Show that λ is an eigenvalue of the B matrix iff λ is an eigenvalue of B'

Alex_Of_Darkness

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Show that λ is an eigenvalue of the B matrix if, and only if λ is an eigenvalue of B'

It's supposed to be really easy to show that but I guess that it is only the case when you know how to do it :p
 
Let's be clear. Please state what B' is. Notation isn't always uniform.

Is \(\displaystyle |\lambda I - B|\) any different from \(\displaystyle |\lambda I - B'|\)
 
Fair enough. Now the answer to my determinant question...
 
Pick any 2x2 matrix and play with it a bit.
Expand your exploration from there.
 
Pick any 2x2 matrix and play with it a bit.
Expand your exploration from there.

So I tried with the matrix B (first row 1,2 ; second row 3,4) so B' (first row 1,3 ; second row 2,4) and it gives me | λ I - B | = | λ I - B' |


But I'm still really not sure how to resolve my problem!
 
You have answered my question for this single matrix. What does it mean for the Eigenvalues if you get EXACTLY the same Characteristic equation?

Does your example extend to an arbitrary square matrix?
 
You have answered my question for this single matrix. What does it mean for the Eigenvalues if you get EXACTLY the same Characteristic equation?

Does your example extend to an arbitrary square matrix?

Yeah I tested a few other 2x2 matrix and it seems to always give the same Eigenvalues when you use B and it's transposed matrix. So yeah I'm pretty sure that this example extend to an arbitrary square matrix!
 
You seem to be done. Is there anything else you need to demonstrate to accomplish your purpose?
 
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