- #1
Dustinsfl
- 2,281
- 5
Show that if [tex]\lambda=1[/tex] is an eigenvalue of A, then the matrix B will be singular.
[tex]\mu(1)=1-2+1=0[/tex]
[tex]B\mathbf{x}=0\mathbf{x}[/tex]
or does this need to be done via determinant?
[tex]det(B-\mu I)=0[/tex] since [tex]\lambda=1, \mu=0[/tex].
[tex]det(B-0I)=det(B)=0[/tex] Hence B is singular.
[tex]\mu(1)=1-2+1=0[/tex]
[tex]B\mathbf{x}=0\mathbf{x}[/tex]
or does this need to be done via determinant?
[tex]det(B-\mu I)=0[/tex] since [tex]\lambda=1, \mu=0[/tex].
[tex]det(B-0I)=det(B)=0[/tex] Hence B is singular.