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UPSC 2020 Maths Optional Paper 1 Q1b — Step-by-Step Solution

10 marks · Section A

Rank and nullity; rank-nullity theorem · Linear Algebra · asked 7× in 13 yrs · Read the full method →

Question

Let $M_2(\mathbb{R})$ be the vector space of all $2\times2$ real matrices. Let $B=\begin{bmatrix}1 & -1\\ -4 & 4\end{bmatrix}$. Suppose $T:M_2(\mathbb{R})\to M_2(\mathbb{R})$ is a linear transformation defined by $T(A)=BA$. Find the rank and nullity of $T$. Find a matrix $A$ which maps to the null matrix.

Technique

Compute $\det B$ to get $\operatorname{rank}(B)$; use $\operatorname{rank}(T)=n\cdot\operatorname{rank}(B)$, or solve $BA=O$ entrywise and apply rank–nullity.

Solution

$\dim M_2(\mathbb{R})=4$. By rank–nullity, $\operatorname{rank}(T)+\operatorname{nullity}(T)=4$.

Step 1 — Structure of $B$

so $B$ is singular. $B\neq O$, so $\operatorname{rank}(B)=1$. The columns of $B$ are $\begin{bmatrix}1\\-4\end{bmatrix}$ and $\begin{bmatrix}-1\\4\end{bmatrix}=-\begin{bmatrix}1\\-4\end{bmatrix}$; the column space of $B$ is the line $\operatorname{span}\{(1,-4)^T\}$.

Step 2 — Compute the null space of $T$

Write $A=\begin{bmatrix}p&q\\r&s\end{bmatrix}$. Then

$T(A)=O$ requires each entry zero:

The last two are $-4$ times the first two, so the independent conditions are $p=r$ and $q=s$. Thus

Step 3 — Nullity and rank

The kernel is $\Big\{\begin{bmatrix}p&q\\p&q\end{bmatrix}\Big\}=\operatorname{span}\left\{\begin{bmatrix}1&0\\1&0\end{bmatrix},\begin{bmatrix}0&1\\0&1\end{bmatrix}\right\}$, which is $2$-dimensional.

A nonzero matrix mapped to $O$

Take $p=q=1$ (i.e. $A=\begin{bmatrix}1&1\\1&1\end{bmatrix}$):

Answer