Congruence and similarity of matrices

At a Glance

• Frequency: 2 sub-parts across 2 of 13 years (2017, 2018)
• Priority tier: T3
• Marks (count): 10 (1), 12 (1)
• Average solve time: ~6 min
• Difficulty mix: easy 2
• Section: A | Dominant type: proof

Why This Chapter Matters

Both UPSC questions on this atom ask for the same proof: similar matrices share their characteristic polynomial (hence the same eigenvalues). The algebraic core is a single factorisation — $B-\lambda I=P^{-1}(A-\lambda I)P$ — followed by multiplicativity of the determinant. The proof takes four lines and earns 10–12 marks. The 2017 version asks about the characteristic polynomial; the 2018 version focuses on the eigenvalues — both reduce to the same argument. Prepare this atom in under 15 minutes and it becomes a guaranteed mark-earner.

Minimum Theory

Similarity. Matrices $A$ and $B$ are similar if there exists an invertible matrix $P$ with $B=P^{-1}AP$. Similarity is an equivalence relation. Geometrically, $A$ and $B$ represent the same linear transformation in different bases.

Characteristic polynomial. For an $n\times n$ matrix $M$, $\chi_M(\lambda)=\det(M-\lambda I)$ is a monic polynomial of degree $n$ in $\lambda$. Its roots (with multiplicity) are the eigenvalues of $M$.

The key factorisation. If $B=P^{-1}AP$, then: $B-\lambda I = P^{-1}AP - \lambda P^{-1}IP = P^{-1}(A-\lambda I)P.$ (The step $\lambda I=P^{-1}(\lambda I)P$ is the crucial move — regroup before applying the determinant.)

Consequence. Using $\det(XY)=\det(X)\det(Y)$ and $\det(P^{-1})\det(P)=1$: $\chi_B(\lambda)=\det(P^{-1})\,\det(A-\lambda I)\,\det(P)=\chi_A(\lambda).$

Congruence. Matrices $A$ and $B$ are congruent if $B=P^TAP$ for some invertible $P$. Congruence preserves the signature (index of positivity) of a bilinear form, but not necessarily the eigenvalues or the characteristic polynomial. UPSC has not directly tested congruence; both corpus questions test similarity.

Question Archetypes

similarity-invariance (2 question(s); 2017, 2018)

Recognition Cues

• “Show that similar matrices have the same characteristic polynomial.”
• “Prove that if $A$ and $B$ are similar, they have the same eigenvalues.”
• Both reduce to the same four-step algebraic proof.

Solution Template

1. Set up. Let $B=P^{-1}AP$ for some invertible $P$. Write $\chi_B(\lambda)=\det(B-\lambda I)$.
2. Key step. Write $\lambda I=P^{-1}(\lambda I)P$ so that $B-\lambda I=P^{-1}(A-\lambda I)P$.
3. Apply $\det$. $\chi_B(\lambda)=\det(P^{-1})\det(A-\lambda I)\det(P)=\chi_A(\lambda)$.
4. Conclusion. Same characteristic polynomial $\Rightarrow$ same roots $\Rightarrow$ same eigenvalues (with multiplicities).

Worked Example(s)

2017 Paper 1, 2017-P1-Q1b (10 marks)

Show that similar matrices have the same characteristic polynomial.

Let $A,B$ be $n\times n$ matrices with $B=P^{-1}AP$ for some invertible $P$.

Step 1 — Setup. $\chi_B(\lambda)=\det(B-\lambda I)$.

Step 2 — Key factorisation. $B-\lambda I=P^{-1}AP-\lambda I=P^{-1}AP-\lambda P^{-1}IP=P^{-1}(A-\lambda I)P.$

Step 3 — Determinant. $\chi_B(\lambda)=\det\!\bigl(P^{-1}(A-\lambda I)P\bigr)=\det(P^{-1})\,\det(A-\lambda I)\,\det(P).$ Since $\det(P^{-1})\det(P)=\det(P^{-1}P)=\det(I)=1$: $\chi_B(\lambda)=\det(A-\lambda I)=\chi_A(\lambda).$

Conclusion. $\boxed{\;B=P^{-1}AP\;\Longrightarrow\;\chi_B(\lambda)=\chi_A(\lambda)\;\text{ for all }\lambda.\;}\qquad\blacksquare$

In particular, similar matrices have the same trace and the same determinant.

2018 Paper 1, 2018-P1-Q2a (12 marks)

Show that if $A$ and $B$ are similar $n\times n$ matrices, then they have the same eigenvalues.

Same proof as above. After showing $\chi_B(\lambda)=\chi_A(\lambda)$, add the eigenvalue conclusion:

Step 4 — Eigenvalues. The eigenvalues of a matrix are the roots of its characteristic polynomial. Since $\chi_A$ and $\chi_B$ are identical polynomials, they have the same roots with the same algebraic multiplicities. Hence $A$ and $B$ have the same eigenvalues.

Remark (eigenvectors). If $Av=\lambda v$ ($v\neq0$), then $B(P^{-1}v)=P^{-1}AP\cdot P^{-1}v=P^{-1}Av=\lambda(P^{-1}v)$. So $P^{-1}v$ is an eigenvector of $B$ for the same eigenvalue $\lambda$ — an independent confirmation.

$\boxed{\;B=P^{-1}AP\;\Longrightarrow\;A\text{ and }B\text{ share all eigenvalues with the same algebraic multiplicities.}\;}\qquad\blacksquare$

Common Traps

• The critical algebraic step is $\lambda I=P^{-1}(\lambda I)P$. Without this regrouping, $B-\lambda I$ does not factor as a similarity. Do not try to expand $\det(P^{-1}AP-\lambda I)$ entry-by-entry.
• State $\det(P^{-1})\det(P)=1$ explicitly. This is the step that makes the scalar factors cancel; simply writing "$=\det(A-\lambda I)$" without justification loses the intermediate marks.
• The converse is false. Equal characteristic polynomials do not imply similarity. For example, $\bigl(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\bigr)$ and the $2\times 2$ zero matrix both have $\chi(\lambda)=\lambda^2$ but are not similar. Only mention the proved direction.
• “Same eigenvalues” means with algebraic multiplicities (since the entire polynomial is preserved). The 2018 question specifically says “same eigenvalues” — include multiplicity in your conclusion.
• Similarity is not the same as congruence ($B=P^TAP$). Similarity preserves eigenvalues and the characteristic polynomial; congruence preserves only the signature (number of positive/negative eigenvalues for symmetric matrices). In this atom, both questions test similarity.

Marks-Aware Writing

10-mark question (characteristic polynomial): The proof has three active mathematical steps (setup, factorisation, determinant) and a conclusion. Write each step on its own line. The factorisation $B-\lambda I=P^{-1}(A-\lambda I)P$ is the pivotal display equation; box it or put it on a line by itself. Total length: about 8–10 lines.

12-mark question (eigenvalues): Same proof, plus an explicit eigenvalue conclusion step. After showing $\chi_A=\chi_B$, add “eigenvalues are roots of the characteristic polynomial; since $\chi_A=\chi_B$, the roots coincide with the same multiplicities.” Include the eigenvector remark as a secondary confirmation for the extra 2 marks.

Practice Set

(No additional practice questions in the corpus for this exact atom; the two worked examples above cover both variants.)