Diagonalization, similarity and matrix powers

Objectives, prerequisites and scope

Decide whether an eigenbasis exists, construct a diagonalization, and use it to compute powers. Prerequisites: eigenvalues, eigenspaces, rank and matrix inverses. Diagonalization is a supporting development of the official eigenvalues/eigenvectors heading, not a separately named GATE 2026 CS syllabus heading. We work over the stated field F; the choice of R or C can change the answer.

1. What diagonalization means

A is diagonalizable over F if A=PDP⁻¹ for an invertible matrix P over F and a diagonal D over F. Equivalently AP=PD or P⁻¹AP=D. Let the columns of P be v₁,…,vₙ and D=diag(λ₁,…,λₙ). Comparing columns of AP=PD gives Avᵢ=λᵢvᵢ. Invertibility of P means that these n eigenvectors are linearly independent. Thus an eigenbasis is both necessary and sufficient.

The order matters: column i of P must correspond to entry i of D. Scaling any eigenvector by a nonzero scalar is allowed; changing its position requires the same permutation of D. P need not be orthogonal. The expression PAP⁻¹ generally differs from the intended P⁻¹AP when P stores the eigenvectors as columns.

2. Construct and check a complete example

For A=[[2,1],[0,3]], take v₁=(1,0)ᵀ for λ₁=2 and v₂=(1,1)ᵀ for λ₂=3. Then P=[[1,1],[0,1]], P⁻¹=[[1,−1],[0,1]] and D=diag(2,3). Compute AP=[[2,3],[0,3]]=PD. Therefore A=PDP⁻¹. Checking AP=PD is often simpler than multiplying all three factors, but you must also establish det(P)≠0.

For any integer k≥0, adjacent P⁻¹P factors cancel to give Aᵏ=PDᵏP⁻¹. In this example Aᵏ=[[2ᵏ,3ᵏ−2ᵏ],[0,3ᵏ]]. At k=0 the off-diagonal entry is zero and the result is I; at k=3 it is [[8,19],[0,27]]. Negative k are allowed only if A is invertible. More generally f(A)=Pf(D)P⁻¹ for any polynomial f.

3. A reliable existence test

Over F, the characteristic polynomial must split into linear factors and the sum of eigenspace dimensions must be n. Equivalently, the characteristic polynomial must split over F and, for each eigenvalue, geometric multiplicity must equal algebraic multiplicity. Compute the dimensions using nullity of A−λI. Eigenvectors from different eigenspaces contribute independently.

Having n distinct eigenvalues in F is sufficient. It is not necessary: Iₙ is diagonalizable even though its sole eigenvalue has multiplicity n. Repeated eigenvalues are a signal to compute eigenspace dimensions, not an automatic failure. For diag(1,1,2), the eigenspace for 1 is two-dimensional and that for 2 is one-dimensional, providing a full basis.

In contrast J=[[2,1],[0,2]] has characteristic polynomial (t−2)² and E₂=span{(1,0)ᵀ}. Its only eigenspace has dimension one, so it is not diagonalizable even though det(J)=4≠0. Conversely diag(0,1) is diagonalizable although singular. Invertibility and diagonalizability answer different questions.

4. Similarity is a change of basis

B=P⁻¹AP represents the same linear map in a different basis. Similar matrices have the same characteristic polynomial, eigenvalues with multiplicities, trace, determinant and rank. Equal trace and determinant alone do not prove similarity. For example 2I₂ and J above share eigenvalues, trace and determinant, but one is diagonalizable and the other is not, so they cannot be similar.

Row reduction is not generally a similarity transformation. Elementary row operations preserve the solution set when applied consistently to an augmented system, but they can change eigenvalues. For example multiplying the first row of I₂ by 2 gives diag(2,1). Do not row-reduce A to find its eigenvalues; row-reduce A−λI after determining λ to find eigenspaces.

5. Symmetric matrices and the field caveat

The real spectral theorem says a real symmetric A has an orthonormal basis of real eigenvectors. Therefore A=QDQᵀ with QᵀQ=I and real diagonal D. Eigenvectors for distinct eigenvalues are orthogonal: if Au=λu and Av=μv, symmetry gives λuᵀv=(Au)ᵀv=uᵀAv=μuᵀv, so λ≠μ implies uᵀv=0. Within a repeated eigenspace, choose an orthonormal basis.

For S=[[2,1],[1,2]], eigenvectors (1,1)ᵀ and (1,−1)ᵀ have eigenvalues 3 and 1. Dividing each by √2 gives the columns of an orthogonal Q; D=diag(3,1). Symmetry does not imply positive eigenvalues or invertibility: −I and the zero matrix are counterexamples.

The real rotation [[0,−1],[1,0]] is not diagonalizable over R because it has no real eigenvalues. Over C its distinct eigenvalues i and −i make it diagonalizable. Always retain the field in an existence claim.

Checklist

Find all eigenvalues with multiplicity, compute each eigenspace basis, count n independent vectors, and form P and matching D. Verify AP=PD and invertibility of P. Use the power formula only after this justification. Distinguish distinct eigenvalues (sufficient), repeated eigenvalues (needs checking), and symmetry (guarantees real orthogonal diagonalization).

Original practice: six single-correct MCQs

These are original questions, not previous-year questions. Choose exactly one option.

LG01. A is n×n over a field F. When is A diagonalizable over F?

A. When det(A)≠0 B. When A has n linearly independent eigenvectors in Fⁿ C. When A has at least one eigenvalue in F D. When every diagonal entry is nonzero

Answer: B

A: Invertibility is neither necessary nor sufficient.

B: These eigenvectors form the invertible change-of-basis matrix P with P⁻¹AP diagonal.

C: One eigenvalue alone does not supply an eigenbasis.

D: Diagonal entries do not establish an eigenbasis.

LG02. Which matrix is NOT diagonalizable over R?

A. [[2,0],[0,2]] B. [[0,0],[0,1]] C. [[2,1],[0,2]] D. [[1,1],[1,1]]

Answer: C

A: This matrix is already diagonal.

B: A zero eigenvalue does not prevent diagonalization.

C: Its only eigenspace is span{(1,0)ᵀ}; one independent vector is insufficient.

D: It is real symmetric and has eigenvalues 2 and 0.

LG03. A=[[2,1],[0,3]], P=[[1,1],[0,1]]. What is P⁻¹AP?

A. diag(3,2) B. diag(2,3) C. [[2,1],[0,3]] D. I

Answer: B

A: The diagonal order must match the eigenvector columns of P.

B: The first column has eigenvalue 2 and the second has eigenvalue 3.

C: P changes the basis to eigenvectors, eliminating the off-diagonal term.

D: Similarity preserves eigenvalues 2 and 3, not both 1.

LG04. For A=[[2,1],[0,3]], what is A³?

A. [[8,1],[0,27]] B. [[8,19],[0,27]] C. [[6,3],[0,9]] D. [[8,27],[0,27]]

Answer: B

A: Taking powers entrywise is invalid for the off-diagonal entry.

B: Using PD³P⁻¹ gives the off-diagonal entry 3³−2³=19.

C: This is 3A, not A³.

D: The off-diagonal entry is 27−8, not 27.

LG05. Which claim holds for every real symmetric matrix?

A. All eigenvalues are positive B. It is invertible C. It has an orthonormal basis of real eigenvectors D. All eigenvalues are distinct

Answer: C

A: For example, −I has negative eigenvalues.

B: The zero matrix is symmetric and singular.

C: The real spectral theorem gives QᵀAQ diagonal with QᵀQ=I.

D: I has repeated eigenvalues and is symmetric.

LG06. A real 3×3 matrix has eigenvalues 1,1,2. The eigenspace for 1 has dimension 2. Is it diagonalizable over R?

A. Yes, because the eigenspaces supply 2+1 independent vectors B. No, because an eigenvalue repeats C. No, because determinant is 2 D. Only if A is already diagonal

Answer: A

A: The simple eigenvalue 2 has a one-dimensional eigenspace, independent of the eigenspace for 1.

B: Repetition is allowed when enough eigenvectors exist.

C: The determinant does not obstruct diagonalization.

D: Being similar to a diagonal matrix does not require already being diagonal.

Source and scope

Supporting lesson under eigenvalues and eigenvectors. Diagonalization and similarity are not separately named in the official 2026 CS syllabus; they are included to explain and apply eigenvectors.

Official GATE 2026 CS syllabus (verified 2026-10-14).