Linear Algebra for GATE CS

Linear algebra is a high-value Engineering Mathematics block because a small set of procedures answers many different-looking questions. The difficulty is usually not a long calculation. It is choosing the right test for rank, consistency or eigenvalues and carrying the arithmetic cleanly.

Use the Engineering Mathematics for GATE course as the main learning path. This guide builds the procedures you should be able to execute by hand.

Rank by row reduction

The rank of a matrix is the number of linearly independent rows or columns. In a numerical, reduce the matrix to row-echelon form using elementary row operations and count the non-zero rows, equivalently the pivot positions.

Consider the 3 x 4 matrix

A = [[1, 2, 1, 3], [2, 4, 0, 6], [1, 2, 2, 3]]

Start with:

Eliminate the first entry of R2:

R2 <- R2 - 2R1 = [0, 0, -2, 0]

Eliminate the first entry of R3:

R3 <- R3 - R1 = [0, 0, 1, 0]

Now R2 is minus twice R3. Remove it:

R2 <- R2 + 2R3 = [0, 0, 0, 0]

After placing the non-zero rows first, an echelon form is:

[[1, 2, 1, 3], [0, 0, 1, 0], [0, 0, 0, 0]]

There are two pivots, so rank(A) = 2.

Check the result independently. The first and third original rows are not scalar multiples, so rank is at least 2. The row operation showed the remaining row is dependent, so rank is at most 2. Both bounds meet at 2.

Two facts remove many options quickly:

• For an m x n matrix, rank(A) <= min(m, n).

• A square n x n matrix has full rank n exactly when det(A) is not zero.

The determinant shortcut applies only to square matrices. For a rectangular matrix like the worked example, use pivots or a suitable non-zero minor.

Classifying systems Ax=b with ranks

For n unknowns, compare the coefficient matrix A with the augmented matrix [A|b]. The Rouché-Capelli test gives three cases:

• rank(A) < rank([A|b]): no solution.

• rank(A) = rank([A|b]) = n: one unique solution.

• rank(A) = rank([A|b]) < n: infinitely many solutions.

Consider this system:

x + y + z = 3

2x + 2y + 2z = 7

x - y + z = 1

Its augmented matrix is:

[[1, 1, 1 | 3], [2, 2, 2 | 7], [1, -1, 1 | 1]]

Apply R2 <- R2 - 2R1:

[2, 2, 2 | 7] - 2[1, 1, 1 | 3] = [0, 0, 0 | 1]

The coefficient part of this row is zero, but the augmented entry is 1. It represents the contradiction 0 = 1. The coefficient matrix has two independent rows, so rank(A) = 2. The contradiction row adds an augmented pivot, so rank([A|b]) = 3. Since 2 < 3, the system has no solution.

The direct contradiction checks the rank conclusion. No values of x, y and z can make the second equation equal twice the first on the left while changing 6 on the right to 7.

Eigenvalues and the property set

For a square matrix, eigenvalues satisfy:

det(A - lambda I) = 0

Take

A = [[4, 1], [2, 3]]

Then:

det(A - lambda I) = (4 - lambda)(3 - lambda) - 2

= lambda^2 - 7lambda + 10

= (lambda - 5)(lambda - 2)

Therefore, the eigenvalues are 5 and 2.

Now check them through standard properties:

• Sum equals trace: 5 + 2 = 7, and trace(A) = 4 + 3 = 7.

• Product equals determinant: 5 x 2 = 10, and det(A) = 12 - 2 = 10.

• Triangular matrix: the eigenvalues of [[3, 4], [0, -1]] are its diagonal entries 3 and -1.

• Power: if A has eigenvalues 5 and 2, then A^2 has eigenvalues 25 and 4.

• Inverse: because neither eigenvalue is zero, A^-1 has eigenvalues 1/5 and 1/2.

Trace and determinant provide a fast arithmetic audit for a 2 x 2 characteristic polynomial. If the root sum or product disagrees, fix the polynomial before moving on.

The polynomial shortcut f(A)

If lambda is an eigenvalue of A, then f(lambda) is an eigenvalue of f(A) for a polynomial f. This avoids multiplying matrices.

Using the same A, find the eigenvalues of:

A^2 - 3A + 2I

Here f(lambda) = lambda^2 - 3lambda + 2. Apply it to each eigenvalue of A:

f(5) = 5^2 - 3(5) + 2 = 25 - 15 + 2 = 12

f(2) = 2^2 - 3(2) + 2 = 4 - 6 + 2 = 0

So the required eigenvalues are 12 and 0. Their product is zero, which also tells us that det(A^2 - 3A + 2I) = 0. That independent consequence is consistent with the result.

How GATE tests linear algebra

Expect row-reduction rank, consistency of linear equations, determinant and trace properties, eigenvalues of a transformed matrix, or a conceptual statement about invertibility. A common trap is to compare rank(A) with the number of equations instead of the number n of unknowns when deciding unique versus infinite solutions.

About 200 published Linear Algebra questions are available in the KnowledgeGate bank. The GATE CS Subject Weightage guide helps place the topic in a revision order, and the GATE CS 6-Month Plan helps space learning and testing. For the current syllabus and official exam-specific instructions, refer to the official GATE portal of the organising IIT. GATE CS Exam Preparation Courses & Test Series keeps the full subject route together.

The short version and next step

For rank, row-reduce and count pivots. For Ax=b, compare rank(A), rank([A|b]) and the number of unknowns. For eigenvalues, form the characteristic polynomial, then check the roots against trace and determinant. For f(A), transform the eigenvalues directly through f.

Once these procedures are stable, move to timed mixed practice in the GATE Test Series. Rechecking each answer with a second property is faster than discovering an arithmetic mistake after completing the whole question.