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: $\det (A-\lambda I)=p(\lambda) = (\lambda - a )(\lambda - b)\cdots (\lambda - z)$)
  • $\begingroup$ What is your definition of a characteristic polynomial? $\endgroup$ Aug 23, 2020 at 6:20
  • $\begingroup$ $p(\lambda)$, the entire first line... they are all equal so they must all be the characteristic polynomial $\endgroup$
    – user3180
    Aug 23, 2020 at 6:24
  • $\begingroup$ I feel that your title is misleading as it sounds like you are asking to prove your definition. $\endgroup$ Aug 23, 2020 at 6:27
  • $\begingroup$ 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. : en.wikipedia.org/wiki/Characteristic_polynomial $\endgroup$
    – user3180
    Aug 23, 2020 at 6:29
  • $\begingroup$ Look at the formal definition section in this page. I kind of understand your confusion and already writing an answer. See if that helps. $\endgroup$ Aug 23, 2020 at 6:31

3 Answers 3

  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 $(A- \lambda I)v = 0 \iff Av = \lambda v$.
  • $\begingroup$ 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? $\endgroup$
    – user3180
    Aug 23, 2020 at 6:52
  • $\begingroup$ Are you convinced that $\det(A - \lambda I)$ gives us an $n$-degree polynomial? $\endgroup$ Aug 23, 2020 at 6:55
  • $\begingroup$ snipboard.io/bRxT3Y.jpg $\endgroup$
    – user3180
    Aug 23, 2020 at 7:01
  • $\begingroup$ Yes, I worked out the cofactor expansion for a 3x3 and I see it $\endgroup$
    – user3180
    Aug 23, 2020 at 7:02
  • $\begingroup$ But what I see is $(a-\lambda)(e-\lambda)(i-\lambda) - fh$ $\endgroup$
    – user3180
    Aug 23, 2020 at 7:03

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$$p(\lambda )=\det (A-\lambda I)=(-1)^n (\lambda - \lambda _1) \cdots (\lambda - \lambda _n).$$


$\dagger$ 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

  • $\begingroup$ 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. $\endgroup$
    – user3180
    Aug 24, 2020 at 5:57
  • $\begingroup$ @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 $\det (tI-A)$. $\endgroup$
    – Later
    Aug 24, 2020 at 6:39
  • $\begingroup$ 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? $\endgroup$
    – user3180
    Aug 25, 2020 at 2:27
  • $\begingroup$ @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. $\endgroup$
    – Later
    Aug 25, 2020 at 4:29
  • $\begingroup$ "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. $\endgroup$
    – user3180
    Aug 25, 2020 at 8:46

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.


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