Linear Algebra

For a square matrix A, a non-zero vector v is an eigenvector with corresponding eigenvalue λ if:

A v = λ v

In words: A's action on v is just scaling (no direction change).

Finding eigenvalues: solve the characteristic equation det(A − λI) = 0.

Finding eigenvectors: for each λ, solve (A − λI)v = 0 (null space).

Properties of eigenvalues:

• Sum of eigenvalues = trace of A (sum of diagonal entries).
• Product of eigenvalues = det(A).
• A is invertible ⟺ no eigenvalue is 0.
• Eigenvalues of A^k are λ^k (with same eigenvectors).
• Eigenvalues of A⁻¹ are 1/λ (with same eigenvectors).
• Triangular matrix: eigenvalues are the diagonal entries.
• Symmetric matrix: all eigenvalues are real.
• Orthogonal matrix: |λ| = 1 for all eigenvalues.

Diagonalization. If A has n linearly independent eigenvectors:

A = PDP⁻¹

where P's columns are eigenvectors and D is diagonal with eigenvalues. Then A^k = PD^k P⁻¹ — exponentiating becomes trivial.

Not every matrix is diagonalizable. Required: algebraic multiplicity = geometric multiplicity for each eigenvalue.

Worked example.

A = [[2, 1], [1, 2]]. Find eigenvalues and eigenvectors.

det(A − λI) = (2−λ)² − 1 = λ² − 4λ + 3 = 0 → λ = 1, 3.

For λ = 1: (A − I)v = [[1,1],[1,1]]v = 0 → v = (1, −1).
For λ = 3: (A − 3I)v = [[−1,1],[1,−1]]v = 0 → v = (1, 1).

Trace check: 2 + 2 = 4 = 1 + 3. ✓
Det check: 4 − 1 = 3 = 1 × 3. ✓

Applications (why eigenvalues matter beyond GATE):

1. Principal Component Analysis (PCA) — eigenvectors of covariance matrix point in directions of maximum variance.
2. PageRank — Google's algorithm finds the principal eigenvector of the web's link matrix.
3. Vibration modes in mechanical / civil engineering.
4. Quantum mechanics — energy levels are eigenvalues of the Hamiltonian.
5. Markov chains — stationary distribution = left eigenvector of transition matrix with eigenvalue 1.

Cayley-Hamilton theorem: every square matrix satisfies its own characteristic equation. For 2×2: A² − (tr A) A + (det A) I = 0.

Useful trick: lets you express A⁻¹ in terms of A: rearranging Cayley-Hamilton, A⁻¹ = [(tr A) I − A] / det A.

Key determinant facts for GATE: det(AB) = det(A)det(B); det(A^T) = det(A); det(kA) = k^n·det(A) for an n×n matrix; det(A^-1) = 1/det(A). A row/column swap flips the sign; adding a multiple of one row to another leaves det unchanged. det(A) = product of eigenvalues. A triangular matrix's determinant is the product of its diagonal entries. Memory aid: 'SWAP-FLIP, SCALE-power-n, INVERSE-reciprocal.' For a 2×2 [[a,b],[c,d]], det = ad − bc. A matrix is singular (non-invertible) iff det = 0, which means rank < n and at least one eigenvalue is 0. These properties let you avoid full cofactor expansion under exam time pressure.

Rank = number of linearly independent rows = number of linearly independent columns = number of non-zero rows in row echelon form = order of the largest non-zero minor. For an m×n matrix, rank ≤ min(m,n). Full row/column rank affects solvability of Ax=b. Shortcut: rank(A) + nullity(A) = n (number of columns), the Rank-Nullity Theorem. A square n×n matrix is invertible iff rank = n iff det ≠ 0. Reduce to echelon form by elementary row operations (which preserve rank) and count pivots. Note rank(AB) ≤ min(rank(A), rank(B)). For a non-zero matrix rank ≥ 1; for an outer product uv^T rank = 1.

Find det(A) where A is 3×3 with trace = 6 and you are told eigenvalues are in arithmetic progression with the smallest being 1. AP with smallest 1: eigenvalues λ, λ+d, λ+2d summing to trace = 6, so 3λ+3d = 6 → λ+d = 2 (the middle term). With smallest = 1, middle = 2, largest = 3. Then det(A) = product of eigenvalues = 1×2×3 = 6. This shows two shortcuts: trace = sum of eigenvalues, det = product of eigenvalues. Always use these to bypass cofactor expansion when eigenvalue information is given.

For Ax = b with A an m×n matrix and augmented matrix [A|b]: (1) If rank(A) ≠ rank([A|b]) → INCONSISTENT, no solution. (2) If rank(A) = rank([A|b]) = n (number of unknowns) → UNIQUE solution. (3) If rank(A) = rank([A|b]) = r < n → INFINITELY many solutions with (n − r) free parameters. Memory aid: 'Ranks differ = none; Ranks equal & = unknowns = one; Ranks equal & < unknowns = infinite.' For homogeneous systems Ax = 0, rank([A|0]) always equals rank(A), so they are always consistent (trivial solution exists); they have non-trivial solutions iff rank(A) < n.

Cramer's Rule (for unique solutions of n×n systems with det(A) ≠ 0): x_i = det(A_i)/det(A), where A_i is A with column i replaced by b. Useful for small systems and conceptual questions, but computationally expensive (O(n!·n) naively). Gaussian elimination is the practical method: reduce [A|b] to row echelon form, then back-substitute; cost is O(n^3). LU decomposition factors A = LU enabling efficient repeated solves for different b. For GATE, remember: Cramer fails when det(A) = 0 (no unique solution); Gaussian elimination with partial pivoting is numerically stable. Time complexity O(n^3) is a frequently tested fact.

Given x + y + z = 6, x + 2y + 3z = 10, x + 2y + kz = μ. For what k does the system have NO unique solution? The coefficient determinant is |[[1,1,1],[1,2,3],[1,2,k]]|. Expand: subtract row 1 from rows 2,3 → [[1,1,1],[0,1,2],[0,1,k-1]]. Det = 1·(1·(k−1) − 2·1) = k − 1 − 2 = k − 3. Unique solution requires det ≠ 0, i.e., k ≠ 3. At k = 3, the system has either no solution or infinitely many depending on μ. This is a classic GATE pattern: set the determinant to zero to find the critical parameter.

For Av = λv (v ≠ 0): characteristic equation is det(A − λI) = 0. Sum of eigenvalues = trace(A); Product of eigenvalues = det(A). For a triangular/diagonal matrix, eigenvalues are the diagonal entries. Eigenvalues of A^k are λ^k; of A^-1 are 1/λ; of A^T are the same as A; of (A + cI) are λ + c. Symmetric real matrices have real eigenvalues and orthogonal eigenvectors. Memory aid: 'Trace = SUM, Det = PRODUCT.' Cayley–Hamilton: every matrix satisfies its own characteristic equation. Eigenvalues of an n×n matrix number n (counting multiplicity). A matrix is singular iff 0 is an eigenvalue.

A is diagonalizable iff it has n linearly independent eigenvectors; then A = PDP^-1 where D is diagonal with eigenvalues and P's columns are eigenvectors. This makes A^k = PD^kP^-1 trivial. Distinct eigenvalues guarantee diagonalizability (sufficient, not necessary). Algebraic multiplicity ≥ geometric multiplicity; defective matrices (geometric < algebraic) are NOT diagonalizable. Cayley-Hamilton theorem: if characteristic polynomial is p(λ), then p(A) = 0; this lets you express A^n and A^-1 in terms of lower powers. Symmetric matrices are always diagonalizable by an orthogonal matrix. Use these to compute high powers of matrices efficiently in exams.

Find eigenvalues of A = [[4,1],[2,3]]. Characteristic equation: det(A − λI) = (4−λ)(3−λ) − (1)(2) = 0. Expand: 12 − 4λ − 3λ + λ^2 − 2 = λ^2 − 7λ + 10 = 0. Factor: (λ − 5)(λ − 2) = 0, so λ = 5, 2. Verify: trace = 4 + 3 = 7 = 5 + 2 ✓; det = 4·3 − 1·2 = 10 = 5 × 2 ✓. Shortcut for 2×2: λ^2 − (trace)λ + det = 0. So you can write λ^2 − 7λ + 10 = 0 directly without expanding. Always cross-check with trace and determinant.

Vectors are linearly independent if c1v1 + ... + ckvk = 0 implies all ci = 0; otherwise dependent. A basis is a linearly independent set that spans the space; every basis of a space has the same number of vectors = the dimension. In R^n any n linearly independent vectors form a basis, and any set of more than n vectors is dependent. To test independence of n vectors in R^n, form a matrix and check det ≠ 0 (independent) or rank = n. The four fundamental subspaces of an m×n matrix A: column space (dim = rank), row space (dim = rank), null space (dim = n − rank), left null space (dim = m − rank). Memory aid: 'Independent + Spanning = Basis.'

For a linear transformation T: V → W (or matrix A, m×n), the Rank-Nullity Theorem states: rank(T) + nullity(T) = dim(domain) = n. Here rank = dim of image/column space, nullity = dim of kernel/null space. T is injective (one-one) iff nullity = 0; T is surjective (onto) iff rank = dim(W). A linear transformation satisfies T(av + bw) = aT(v) + bT(w). The matrix of T depends on chosen bases. Key shortcut: for a square matrix, injective ⟺ surjective ⟺ bijective ⟺ invertible ⟺ det ≠ 0 (all equivalent in finite dimensions). Remember: nullity counts free variables in Ax = 0.

Are v1 = (1,2,3), v2 = (2,4,6), v3 = (1,0,1) linearly independent in R^3? Form matrix rows and compute det of [[1,2,3],[2,4,6],[1,0,1]]. Notice row 2 = 2 × row 1, so two rows are proportional → det = 0 → the set is linearly DEPENDENT. Rank = 2 (v1 and v3 are independent, v2 is redundant). Thus they span only a 2-dimensional subspace (a plane), not all of R^3. Shortcut: whenever one vector is a scalar multiple of another, the set is immediately dependent without computing the full determinant. Dimension of span = rank = 2.