GATE CS 2008 Q28 — Eigenvalues and Eigenvectors, LU Decomposition

To find how many of the given matrices have an eigenvalue of $1$, we can analyze each matrix individually:

1. First Matrix: $A_1 = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$

   This is a diagonal matrix. The eigenvalues of a diagonal matrix are its diagonal entries. Therefore, the eigenvalues are $\lambda = 1$ and $\lambda = 0$. This matrix has an eigenvalue of $1$.

2. Second Matrix: $A_2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$

   This is an upper triangular matrix. The eigenvalues of a triangular matrix are its diagonal entries. Therefore, the eigenvalues are $\lambda = 0$ and $\lambda = 0$. This matrix does not have an eigenvalue of $1$.

3. Third Matrix: $A_3 = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}$

   We find the eigenvalues using the characteristic equation $\det(A_3 - \lambda I) = 0$:

   $$\det\begin{bmatrix} 1-\lambda & -1 \\ 1 & 1-\lambda \end{bmatrix} = 0 \implies (1-\lambda)^2 - (-1)(1) = 0$$ $$(1-\lambda)^2 + 1 = 0 \implies (1-\lambda)^2 = -1 \implies 1-\lambda = \pm i \implies \lambda = 1 \pm i$$

   The eigenvalues are complex numbers ($1+i$ and $1-i$). This matrix does not have an eigenvalue of $1$.

4. Fourth Matrix: $A_4 = \begin{bmatrix} -1 & 0 \\ 1 & -1 \end{bmatrix}$

   This is a lower triangular matrix. Its eigenvalues are its diagonal entries: $\lambda = -1$ and $\lambda = -1$. This matrix does not have an eigenvalue of $1$.

Thus, only one matrix ($A_1$) has an eigenvalue of $1$.