Question

If \(\begin{pmatrix} 2 \\ a \end{pmatrix}\) is an eigenvector corresponding to the smallest eigenvalue of the matrix \(\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}\), the value of \(a\) is (Answer in integer)

Hint:

Find the eigenvalues of the matrix first, pick the smaller one, then solve \((M-\lambda I)v=0\) for the eigenvector direction and match it to \((2,a)\).

Answer 1

Correct Answer: -2

Step 1: Set up the characteristic equation.
For a matrix \(M\), the eigenvalues \(\lambda\) satisfy \(\det(M - \lambda I) = 0\). Here \(M = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}\), so:
\[ \det \begin{pmatrix} 1-\lambda & 2 \\ 2 & 1-\lambda \end{pmatrix} = 0 \]
\[ (1-\lambda)^2 - 4 = 0 \]

Step 2: Solve for the eigenvalues.
\[ (1-\lambda)^2 = 4 \implies 1-\lambda = \pm 2 \]
This gives \(\lambda = 1-2 = -1\) or \(\lambda = 1+2 = 3\). The smallest eigenvalue is \(\lambda = -1\).

Step 3: Find the eigenvector for the smallest eigenvalue.
An eigenvector \(v = \begin{pmatrix} x \\ y \end{pmatrix}\) for \(\lambda = -1\) satisfies \((M - (-1)I)v = 0\), i.e. \((M+I)v = 0\):
\[ \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \]
Both rows give the same equation: \(2x + 2y = 0\), so \(y = -x\).

Step 4: Match with the given eigenvector.
The given eigenvector is \(\begin{pmatrix} 2 \\ a \end{pmatrix}\), so \(x = 2\) and \(y = a\). Using \(y = -x\):
\[ a = -2 \]

Final Answer:
The value of \(a\) is \(-2\). \[ \boxed{a = -2} \]