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I have no clue how to prove this equality, first shown here: Show that the determinant of $A$ is equal to the product of its eigenvalues

$\det (A-\lambda I)=p(\lambda) = (-1)^n (\lambda - \lambda_1 )(\lambda - \lambda_2)\cdots (\lambda - \lambda_n)$

Can someone show how this is derived from either the three core properties of the determinant, cofactor expansion, or the big formula (Strang's term for permutation form of the determinant)?

I know how to show that the $\det(A- \lambda I)$ is a polynomial of dimension n by using cofactor expansion. And then, using the fundamental theorem of algebra, we can rewrite this polynomial in terms of n roots. HOWEVER, two problems.

  1. The $(-1)^n$ is unexplained.
  2. The fundamental theorem of linear algebra just says there are n roots, it doesn't tell us that the roots have to be the eigenvalues (i.e. it tells us this: $\det (A-\lambda I)=p(\lambda) = (\lambda - a )(\lambda - b)\cdots (\lambda - z)$)
Later
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user3180
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  • What is your definition of a characteristic polynomial? – Ajay Kumar Nair Aug 23 '20 at 06:20
  • $p(\lambda)$, the entire first line... they are all equal so they must all be the characteristic polynomial – user3180 Aug 23 '20 at 06:24
  • I feel that your title is misleading as it sounds like you are asking to prove your definition. – Ajay Kumar Nair Aug 23 '20 at 06:27
  • Actually, it's the 2nd and 3rd expressions in the equality that are the characteristic polynomial. The determinant is equal to the characteristic polynomial but is not the c.p. : https://en.wikipedia.org/wiki/Characteristic_polynomial – user3180 Aug 23 '20 at 06:29
  • Look at the formal definition section in this page. I kind of understand your confusion and already writing an answer. See if that helps. – Ajay Kumar Nair Aug 23 '20 at 06:31

3 Answers3

2
  1. When you calculate the determinant of $A - \lambda I$, you will notice that the sign of coefficient of the highest power term in the polynomial depends on the order of the matrix. That explains the $(-1)^n$. This sign is insignificant as we only need to find the roots.
  2. $\lambda$ satisfies $\det(A - \lambda I) = 0 \iff A - \lambda I$ is not invertible $\iff \exists v(\neq 0) \in V$ such that $(A- \lambda I)v = 0 \iff Av = \lambda v$.
Ajay Kumar Nair
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  • Thanks. I still don't see it though. Can you show explicitly how we can arrive at the polynomial form from cofactor expansion or one of the other methods? – user3180 Aug 23 '20 at 06:52
  • Are you convinced that $\det(A - \lambda I)$ gives us an $n$-degree polynomial? – Ajay Kumar Nair Aug 23 '20 at 06:55
  • https://snipboard.io/bRxT3Y.jpg – user3180 Aug 23 '20 at 07:01
  • Yes, I worked out the cofactor expansion for a 3x3 and I see it – user3180 Aug 23 '20 at 07:02
  • But what I see is $(a-\lambda)(e-\lambda)(i-\lambda) - fh$ – user3180 Aug 23 '20 at 07:03
  • This looks quite different from the first equality in my question – user3180 Aug 23 '20 at 07:03
  • You get an $n-$ degree polynomial by just calculating the determinant. The you use the Fundamental theorem of Algebra to get it into the form of linear factors. These linear factors give you the roots which are our eigenvalues. – Ajay Kumar Nair Aug 23 '20 at 07:07
  • Sure, but my problem is with the $(-1)^n$ term then. if you look at my calculations I'm not sure how that pops out, such that you can have $det(A- \lambda I) = (-1)^n$ * (some polynomial of degree n) where you only apply fundamental theorem of algebra to (some polynomial of degree n) – user3180 Aug 23 '20 at 07:09
  • Hi ajay, I have a problem with your part 2 – user3180 Aug 25 '20 at 02:02
  • Part 2 of your answer argues that for every root of $\det(A-\lambda I)$ = 0, that root has a corresponding eigenvector. But that is not the claim we have to make. The claim we have to make is the multiset of all the roots of $\det(A-\lambda I)$ = 0 is the same as the multiset of all the eigenvalues with multiplicities – user3180 Aug 25 '20 at 02:28
  • Imagine the true multiset of eigenvalues with multiplicity is (1,3,3,4). Your part 2 argument would allow $\det(A-\lambda I) = 0$ to produce a multiset of roots (1,3,4,4), which is not correct – user3180 Aug 25 '20 at 02:29
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According to the definition, a scalar $\lambda$ is called an eigenvalue of an $n\times n$ matrix $A$ if and only if there is a nonzero vector $\mathbf{v}$ such that$$A\mathbf{v}=\lambda \mathbf{v},$$or, equivalently$$(\lambda I - A)\mathbf{v}=\mathbf{0},$$ where $I$ is the $n \times n$ identity matrix. Since $\bf{v}$ is a nonzero vector, we must have$$\det (\lambda I - A) =0\tag{*}\label{*}$$(Otherwise, the matrix $\lambda I -A$ would be invertible, which implies that we could multiply the equation by $\det (\lambda I- A)^{-1}$ and so$$(\lambda I -A)^{-1}(\lambda I -A)\mathbf{v}=(\lambda I - A)^{-1}\mathbf{0} \quad \Rightarrow \quad I \mathbf{v}=\mathbf{0} \quad \Rightarrow \quad \mathbf{v}=\mathbf{0},$$ which is a contradiction).

Conversely, An equation of the form \ref{*} in general is a polynomial of degree $n$, so by the fundamental theorem of algebra it has $n$ roots $\lambda _1 , ..., \lambda _n$. So we can write it as$$\det (\lambda I -A) =(\lambda - \lambda _1 ) \cdots (\lambda - \lambda _n). \tag{**}\label{**}$$

Each $\lambda _i$ must be an eigenvalue of the matrix $A$ because from $\det (\lambda _i I -A)=0$ we conclude that the matrix $\lambda _i I -A$ is non-invertible$\dagger$ and so there must be a nonzero $\mathbf{v_i}$ such that $(\lambda _i I -A)(\mathbf{v_i})=\mathbf{0}$$\ddagger$, which is equivalent to $A \mathbf{v_i}=\lambda _i \mathbf{v_i}$

Now, if we define the characteristic polynomial of a matrix $A$ as$$p(t)=\det ( A-tI),$$then we have$$p(\lambda )=\det (A-\lambda I)=\det (-(\lambda I-A))=(-1)^n\det(\lambda I -A)$$(We used the basic property of the determinant stating that$\det (cA)=c^n \det A$).

Thus, by applying \ref{**} we conclude that$$p(\lambda )=\det (A-\lambda I)=(-1)^n (\lambda - \lambda _1) \cdots (\lambda - \lambda _n).$$

Footnote

$\dagger$ This follows from the fact that if a matrix $A$ is invertible then there exists some matrix $B$ such that $AB=I$ so $$\det (AB)= \det (A) \det (B) = \det (I) =1 \quad \Rightarrow \quad \det (A) = \frac{1}{\det (B)} \neq 0,$$ so if the determinant of a matrix is zero then the matrix must be non-invertible.

$\ddagger$ This follows from the fact that if $Ax=0$ has only the trivial solution then the matrix $A$ must be invertible, so if a matrix $A$ is non-invertible, there must be a nontrivial soluttion for $Ax=0$. For more information, please see this post

Later
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  • Very interesting. I understand this proof. I will note though, that $\det(A - \lambda I) = (\lambda - \lambda_1) ... (\lambda -\lambda_n)$ is also true according to the fundamental theorem of algebra right? But we have to use $\det(\lambda I - A)$ to get the final result we want. – user3180 Aug 24 '20 at 05:57
  • @user3180 The fundamental theorem of algebra is used to show that a polynomial of degree $n$ has $n$ roots. The factorization of the polynomial into the linear factors $(\lambda - \lambda _i)$ comes from Euclidean division of polynomials; for more details, you can see this post, for example. Please also note that the $(-1)^n$ appears in the final result because some authors prefer to define $p(t)$ as $\det (A-tI)$, rather than $\det (tI-A)$. – Later Aug 24 '20 at 06:39
  • We have to somehow argue that the $\lambda_i$'s produced in your polynomial are equivalent to the multiset of eigenvalues with multiplicities. This is not shown even with part 2 of Ajay's answer above. Is there a way we can complete the argument correctly? – user3180 Aug 25 '20 at 02:27
  • @user3180 Sorry, I cannot understand what you mean. When you solve $\det (\lambda I -A)$ for $\lambda$, you will get all eigenvalues with multiplicities, which is the same as finding the multiset of eigenvalues with multiplicities; no proof is needed. – Later Aug 25 '20 at 04:29
  • "When you solve det(λI−A) for λ, you will get all eigenvalues with multiplicities" This must be proved... in your above proof, you use fundamental theorem of algebra. Fundamental theorem of algebra says there are roots; it does not say the roots have to be eigenvalues nor does it say the roots have to equal the multiset of eigenvalues with multiplicities. – user3180 Aug 25 '20 at 08:46
  • "the fundamental theorem of algebra it has n roots λ1,...,λn" When you label the roots this way, as lambdas with subscripts, you are saying the roots are the multiset of eigenvalues with multiplicities. This jump can't be made without proof, because the fundamental theorem does not directly imply the roots are lambdas with subscripts. – user3180 Aug 25 '20 at 08:48
  • "The equation * in general is a polynomial of degree n, so by the fundamental theorem of algebra it has n roots a,...,z. " The only statement you can make is that fundamental theorem gives some arbitrary roots a, b, c.. etc. There is no logic here implying that these roots are eigenvalues. – user3180 Aug 25 '20 at 08:50
  • Please see my latest comment under Ajay's answer for more details – user3180 Aug 25 '20 at 08:53
  • "When you solve det(λI−A) for λ, you will get all eigenvalues with multiplicities" This part must be proved. – user3180 Aug 25 '20 at 08:54
  • There's a lot of distraction here. The OP's eigenvalues are listed with algebraic multiplicities, as otherwise the claim would not be true. But the only way to define algebraic multiplicities is as root multiplicities of the characteristic polynomial; thus, $\lambda_1, \lambda_2, \ldots, \lambda_n$ have to be defined as the roots of $\det\left(A-tI\right)$ listed with their root multiplicities. Defining them via eigenvectors does not work, since that definition does not yield the right multiplicities (in general). – darij grinberg Aug 25 '20 at 10:55
  • Why not? The dimension of the eigenspace provides the multiplicity, no? – user3180 Aug 25 '20 at 11:07
  • @darijgrinberg I agree that algebraic multiplicities are defined as root multiplicities of the characteristic polynomial, but the two definitions of eigenvalues are equivalent, as I proved in my answer. – Later Aug 25 '20 at 14:26
  • @user3180 Does my updated answer prove what you wanted? – Later Aug 25 '20 at 14:33
  • @Later No, due to the same reason Ajay's answer above is not sufficient – user3180 Aug 25 '20 at 19:54
  • "Imagine the true multiset of eigenvalues with multiplicity is (1,3,3,4). Your argument would allow det(A−λI)=0 to produce a multiset of roots (1,3,4,4), which is not correct " – user3180 Aug 25 '20 at 19:55
  • @user3180 If the multiset of eigenvalues is $(1,3,3,4)$, then the characteristic polynomial must be $\det (\lambda I - A)= (\lambda - 1)(\lambda -3)(\lambda -3)(\lambda -4)$ and so will never produce a multiset of roots $(1, 3, 4, 4)$, which is related to the different characteristic polynomial $\det (\lambda I - A)= (\lambda - 1)(\lambda -3)(\lambda -4)(\lambda -4)$. – Later Aug 26 '20 at 04:08
  • "If the multiset of eigenvalues is (1,3,3,4), then the characteristic polynomial must be det(λI−A)=(λ−1)(λ−3)(λ−3)(λ−4)" This is what we are trying to prove (more or less). You cannot prove a claim using the claim itself. – user3180 Aug 26 '20 at 04:52
  • The point is I want to use a distinct definition of the determinant, a distinct definition of the eigenvalues, and show they come together such that the roots of $\det(\lambda I - A)$ equals the eigenvalues with multiplicities. You are defining the determinant in terms of the eigenvalues above, you are not using two distinct definitions. – user3180 Aug 26 '20 at 04:57
  • For example, from the cofactor expansion definition of the determinant, we can show that there is an n-degree polynomial equal to $\det(\lambda I-A)$. Then from fundamental theorem we know there are n-roots. But at this point we need to explicitly prove that these roots are the eigenvalues with multiplicities. You cannot just say that's what these roots are without proof. – user3180 Aug 26 '20 at 04:58
  • @user3180 You agree that $\det (\lambda I -A)$ is a an $n$-degree polynomial which, by the fundamental theorem of algebra, has $n$-roots, right? Now, I have proved in my answer (the paragraph starting with "Conversely") that the set of the roots forms the multiset of the eigenvalues. – Later Aug 26 '20 at 05:52
  • "Each λi must be an eigenvalue of the matrix A because from det(λiI−A)=0 we conclude that the matrix λiI−A is non-invertible† and so there must be a nonzero vi such that (λiI−A)(vi)=0‡, which is equivalent to Avi=λivi" This paragraph – user3180 Aug 26 '20 at 06:00
  • Does not prove that the multiset of $\lambda_i$s in the polynomial expression form the multiset of eigenvalues with multiplicities. It only proves that each $\lambda_i$ is an eigenvalue. These are two very different claims – user3180 Aug 26 '20 at 06:01
  • And you must prove the former for a correct proof – user3180 Aug 26 '20 at 06:01
  • @user3180 $\det (\lambda _i I -A)$ is a polynomial of the form $(\lambda- \lambda _1)^{k_1} \cdots (\lambda - \lambda _m)^{k_m}$, where $\lambda _i$'s are distinct. So, we know the multiplicity of each eigenvalue. Thus, the multiset of eigenvalues with multiplicities is known. – Later Aug 26 '20 at 06:14
  • No, no... you can't say that. Unless you are using fundamental theorem in a way that is giving the exponents $k_i$ – user3180 Aug 26 '20 at 06:32
  • @user3180 The original problem is to prove that $\det (\lambda I - A)=(\lambda - \lambda _1) \cdots (\lambda - \lambda _n)$, where $\lambda _i$ is an eigenvalue, right? You already accepted that $\det (\lambda I -A)$ is an $n$-degree polynomial so by the fundamental theorem of algebra it has $n$ roots (counting multiplicities). I also proved that each roots must be an eigenvalue. So, the polynomial can be written as the product of $(\lambda - \lambda _i)$. So, what is unclear? – Later Aug 26 '20 at 10:12
  • I explained above what the problem is. "You already accepted that det(λI−A) is an n-degree polynomial so by the fundamental theorem of algebra it has n roots (counting multiplicities). I also proved that each roots must be an eigenvalue." These two statements are absolutely not sufficient to show that the roots of $\det(λI−A)$ produce the multiset of eigenvalues with multiplicities. – user3180 Aug 26 '20 at 10:28
  • Again, imagine the true multiset of eigenvalues is (1,3,3,4). Under your proof, we can have 4 roots. Then, you also prove that each root must be an eigenvalue. But under your proof (1,3,4,4) and (1,3,3,4) and even (1,1,3,4) are all valid multisets of eigenvalues with multiplicities, when in reality only (1,3,3,4) is valid. – user3180 Aug 26 '20 at 10:31
  • @user3180 Why do you need to show "the roots of $\det (\lambda I -A)$ produce the multiset of eigenvalues with multiplicities" for proving $\det (\lambda I -A)= (\lambda - \lambda _1) \cdots (\lambda - \lambda _n)$? – Later Aug 26 '20 at 13:38
  • Because those are identical statements. – user3180 Aug 27 '20 at 03:28
  • @user3180 "Commenting" has not been designed for chatting between users; unfortunately, we are violating some SE rule here. If you want to discuss more, we can talk in this room. – Later Aug 27 '20 at 08:40
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Actually, other than the definition of characteristic polynomial (CP) in wikipedia, there is another equivalent definition: $p(t) = \prod (t-\lambda_i)^{d_i}$, where $d_i$ is the algebraic multiplicity of eigenvalue $\lambda_i$. The latter definition is adopted in the book "Axler, Sheldon. Linear algebra done right. springer, 2015."

Building on top of the incomplete solution of @Ajay, we need to further show that the algebraic multiplicity is always larger than the geometric multiplicity, for the case the geometric multiplicity is larger than 1 for some eigenvalue. The proof can be found in here.

Regarding to the $(-1)^n$, please look at here.

GZ92
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