Question

Let \(A\) be a real symmetric matrix. \(\lambda_1,\lambda_2\;(\lambda_1\neq\lambda_2)\) be two eigen values of \(A\) and \(X_1,X_2\) are respectively the eigen vectors of \(A\) corresponding to \(\lambda_1\) and \(\lambda_2\), then \(X_1^TX_2=\)

• A. \(X_2^TX_1\neq 0\)
• B. \(X_1X_2^TX_1=0\)
• C. \(X_2X_1^TX_2=0\)
• D. \(X_2^TX_1=0\)

Hint:

For every real symmetric matrix, eigenvectors corresponding to distinct eigenvalues are always orthogonal. This property is frequently used in linear algebra and matrix theory problems.

Answer 1

The Correct Option is D

Concept: A fundamental property of a real symmetric matrix is that eigenvectors corresponding to distinct eigenvalues are mutually orthogonal. If \[ AX_1=\lambda_1X_1,\qquad AX_2=\lambda_2X_2, \] where \(\lambda_1\neq\lambda_2\), then \[ X_1^TX_2=0. \] Hence, the dot product of the two eigenvectors is zero.

Step 1: Write the given eigenvalue equations. Since \(X_1\) and \(X_2\) are eigenvectors of the symmetric matrix \(A\), \[ AX_1=\lambda_1X_1, \] and \[ AX_2=\lambda_2X_2. \]

Step 2: Premultiply the second equation by \(X_1^T\). We obtain \[ X_1^TAX_2=\lambda_2X_1^TX_2. \] Since \(A\) is symmetric, \[ A^T=A. \] Also, \[ X_1^TAX_2=(AX_1)^TX_2. \] Using \[ AX_1=\lambda_1X_1, \] we get \[ X_1^TAX_2 =\lambda_1X_1^TX_2. \] Therefore, \[ \lambda_1X_1^TX_2 = \lambda_2X_1^TX_2. \]

Step 3: Use the fact that the eigenvalues are distinct. Thus, \[ (\lambda_1-\lambda_2)X_1^TX_2=0. \] Since \[ \lambda_1\neq\lambda_2, \] it follows that \[ X_1^TX_2=0. \] Hence, \[ X_2^TX_1=0. \] Therefore, the correct option is \[ \boxed{(D)\;X_2^TX_1=0.} \]