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Assume $A$ is an $n \times n$ diagonalizable matrix with eigenvalues $\lambda_1, ... , \lambda_n$. Prove that $A - \lambda_1I$ is also diagonalizable and has eigenvalues, $0, \lambda_2 - \lambda_1, ... , \lambda_n - \lambda_1$

Let $B = A - \lambda_1 I$

The character polynomial for $A$ is

$C_1(\lambda) = \det(A - \lambda I) = (\lambda - \lambda_1)...(\lambda - \lambda_n)$, the characteristic polynomial for $B$ is $C_2(\lambda) = \det(A - \lambda_1 I - \lambda I) = \det(A - (\lambda_1 + \lambda)I) = C_1(\lambda_1 + \lambda)$

$= (\lambda - 0)(\lambda + \lambda_1 - \lambda_2)...(\lambda + \lambda_1 - \lambda_n)$

Thus the eigenvalues for $B$ are $0, \lambda_2 - \lambda_1, ... , \lambda_n - \lambda_1$

But how do I show $B$ is diagonalizable?

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Let $A$ have the diagonalization $A = VDV^T$, with $V$ orthogonal and $D$ diagonal. Then $$A-\lambda_1I = VDV^T - \lambda_1VV^T = V(D- \lambda_1I)V^T.$$ Since the diagonal entries of $D$ are exactly the eigenvalues of $A$, then the eigenvalues of $A - \lambda_1I$ are the entries of $D - \lambda_1I$ or $0,\lambda_2-\lambda_1,$ and so on.

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$$A-\lambda I = PDP^{-1} - P\lambda I P^{-1} = P(D-\lambda I)P^{-1}$$

Now simply work out how $D-\lambda I$ looks like and you got your eigenvalues.

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  • $\begingroup$ How do you get $\lambda I = P\lambda I P^{-1}$? $\endgroup$ – Marodian Apr 6 '17 at 19:29
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    $\begingroup$ $\lambda$ is a scalar and $I$ the identity. since $MaN = aMN$ for matrices $M$ and $N$ and a scalar $a$ and $MM^{-1} = I$ for any invertible matrix, the claim follows. Note that this is equivalent to my answer. $\endgroup$ – SZN Apr 6 '17 at 21:13

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