Follow up to B=I-2A-A^2 where mu=1-2lambda+(lambda)^2

[Dustinsfl]

Show that if \lambda=1 is an eigenvalue of A, then the matrix B will be singular.

\mu(1)=1-2+1=0

B\mathbf{x}=0\mathbf{x}

or does this need to be done via determinant?

det(B-\mu I)=0 since \lambda=1, \mu=0.

det(B-0I)=det(B)=0 Hence B is singular.

 

[Mark44]

Show that if \lambda=1 is an eigenvalue of A, then the matrix B will be singular.

\mu(1)=1-2+1=0

B\mathbf{x}=0\mathbf{x}

This is true, but how did you arrive at it? Also, what's the significance of this equation in terms of what you're supposed to show?

or does this need to be done via determinant?

det(B-\mu I)=0 since \lambda=1, \mu=0.

det(B-0I)=det(B)=0 Hence B is singular.

This is another way to show it.