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

Dustinsfl

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.

Mark44

Quote by Dustinsfl

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]

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?

Quote by Dustinsfl

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.

This is another way to show it.

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