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Suppose $A$ and $B$ are two commuting square matrices over an algebraically closed field. It is true that the spectrum of $A+B$ is contained in the set $\{\lambda_1+\lambda_2:\lambda_1 \in \sigma(A), \lambda_2 \in \sigma(B)\}$ where $\sigma(A)$ denotes the spectrum of $A$?

If so, can you provide a reference for this? can more be said?

Thanks

the L
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4 Answers4

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Diagonalizable matrices

Let's say that $A$ and $B$ are diagonalizable. Then, since they commute, there exists an invertible matrix $T$ such that both $T^{-1}AT = D_A$ and $T^{-1}BT = D_B$ are diagonal matrices (Horn & Johnson, Matrix Analysis, page 52). The elements of the diagonal matrices are the eigenvalues. Thus we get:

$$A + B = TD_AT^{-1} + TD_BT^{-1} = T(D_A + D_B)T^{-1}$$

and so the eigenvalues of $A+B$ have the desired form.

General case

In the general case you can use simultaneous triangularization. The idea is the same, since $A$ and $B$ commute, there exists a unitary $U$ such that $U^*AU = T_A$ and $U^*BU = T_B$ where $T_A$ and $T_B$ are upper triangular matrices with the eigenvalues on the diagonal (Horn & Johnson, Matrix Analysis, page 81). Now we get:

$$A+B = U(T_A+T_B)U^*$$

and thus the eigenvalues have the desired form.

Calle
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    The OP is about matrices "over an algebraically closed field", not over $\mathbb C$. So replace "unitary" and "$^*$" by "invertible" and "$^{-1}$" (and probably use another reference that actually proves this). – darij grinberg Mar 24 '19 at 22:56
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Frobenius has proved that commuting matrices over an algebraically closed field can be simultaneously brought into upper triangular form (this holds more generally for a solvable Lie algebras of matrices by Lie's theorem). To prove this for commuting matrices, show that they have a simultaneous eigenvector and proceed by induction on the dimension.

One way to prove this last fact (again for commuting matrices) is to use the Hilbert Nullstellensatz, see here, a nice proof of which you can find online e.g. on p.116 of Pete L. Clark's notes.

Community
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t.b.
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    @anonymous: I'm slightly confused: Why do you ask your question in terms of algebraically closed fields if you're then accepting an answer that presupposes that you work over the complex numbers? – t.b. Jan 29 '11 at 17:33
  • i tried looking at page number 116 of your reference but it doesn't look like its on that page ? can u check and let me know . Thanks . – Theorem Jan 21 '13 at 20:25
  • @Theorem: Section 11 in the current version is called Nullstellensätze and in any case the pdf is fully searchable... – t.b. Feb 03 '13 at 09:41
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Yes.

Matrix Analysis by Horn and Johnson has a proof.

Brian
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  • The proof uses the fact that commuting matrices can be simultaneously upper-triangularized. They follow with an example where the other direction doesn't hold. – Brian Jan 29 '11 at 17:03
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Yes see this(article about commutating matrices). In particular, if $\alpha_i$ and $\beta_j$ are the eigenvalues of $A$ and $B$, then one can choose an eigenbasis such that the eigenvalues of $A+B$ are $\alpha_i+ \beta_j$.

NebulousReveal
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  • thanks, I saw this. The problem is that the reference for the fact ultimitely leads to an article of Frobenius, which is 1. In german. 2. I could not locate an online version of it. I wonder if there is a more accessible reference for this fact. – the L Jan 29 '11 at 16:54
  • Can we say if, Spectrum of $A+ B$ is same as ${ λ_1+ λ_2 : λ_1 ∈ σ(A) , λ_2 ∈ σ(B) }$ ? In which case this is true? – Akash Patalwanshi Sep 24 '17 at 14:17
  • Eigenbasis? There isn't always an eigenbasis. – darij grinberg Mar 24 '19 at 22:55