← 2021 Paper 1

UPSC 2021 Maths Optional Paper 1 Q3c-ii — Step-by-Step Solution

7 marks · Section A

Algebra of matrices · Linear Algebra · asked 5× in 14 yrs · Read the full method →

Question

For two square matrices $A$ and $B$ of order 2, show $\operatorname{trace}(AB)=\operatorname{trace}(BA)$. Hence show $AB-BA\ne I_2$.

Technique

Direct manipulation of trace formula; swap dummies.

Solution

Step 1 — trace$(AB)=$trace$(BA)$

Let $A=(a_{ij})$, $B=(b_{ij})$. Then $(AB)_{ii}=\sum_k a_{ik}b_{ki}$, so $\operatorname{trace}(AB)=\sum_i\sum_k a_{ik}b_{ki}$.

Similarly $(BA)_{ii}=\sum_k b_{ik}a_{ki}$, so $\operatorname{trace}(BA)=\sum_i\sum_k b_{ik}a_{ki}$.

Swap dummies $i\leftrightarrow k$ in the second: $\operatorname{trace}(BA)=\sum_k\sum_i b_{ki}a_{ik}=\sum_i\sum_k a_{ik}b_{ki}=\operatorname{trace}(AB)$ ✓.

This works for any square matrices of any order — not just $n=2$.

Step 2 — Hence $AB-BA\ne I_2$

If $AB-BA=I_2$, then $\operatorname{trace}(AB-BA)=\operatorname{trace}(I_2)=2$.

But $\operatorname{trace}(AB-BA)=\operatorname{trace}(AB)-\operatorname{trace}(BA)=0$ (from Step 1).

Contradiction: $0=2$.

So $AB-BA\ne I_2$.

Answer