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UPSC 2026 Maths Optional Paper 1 Q1b — Step-by-Step Solution

10 marks · Section A

Congruence and similarity of matrices · Linear Algebra · asked 3× in 14 yrs · Read the full method →

Question

If $A$ and $B$ are similar matrices, then show that $A$ and $B$ have the same rank, trace, characteristic polynomial and eigenvalues.

Technique

Work from the definition $B=P^{-1}AP$ with $P$ invertible. Rank: an invertible factor changes neither the column space nor the null space. Trace: the commutation identity $\operatorname{tr}(XY)=\operatorname{tr}(YX)$. Characteristic polynomial: multiplicativity of the determinant, $\det(P^{-1}MP)=\det M$. Eigenvalues then follow as a corollary of the characteristic polynomial.

Solution

Setup. $A,B$ are $n\times n$ matrices over a field $F$. By definition, $A$ and $B$ are similar if there exists an invertible $P\in M_n(F)$ with

(The relation is symmetric: putting $Q=P^{-1}$ gives $A=Q^{-1}BQ$, so “the same” statements below are two-sided.)

Step 1 — Same rank.

We prove two lemmas, each using only invertibility of $P$.

(i) $\operatorname{rank}(AP)=\operatorname{rank}(A)$. Compare column spaces. Clearly $C(AP)=\{AP\mathbf x:\mathbf x\in F^n\}\subseteq C(A)$. Conversely, for any $A\mathbf y\in C(A)$ write $\mathbf y = P(P^{-1}\mathbf y)$, so $A\mathbf y = (AP)(P^{-1}\mathbf y)\in C(AP)$. Hence $C(AP)=C(A)$ and therefore

(ii) $\operatorname{rank}(P^{-1}M)=\operatorname{rank}(M)$ for any $M$. Compare null spaces. Since $P^{-1}$ is invertible, $P^{-1}\mathbf v=\mathbf 0 \iff \mathbf v=\mathbf 0$. Hence

so the nullities agree, and by the rank–nullity theorem ($\operatorname{rank}+\operatorname{nullity}=n$) the ranks agree.

Applying (ii) with $M=AP$ and then (i):

Step 2 — Same trace.

First the commutation identity. For $n\times n$ matrices $X=(x_{ij})$, $Y=(y_{ij})$,

Now take $X=P^{-1}$ and $Y=AP$:

Step 3 — Same characteristic polynomial.

Write the characteristic polynomial of $M$ as $\chi_M(\lambda)=\det(\lambda I - M)$. Since $\lambda I = P^{-1}(\lambda I)P$,

Taking determinants and using multiplicativity $\det(XY)=\det X\det Y$ together with $\det(P^{-1})=\dfrac{1}{\det P}$ (valid as $\det P\neq0$):

Step 4 — Same eigenvalues (corollary of Step 3).

The eigenvalues of a matrix are precisely the roots of its characteristic polynomial. Since $\chi_A\equiv\chi_B$ as polynomials, $A$ and $B$ have exactly the same eigenvalues, with the same algebraic multiplicities.

Their geometric multiplicities also agree: for each scalar $\lambda$,

so by Step 1 (applied to the matrix $\lambda I-A$) $\operatorname{rank}(\lambda I-B)=\operatorname{rank}(\lambda I-A)$, hence $\dim N(\lambda I-B)=\dim N(\lambda I-A)$. $\blacksquare$

Remark. Steps 2 and 4 are consistent: $\operatorname{tr}$ is the sum of the eigenvalues (coefficient of $\lambda^{n-1}$ in $\chi$, up to sign), so equality of characteristic polynomials already forces equality of traces — but the direct proof in Step 2 is the one that costs nothing and is expected.

Step 5 — What is not the same: the eigenvectors.

This is the standard trap and it is worth stating explicitly. Similar matrices share eigenvalues but in general do not share eigenvectors; the eigenvectors correspond under $P^{-1}$:

Let $A\mathbf v=\lambda\mathbf v$ with $\mathbf v\neq\mathbf 0$. Put $\mathbf w = P^{-1}\mathbf v$. Then $\mathbf w \neq \mathbf 0$ (as $P^{-1}$ is invertible) and

Conversely, if $B\mathbf w=\lambda\mathbf w$ then $A(P\mathbf w)=\lambda(P\mathbf w)$. Hence

and the map $\mathbf v \mapsto P^{-1}\mathbf v$ is an isomorphism of the eigenspace $E_\lambda(A)$ onto $E_\lambda(B)$. The eigenvectors coincide only in the special case $P\mathbf v = c\,\mathbf v$.

Interpretation (why all of this is inevitable). $A$ and $B$ represent the same linear operator $T$ on $F^n$ with respect to two different bases, $P$ being the change-of-basis matrix. Rank, trace, characteristic polynomial and eigenvalues are attributes of $T$ itself, hence basis-free; the coordinate column of an eigenvector is not, which is exactly why it transforms by $P^{-1}$.

Answer