GO Classes CS/DA 2025 | Weekly Quiz 5 | Linear Algebra | Question: 6

Answer 1

Option B

if \( A \) is a nilpotent matrix with \( k = 2 \) (meaning \( A^2 = 0 \)), then \( A \) cannot have a non-zero eigenvalue. 

To understand why, let's suppose that \( A \) has a non-zero eigenvalue \( \lambda \). Then there exists a non-zero vector \( v \) such that \( Av = \lambda v \). 

Now, let's consider \( A^2v = A(Av) = A(\lambda v) = \lambda(Av) = \lambda(\lambda v) = \lambda^2v \).

Since \( A^2 = 0 \), we have \( \lambda^2v = 0 \). But since \( v \) is non-zero, this implies that \( \lambda^2 = 0 \). However, if \( \lambda^2 = 0 \), then \( \lambda = 0 \). Thus, any eigenvalue of \( A \) must be zero if \( A \) is nilpotent with \( k = 2 \).

Answer 2

• A. Distinct eigenvalues have LI eigenvectors but same eigenvalue can also have LI eigenvectors.
  
  
• B. $A^2 = 0$ then $A$ has only $0$ as its eigenvalue.
  
  
• C. True
  
  
• D. True

Answer 3

B) $A^K=0, \quad K=2$
$$
\begin{aligned}
& \Rightarrow \quad A^2=0 . \\
& \Rightarrow \quad \lambda^2=0 \quad \text { (cayley-Hamillon Theorem) } \\
& \Rightarrow \lambda=0 \quad(\therefore \text { nonzero eigen value } \\
& \text { is not possible). }
\end{aligned}
$$
c)
$$
\begin{aligned}
& A x=\lambda x \\
& \Rightarrow|A-\lambda I|=0 .
\end{aligned}
$$

Suppose the characteristic equation is
$$
\begin{array}{r}
C_0 \lambda^n+c_1 \lambda^{n-1}+C_2 \lambda^{n-2}+\cdots+C_n \lambda+C_{n+1}=0 \\
-(1) \\
\Rightarrow C_0 A^n+C_1 A^{n-1}+C_2 A^{n-2}+\cdots+C_n A+C_{n+1}=0 \\
\text { (Cayley-Hamilton) }
\end{array}
$$
applying Transpose on the equation.
$$
\begin{aligned}
& C_0\left(A^{\top}\right)^n+C_1\left(A^{\top}\right)^{n-1}+\ldots+C_n A^{\top}+C_{n+1}=0 . \\
& C_0\left(\lambda^1\right)^n+c_1\left(\lambda^1\right)^{n-1}+\ldots+C_n \lambda^{\prime}+C_{n+1}=0 \text{ }...(2)
\end{aligned}
$$
equation (1) and (2) are same
$$
\therefore \lambda^{\prime}=\lambda
$$