Show that if [tex]\lambda=1[/tex] is an eigenvalue of A, then the matrix B will be singular.(adsbygoogle = window.adsbygoogle || []).push({});

[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.

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

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