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

This matrix has a special structure worth recognizing: it is a $2\times2$ symmetric matrix with equal diagonal entries, of the form $\begin{pmatrix} a & b \\ b & a \end{pmatrix}$ with $a = 1$ and $b = 2$. Matrices of this exact shape always have the same pair of eigenvectors, $(1, 1)$ and $(1, -1)$, no matter what $a$ and $b$ are, so there is no need to solve the full characteristic equation from scratch.

For this pattern, the eigenvector $(1, 1)$ corresponds to eigenvalue $a + b$, and the eigenvector $(1, -1)$ corresponds to eigenvalue $a - b$. Check this directly: multiplying $\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$ by $(1, 1)$ gives $(3, 3) = 3(1,1)$, so eigenvalue $3$ goes with $(1,1)$. Multiplying by $(1, -1)$ gives $(-1, 1) = -1(1,-1)$, so eigenvalue $-1$ goes with $(1, -1)$.

Comparing the two eigenvalues, $3$ and $-1$, the smallest one is $\lambda = -1$, which pairs with the eigenvector direction $(1, -1)$ (or any scalar multiple of it, such as $(2, -2)$).

The question's eigenvector is $(2, a)$, which must be a scalar multiple of $(1, -1)$. Since the first component $2$ is twice $1$, the second component must also be twice $-1$:

\[ a = 2 \times (-1) = -2 \]

Let's summarize:

• A symmetric matrix $\begin{pmatrix} a & b \\ b & a \end{pmatrix}$ always has eigenvectors $(1,1)$ and $(1,-1)$, with eigenvalues $a+b$ and $a-b$.
• Here the eigenvalues are $3$ and $-1$, so the smallest, $-1$, belongs to the $(1,-1)$ direction.
• Scaling $(1,-1)$ so its first entry becomes $2$ forces the second entry to be $-2$.

The value of $a$ is $-2$.