GATE 2018 | MATHS | Q-52

Question

Consider the matrix \( A = I_9 - 2u^T u \) with \( u = \frac{1}{3}[1, 1, 1, 1, 1, 1, 1, 1, 1] \), where \( I_9 \) is the \(9 \times 9\) identity matrix and \( u^T \) is the transpose of \( u \). If \( \lambda \) and \( \mu \) are two distinct eigenvalues of \( A \), then \[ | \lambda - \mu | =  \] _________

Answer 1

\( A \) is a symmetric matrix because it's equal to its transpose (\( A = A^T \)).
Symmetric matrices have real eigenvalues.

The matrix \( uu^T \) has rank 1, meaning it has only one non-zero eigenvalue.
Subtracting 2 times this rank-1 matrix from the identity matrix \( I_9 \) results in \( A \) having:
One eigenvalue of 1 (from the 8-dimensional eigenspace of \( uu^T \) with eigenvalue 0).
Eight eigenvalues of -1 (from the 1-dimensional eigenspace of \( uu^T \) with eigenvalue 1, scaled by -2).

Since the distinct eigenvalues are 1 and -1, their absolute difference is \( |1 - (-1)| = 2 \).

Therefore, \( |\lambda - \mu| = 2 \) for any two distinct eigenvalues \( \lambda \) and \( \mu \) of matrix \( A \).

Answer 2

$$uu^T = \frac{1}{3}\cdot\frac{1}{3}\cdot [1,1,1,1,1,1,1,1,1][1,1,1,1,1,1,1,1,1]^T = \frac{1}{9}\cdot [9]=[1]$$ which has eigen value of $1$ so we know that AB and BA share there non zero eigen values always so $$\text{non-zero-eigenvalues}(u^Tu) = \text{non-zero-eigenvalues}(uu^T)=1$$

since we know that $$\text{number of eigen values = dimension of matrix  = 9}$$ so other 8 eigen values must be 0 since all non-zero eigen values need to be shared

so 2 distinct eigen values of $u^Tu$ are $0,1$

$$\lambda_A = \lambda_I – 2\lambda_{u^Tu} = 1 – (0 \ \text{or} \  2) = (1 \ \text{or} \  -1)$$

so ans = $$|1-(-1)|=2$$