- #1
Whatever123
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Homework Statement
Prove that if two matrices are similar then they have the same eigenvalues with the same algebraic and geometric multiplicity.
Homework Equations
Matrices A,B are similar if A = C[tex]\breve{}[/tex]BC for some invertible C (and C inverse is denoted C[tex]\breve{}[/tex] because I tried for a long time to figure out how to get an inverse sign in latex but couldn't figure it out...).
The Attempt at a Solution
To show that two similar matrices have the same eigenvalues with the same geometric multiplicities, I need to show that their characteristic polynomials are the same.
Let A,B be similar matrices. Then,
A = C(inverse)BC
A-[tex]\lambda[/tex]I = C[tex]\breve{}[/tex]BC - [tex]\lambda[/tex]I
A-[tex]\lambda[/tex]I = C[tex]\breve{}[/tex]BC - [tex]\lambda[/tex]C[tex]\breve{}[/tex]C since C[tex]\breve{}[/tex]C=I
A-[tex]\lambda[/tex]I = C[tex]\breve{}[/tex][B-[tex]\lambda[/tex]]C
det(A-[tex]\lambda[/tex]I) = det(C[tex]\breve{}[/tex][B-[tex]\lambda[/tex]I]C)
det(A-[tex]\lambda[/tex]I) = det(C[tex]\breve{}[/tex]det(B-[tex]\lambda[/tex]I)det(C)
det(A-[tex]\lambda[/tex]I) = det(C[tex]\breve{}[/tex])det(C)det(B-[tex]\lambda[/tex]I)
Therefore, det(A-[tex]\lambda[/tex]I) = det(B-[tex]\lambda[/tex]I)
So, A, B have the same characteristic polynomials. This implies that they have the same eigenvalues with the same algebraic multiplicity. However, I do not think that this implies that they have the same geometric multiplicity because it doesn't same anything about the dimension of the eigenspace. Does it? Any suggestions on how I should try to show that the geometric multiplicity is the same?
So does anyone have any advice? I spent a while to get the first part, I just could use a nudge in the right direction for the next part... Thanks in advance.