Why does $\lambda(zA)=z\lambda(A)$?

Question

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 $\lambda(zA)=z\lambda(A)$?

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 $\lambda z=z\lambda$. — Jyrki Lahtonen, Sep 01 '16 at 10:17
@JyrkiLahtonen Isn't it intending to ask "The eigenvalues of $zA$ are exactly the eigenvalues of $A$ times $z$"? — rschwieb, Sep 01 '16 at 10:20
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 $A$. — An.Ditlev, Sep 01 '16 at 10:21
@rschwieb That interpretation makes much more sense! A bit odd notation, but we've seen stranger things. — Jyrki Lahtonen, Sep 01 '16 at 10:48
Possible duplicate of How to prove "eigenvalues of polynomial of matrix $A$ = polynomial of eigenvalues of matrix $A$ " — Surb, Sep 01 '16 at 12:50

An.Ditlev · Answer 1 · 2016-09-01T11:01:02.097

2

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 $A$?

edited Sep 01 '16 at 11:01

answered Sep 01 '16 at 10:24

An.Ditlev

1,029

Surb · Accepted Answer · 2016-09-01T12:48:14.440

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 $z_1,\ldots,z_n\in \Bbb Z$.

@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. — Surb, Sep 01 '16 at 10:59

score 1 · Answer 3 · answered Sep 01 '16 at 10:25

1

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 $\lambda(zA)=z\lambda(A)$.

answered Sep 01 '16 at 10:25

Josu Etxezarreta Martinez

1,152

1

That only works for diagonalizable matrices – Ben Grossmann Sep 01 '16 at 11:19

Why does $\lambda(zA)=z\lambda(A)$?

3 Answers3