Q.55 Consider the matrix [] where λ is a positive real number. Which of the following vectors is/are...

Q.55

Consider the matrix

[ 2   1 ]
[ 1   2 ]

where λ is a positive real number. Which of the following vectors is/are eigenvector(s) of this matrix?

(A) 1
―――
√(2λ)

(B) 1
――
√2

(C) √2
1

(D) √2
λ

Answer: Option B, C

Solution

Given matrix:

\[ A=\begin{bmatrix}2 & 1\\1 & 2\end{bmatrix} \]

Step 1: Eigenvalues

Characteristic equation:

\[ |A-\lambda I|=0 \]

\[ \begin{vmatrix} 2-\lambda & 1\\ 1 & 2-\lambda \end{vmatrix}=0 \]

\[ (2-\lambda)^2 - 1 = 0 \]

\[ \lambda^2 - 4\lambda + 3 = 0 \]

\[ (\lambda-1)(\lambda-3)=0 \]

Therefore eigenvalues are:

\[ \lambda_1=3,\quad \lambda_2=1 \]

---

Step 2: Eigenvector for λ = 3

\[ (A-3I)v=0 \]

\[ \begin{bmatrix} -1 & 1\\ 1 & -1 \end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}=0 \]

This gives:

\[ x=y \]

Eigenvector:

\[ \begin{bmatrix}1\\1\end{bmatrix} \]

Normalized form:

\[ \frac{1}{\sqrt2} \begin{bmatrix}1\\1\end{bmatrix} \]

---

Step 3: Eigenvector for λ = 1

\[ (A-I)v=0 \]

\[ \begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}=0 \]

This gives:

\[ x=-y \]

Eigenvector:

\[ \begin{bmatrix}1\\-1\end{bmatrix} \]

Normalized form:

\[ \frac{1}{\sqrt2} \begin{bmatrix}1\\-1\end{bmatrix} \]

---

Final Answer

Eigenvectors are:

\[ \frac{1}{\sqrt2} \begin{bmatrix}1\\1\end{bmatrix} \quad \text{and} \quad \frac{1}{\sqrt2} \begin{bmatrix}1\\-1\end{bmatrix} \]

Correct Options: (B) and (C)

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