UPSC 2024 Maths Optional Paper 1 Q3a — Step-by-Step Solution

15 marks · Section A

Matrix of a linear transformation · Linear Algebra · asked 10× in 13 yrs · Read the full method →

Question

Let $V=M_{2\times2}(\mathbb{R})$ denote a vector space over the field of real numbers. Find the matrix of the linear mapping $\phi:V\to V$ given by $\phi(v)=\begin{pmatrix}1 & 2\\3 & -1\end{pmatrix}v$ with respect to the standard basis of $M_{2\times2}(\mathbb{R})$ , and hence find the rank of $\phi$ . Is $\phi$ invertible? Justify your answer.

Technique

Compute $\phi(E_{ij})=AE_{ij}$ for each standard basis matrix; assemble the $4\times4$ representation matrix; use $\det[\phi]=(\det A)^2$ .

Solution

Step 1 — Standard basis of $V$ .

E_{11}=\begin{pmatrix}1 & 0\\0 & 0\end{pmatrix},\;E_{12}=\begin{pmatrix}0 & 1\\0 & 0\end{pmatrix},\;E_{21}=\begin{pmatrix}0 & 0\\1 & 0\end{pmatrix},\;E_{22}=\begin{pmatrix}0 & 0\\0 & 1\end{pmatrix}.

Step 2 — Compute $\phi$ on each basis element. Let $A=\begin{pmatrix}1 & 2\\3 & -1\end{pmatrix}$ .

AE_{11}=\begin{pmatrix}1 & 0\\3 & 0\end{pmatrix}=E_{11}+3E_{21},\quad AE_{12}=\begin{pmatrix}0 & 1\\0 & 3\end{pmatrix}=E_{12}+3E_{22},

AE_{21}=\begin{pmatrix}2 & 0\\-1 & 0\end{pmatrix}=2E_{11}-E_{21},\quad AE_{22}=\begin{pmatrix}0 & 2\\0 & -1\end{pmatrix}=2E_{12}-E_{22}.

Step 3 — Assemble $[\phi]$ .

In the ordering $(E_{11},E_{12},E_{21},E_{22})$ , each column of $[\phi]$ is the coordinate vector of $\phi(E_{ij})$ :

[\phi]=\begin{pmatrix}1 & 0 & 2 & 0\\0 & 1 & 0 & 2\\3 & 0 & -1 & 0\\0 & 3 & 0 & -1\end{pmatrix}.

Step 4 — Rank and invertibility.

Reordering the basis as $(E_{11},E_{21},E_{12},E_{22})$ gives a block-diagonal form with two copies of $A$ . Therefore

\det[\phi]=(\det A)^2,\qquad \det A=(1)(-1)-(2)(3)=-7,\qquad \det[\phi]=49\ne 0.

So $\operatorname{rank}\phi=4$ and $\phi$ is invertible.

Answer

\boxed{\;[\phi]=\begin{pmatrix}1 & 0 & 2 & 0\\0 & 1 & 0 & 2\\3 & 0 & -1 & 0\\0 & 3 & 0 & -1\end{pmatrix},\quad\operatorname{rank}\phi=4,\quad\phi\text{ is invertible.}\;}