin x = x - \frac{x^3}{3!} + \cdots$ ; $\displaystyle \cos x = 1 - \frac{x^2}{2!} + \cdots$	$\sin$ odd powers, $\cos$ even
$\displaystyle \ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$	Only for $-1 < x \le 1$
$\displaystyle \frac{1}{1 - x} = \sum x^n$	Only for $\lvert x \rvert < 1$
$\displaystyle f(x + h) \approx f(x) + hf'(x) + \frac{h^2}{2}f''(x)$	Halve the second-order term

Example: $e^{0.1} \approx 1 + 0.1 + 0.005 = 1.105$ , against the true 1.10517.

Maxima, minima and optimisation

Formula	Watch out for
Candidates: $f'(c) = 0$ , points where $f'$ does not exist, interval ends	Do not forget the ends
$f''(c) > 0$ minimum; $f''(c) < 0$ maximum	$f''(c) = 0$ decides nothing
First non-zero derivative of even order: extremum; odd order: inflection	Minimum if that derivative is positive
$f'' \ge 0$ means convex; a local minimum of a convex function is global	Strictly convex: at most one minimiser
Mean value theorem: $\displaystyle f'(c) = \frac{f(b) - f(a)}{b - a}$ for some $c \in (a, b)$	Rolle is the case $f(a) = f(b)$
Gradient descent $x \leftarrow x - \eta f'(x)$ ; on $ax^2$ converges iff $\displaystyle 0 < \eta < \frac{1}{a}$	$a > 0$
Newton for a minimum: $\displaystyle x \leftarrow x - \frac{f'(x)}{f''(x)}$	One step on a quadratic

Examples: $f(x) = x^3 - 3x$ has $f'(x) = 3x^2 - 3$ , so $x = \pm 1$ . Since $f''(x) = 6x$ , $x = 1$ is a minimum and $x = -1$ a maximum. On $[0, 3]$ , compare $f(0) = 0$ , $f(1) = -2$ and $f(3) = 18$ : the maximum is 18 at the end point. For $f(x) = x^4$ at 0, the first non-zero derivative is the fourth, so 0 is a minimum.

In one line: A closed-interval question is a comparison of values, and the winner is often an end point.

Using the sheet in the exam

Matrix and series questions often end in a decimal. Practise them on the GATE virtual calculator. Read the MCQ, MSQ and NAT marking scheme, common to every GATE paper too: an MSQ has no negative marks, but no partial credit either. If you know the CS syllabus, GATE CS vs GATE DA shows where the linear algebra overlaps and that optimisation is DA's own.

Remember: No counted paper (2024 to 2026) has examined LU decomposition, but the syllabus names it. An afternoon on it is cheap insurance.

More formula sheets: all of GATE DA · DBMS and Algorithms · Machine Learning · Probability and Statistics

Quick revision

$\operatorname{rank} A + \operatorname{nullity} A$ equals the number of columns.
$\sum \lambda = \operatorname{tr} A$ , $\prod \lambda = \det A$ , and $\det(kA) = k^n \det A$ .
$A^k$ , $A^{-1}$ and $A + cI$ have $\lambda^k$ , $\displaystyle \frac{1}{\lambda}$ and $\lambda + c$ .
Projections are symmetric and idempotent, with rank equal to trace.
$\sigma_i^2$ are the eigenvalues of $A^\top A$ , and $\lVert A \rVert_2 = \sigma_1$ .
$\displaystyle \frac{1 - \cos x}{x^2} \to \frac{1}{2}$ , and $\sigma'$ peaks at $\displaystyle \frac{1}{4}$ .
$f''(c) = 0$ decides nothing; go to the first non-zero derivative.
Gradient descent on $ax^2$ converges only for $\displaystyle 0 < \eta < \frac{1}{a}$ .

Frequently asked questions

What is the relation between eigenvalues, trace and determinant?

The sum of the eigenvalues of a square matrix equals its trace, and their product equals its determinant, counting repeated eigenvalues. For a $2 \times 2$ matrix this gives the characteristic equation $\lambda^2 - (\operatorname{tr} A)\lambda + \det A = 0$ . A zero eigenvalue therefore means a zero determinant, so the matrix is singular.

What is the formula for a projection matrix?

The projection onto the column space of $A$ , whose columns are independent, is $P = A(A^\top A)^{-1}A^\top$ . Onto a single unit vector $u$ it is $uu^\top$ . Every projection matrix is symmetric and idempotent, so $P^2 = P$ , its eigenvalues are only 0 and 1, and its rank equals its trace.

How do you check whether a matrix is positive definite?

A symmetric matrix is positive definite exactly when every eigenvalue is positive, or equivalently when every leading principal minor is positive. For positive semidefinite you need every eigenvalue to be non-negative. Checking only that the leading minors are non-negative is not enough for semidefiniteness, which is a classic trap.

When does gradient descent converge on a quadratic?

For $f(x) = ax^2$ with $a > 0$ , the update $x \leftarrow x - \eta f'(x)$ multiplies $x$ by $1 - 2a\eta$ at each step. It converges exactly when $\displaystyle 0 < \eta < \frac{1}{a}$ . At $\displaystyle \eta = \frac{1}{2a}$ it reaches the minimum in one step, and beyond $\displaystyle \frac{1}{a}$ the iterates grow without bound.

Does continuity imply differentiability?

No. Differentiability implies continuity, but not the other way round. The functions $\lvert x \rvert$ and ReLU, $\max(x, 0)$ , are continuous at 0 but have a corner there, so they are not differentiable at 0. For a piecewise function, match the values at the join for continuity, and also match the one-sided derivatives for differentiability.

Sources

Dates, fees and the syllabus are set by the GATE 2027 organising institute and can change. Always confirm at gate2027.iitm.ac.in.