UPSC 2023 Maths Optional Paper 1 Q2a — 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

If the matrix of a linear transformation $T:\mathbb{R}^3\to\mathbb{R}^3$ relative to the basis $\{(1,0,0),(0,1,0),(0,0,1)\}$ is

\begin{bmatrix}1 & 1 & 2\\-1 & 2 & 1\\0 & 1 & 3\end{bmatrix},

then find the matrix of $T$ relative to the basis $\{(1,1,1),(0,1,1),(0,0,1)\}$ .

Technique

Standard change-of-basis formula $[T]_B=P^{-1}[T]_E P$ where $P$ has the new basis vectors as columns. The triangular structure of $P$ makes inversion easy.

Solution

Step 1 — Set up the change-of-basis matrix.

Let $E=\{e_1,e_2,e_3\}$ be the standard basis and $B=\{b_1,b_2,b_3\}$ where $b_1=(1,1,1),\,b_2=(0,1,1),\,b_3=(0,0,1)$ .

The change-of-basis matrix $P$ has the new basis vectors as columns (expressed in old basis):

P=\begin{bmatrix}1 & 0 & 0\\1 & 1 & 0\\1 & 1 & 1\end{bmatrix}.

This is lower-triangular with $1$ ‘s on diagonal, so easily invertible.

Step 2 — Compute $P^{-1}$ .

Solve $P\mathbf x=\mathbf c$ for $\mathbf x$ given $\mathbf c=(c_1,c_2,c_3)^T$ : $x_1=c_1,\,x_1+x_2=c_2\Rightarrow x_2=c_2-c_1,\,x_1+x_2+x_3=c_3\Rightarrow x_3=c_3-c_2.$

P^{-1}=\begin{bmatrix}1 & 0 & 0\\-1 & 1 & 0\\0 & -1 & 1\end{bmatrix}.

Step 3 — Compute $[T]_E P$ .

$[T]_E=\begin{bmatrix}1 & 1 & 2\\-1 & 2 & 1\\0 & 1 & 3\end{bmatrix}$ , $P=\begin{bmatrix}1 & 0 & 0\\1 & 1 & 0\\1 & 1 & 1\end{bmatrix}$ .

$[T]_E\cdot$ col 1 of $P$ = $(1,1,2)+(-1,2,1)+(2,1,3)$ ? Wait: col 1 of $P$ is $(1,1,1)^T$ . So $[T]_E(1,1,1)^T =$ row by row:

Row 1: $1\cdot 1+1\cdot 1+2\cdot 1=4.$
Row 2: $-1+2+1=2.$
Row 3: $0+1+3=4.$ First col of $[T]_E P$ is $(4,2,4)^T$ .

Col 2 of $P$ = $(0,1,1)^T$ :

Row 1: $0+1+2=3$ .
Row 2: $0+2+1=3$ .
Row 3: $0+1+3=4$ . Second col: $(3,3,4)^T$ .

Col 3 of $P$ = $(0,0,1)^T$ :

Row 1: $0+0+2=2$ .
Row 2: $0+0+1=1$ .
Row 3: $0+0+3=3$ . Third col: $(2,1,3)^T$ .

[T]_E P=\begin{bmatrix}4 & 3 & 2\\2 & 3 & 1\\4 & 4 & 3\end{bmatrix}.

Step 4 — Compute $P^{-1}([T]_E P)$ .

Row 1 of $P^{-1}=(1,0,0)$ — copies row 1 of $[T]_E P$ : $(4,3,2)$ . Row 2 of $P^{-1}=(-1,1,0)$ — gives $-\text{row 1}+\text{row 2}$ : $(-4+2,-3+3,-2+1)=(-2,0,-1)$ . Row 3 of $P^{-1}=(0,-1,1)$ — gives $-\text{row 2}+\text{row 3}$ : $(-2+4,-3+4,-1+3)=(2,1,2)$ .

Answer

\boxed{\;[T]_B=\begin{bmatrix}4 & 3 & 2\\-2 & 0 & -1\\2 & 1 & 2\end{bmatrix}.\;}