MHB Null and non null eigenvalues (Oinker's question at Yahoo Answers)

[Fernando Revilla]

Here is the question:

(a) Find the eigenvalues of the matrix D below. For each eigenvalue, find a corresponding eigenvector.

D =
2 3
4 1(b) Suppose B is an n×n matrix, and λ = 0 is an eigenvalue of B. Prove that B is not invertible.

(c) Suppose B is an n × n non-invertible matrix. Prove that λ = 0 is an eigenvalue of B.

(d) Suppose B is an invertible n × n matrix, and λ is an eigenvalue of B. Prove that 1 is
an eigenvalue of B−1.

Here is a link to the question:

Matrix Question?? a.b.c.d.? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Fernando Revilla]

Hello Oinker,

$(a)\;$ Eigenvalues of $A=\begin{bmatrix}{2}&{3}\\{4}&{1}\end{bmatrix}$: $$\chi_A(\lambda)=\lambda^2-(\mbox{trace }A)\;\lambda+\det A=\lambda^2-3\lambda-10=0\\\Leftrightarrow \lambda=5\mbox{ (simple) }\vee\;\lambda=-2\mbox{ (simple) }$$ Eigenvectors and corresponding basis: $$\ker(A-5I)\equiv\left \{ \begin{matrix} -3x_1+3x_2=0\\4x_1-4x_2=0\end{matrix}\right.,\qquad B_5=\{(1,1)^T\}$$ $$\ker(A+2I)\equiv\left \{ \begin{matrix} 4x_1+3x_2=0\\4x_1+3x_2=0\end{matrix}\right.,\qquad B_{-2}=\{(3,-4)^T\}$$
$(b)\;$ If $\lambda=0$ is eigenvalue of $B$, then $\det (B-0I)=0$ so, $\det B=0$. As a consequence, $B$ is not invertible.

$(c)\;$ If $B$ is not invertible, $0=\det (B)=\det(B-0I)$ which implies that $\lambda=0$ is an eigenvalue of $B$.

$(d)\;$ According to $(b)$ and $(c)$, if $\lambda$ is an eigenvalue of $B$, necessarily $\lambda\neq 0$ and there exists $0\ne x\in K^n$ suach that $Bx=\lambda x$. Then, $$Bx=\lambda x\Rightarrow B^{-1}(Bx)=B^{-1}(\lambda x)\Rightarrow Ix=\lambda (B^{-1}x)\Rightarrow B^{-1}x=\frac{1}{\lambda}x$$ which implies that $1/\lambda$ is an eigenvalue of $B^{-1}$.