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Let $A\in M_n$ be a square matrix and $z\in \mathbb{C}$. Let $\lambda(A)$ denotes the eigenvalues of $A$.

How to prove that $\lambda(zA)=z\lambda(A)$?

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closed as off-topic by Watson, rschwieb, Math1000, Surb, Daniel W. Farlow Sep 1 '16 at 14:20

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    $\begingroup$ Something is a bit strange. That identity holds irrespective of whether $\lambda$ is an eigenvalue of not. All you need is that the matrices form a vector space and that $\lambda z=z\lambda$. $\endgroup$ – Jyrki Lahtonen Sep 1 '16 at 10:17
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    $\begingroup$ @JyrkiLahtonen Isn't it intending to ask "The eigenvalues of $zA$ are exactly the eigenvalues of $A$ times $z$"? $\endgroup$ – rschwieb Sep 1 '16 at 10:20
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    $\begingroup$ Are you asking why the eigenvalues of $zA$ are $z$ times the eigenvalues of $A$. That is you mean that $\lambda(A)$ is a function that returns the eigenvalues of $A$. $\endgroup$ – An.Ditlev Sep 1 '16 at 10:21
  • $\begingroup$ @rschwieb That interpretation makes much more sense! A bit odd notation, but we've seen stranger things. $\endgroup$ – Jyrki Lahtonen Sep 1 '16 at 10:48
  • $\begingroup$ Possible duplicate of How to prove "eigenvalues of polynomial of matrix $A$ = polynomial of eigenvalues of matrix $A$ " $\endgroup$ – Surb Sep 1 '16 at 12:50
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For $z\neq 0$ we have

$$\det(\lambda I-A)=\det\big(\tfrac{1}{z}(z\lambda I-zA)\big)=\tfrac{1}{z^n}\det\big((z\lambda)I-zA\big)$$

Edit: It follows that $\det(\lambda I−A)=0$ if and only if $\det((z\lambda)I−zA)=0$, i.e. $\lambda$ is an eigenvalue of $A$ if and only if $z\lambda$ is an eigenvalue of $zA$. The advantage of this approach is that it also shows that the (algebraic) multiplicity of the eigenvalues are equals.

However, note that the approach of An.Ditlev is simpler and can be nicely generalized as follow:

Let $z_1,\ldots,z_n\in \Bbb C$ and $\alpha_1,\ldots,\alpha_n\in \Bbb N$ and let $v\neq 0$ be such that $Av=\lambda v$, then we have

$$\Big(\sum_{k=1}^nz_kA^{k}\Big)v = \sum_{k=1}^nz_k(A^{k}v)= \sum_{k=1}^nz_k(\lambda^{k}v)=\Big(\sum_{k=1}^nz_k\lambda^{k}\Big)v$$

showing that if $\lambda$ is an eigenvalue of $A$, then $\sum_{k=1}^nz_k\lambda^{k}$ is an eigenvalue of $\sum_{k=1}^nz_kA^{k}$. Moreover, note that if $A$ is invertible, then the same holds also for exponent $z_1,\ldots,z_n\in \Bbb Z$.

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    $\begingroup$ @Undersky Well it follows that $\det(\lambda I-A)=0$ if and only if $\det\big((z\lambda)I-zA\big)=0$, i.e. $\lambda$ is an eigenvalue of $A$ if and only if $z\lambda$ is an eigenvalue of $zA$. Now, the advantage of this approach is that it also shows that the (algebraic) multiplicity of the eigenvalues are equals. $\endgroup$ – Surb Sep 1 '16 at 10:59
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If I understand your question correctly consider the following.

Let $\lambda$ be an eigenvalue of $A$ with corresponding eigenvector $e$ then for $zA$ we have.

$(zA)e=z(Ae)=z(\lambda e)=(z\lambda)e$

Thus $z \lambda$ is an eigenvalue for $zA$. Now can you see why an eigenvalue for $zA$ is an eigenvalue for $A$?

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This happens for the nex reason, if A is diagonaliyable, then:

$A=Q\Lambda Q^{-1}$. So then, diagonalizing $zA$:

$zA=zQ\Lambda Q^{-1}=Q(z\Lambda)Q^{-1}=Q\Lambda_z Q^{-1}\rightarrow \Lambda_z=z\Lambda$, implying that $\lambda(zA)=z\lambda(A)$.

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    $\begingroup$ That only works for diagonalizable matrices $\endgroup$ – Omnomnomnom Sep 1 '16 at 11:19

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