UPSC 2016 Maths Optional Paper 1 Q2b-i — Step-by-Step Solution

8 marks · Section A

Eigenvalues and eigenvectors · Linear Algebra · asked 9× in 13 yrs · Read the full method →

Question

If $A=\begin{bmatrix}1 & 1 & 0\\ 1 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$ , then find the eigenvalues and eigenvectors of $A$ .

Technique

Exploit the block-diagonal structure to factor the characteristic polynomial; solve $(A-\lambda I)\mathbf v=0$ for each $\lambda$ .

Solution

Step 1 — Characteristic polynomial

$A$ is block-diagonal: a $2\times2$ block $\left(\begin{smallmatrix}1&1\\1&1\end{smallmatrix}\right)$ and a $1\times1$ block $[1]$ .

\det(A-\lambda I)=\det\!\begin{pmatrix}1-\lambda&1\\1&1-\lambda\end{pmatrix}\cdot(1-\lambda)=\big[(1-\lambda)^2-1\big](1-\lambda).

$(1-\lambda)^2-1=\lambda^2-2\lambda=\lambda(\lambda-2)$ , so

\det(A-\lambda I)=\lambda(\lambda-2)(1-\lambda)=0.

Step 2 — Eigenvalues

\boxed{\;\lambda=0,\ 1,\ 2.\;}

Three distinct eigenvalues $\Rightarrow$ $A$ is diagonalizable.

Step 3 — Eigenvectors

$\lambda=0$ : $(A-0I)\mathbf v=0$ : $x+y=0,\ z=0$ . Take $\mathbf v_0=(1,-1,0)^{\top}$ .

$\lambda=1$ : $(A-I)\mathbf v=0$ : $\left(\begin{smallmatrix}0&1&0\\1&0&0\\0&0&0\end{smallmatrix}\right)\mathbf v=0\Rightarrow x=0,\ y=0,\ z$ free. Take $\mathbf v_1=(0,0,1)^{\top}$ .

$\lambda=2$ : $(A-2I)\mathbf v=0$ : $-x+y=0,\ -z=0\Rightarrow x=y,\ z=0$ . Take $\mathbf v_2=(1,1,0)^{\top}$ .

Answer

\boxed{\;\lambda=0:(1,-1,0)^{\top},\quad \lambda=1:(0,0,1)^{\top},\quad \lambda=2:(1,1,0)^{\top}.\;}