Show that the determinant of $A$ is equal to the product of its eigenvalues

Show that the determinant of a matrix $A$ is equal to the product of its eigenvalues $\lambda_i$.

So I'm having a tough time figuring this one out. I know that I have to work with the characteristic polynomial of the matrix $\det(A-\lambda I)$. But, when considering an $n \times n$ matrix, I do not know how to work out the proof. Should I just use the determinant formula for any $n \times n$ matrix? I'm guessing not, because that is quite complicated. Any insights would be great.

• This is only true if there are $n$ distinct eigenvalues. In that case, you will have a diagonalisation of the matrix, so it is immediate from the multiplicative property of $\det$. – user1537366 Jan 22 '15 at 5:45
• @user1537366 are you saying that this is not necessarily true in cases where the eigenvalues have multiplicity > 1? – Hunle Nov 1 '17 at 4:29
• @user1537366, is it the product of all eigenvalues, or only a product of the set of distinct eigenvalues? thanks you. – Hunle Aug 26 '18 at 22:47
• The statement in the question was correct. The product of all eigenvalues (repeated ones counted multiple times) is equal to the determinant of the matrix. – inavda Mar 23 at 20:40
• @inavda Why can you say that the determinant is the product of the eigenvalues? consider $\begin{pmatrix} 0 & -1 \\ 0 & 1 \end{pmatrix}$ over $\mathbb{R}$ which doens't have any eigenvalues but determinant 1. I guess we have to require the underlying field to be algebraically closed. – Viktor Glombik Jun 1 at 17:39

I think I got it...

Suppose that $\lambda_1, \ldots, \lambda_n$ are the eigenvalues of $A$. Then the $\lambda$s are also the roots of the characteristic polynomial, i.e.

$$\begin{array}{rcl} \det (A-\lambda I)=p(\lambda)&=&(-1)^n (\lambda - \lambda_1 )(\lambda - \lambda_2)\cdots (\lambda - \lambda_n) \\ &=&(-1) (\lambda - \lambda_1 )(-1)(\lambda - \lambda_2)\cdots (-1)(\lambda - \lambda_n) \\ &=&(\lambda_1 - \lambda )(\lambda_2 - \lambda)\cdots (\lambda_n - \lambda) \end{array}$$

The first equality follows from the factorization of a polynomial given its roots; the leading (highest degree) coefficient $(-1)^n$ can be obtained by expanding the determinant along the diagonal.

Now, by setting $\lambda$ to zero (simply because it is a variable) we get on the left side $\det(A)$, and on the right side $\lambda_1 \lambda_2\cdots\lambda_n$, that is, we indeed obtain the desired result

$$\det(A) = \lambda_1 \lambda_2\cdots\lambda_n$$

So the determinant of the matrix is equal to the product of its eigenvalues.

• Interesting, but don't you also have to show that the leading coefficient of the polynomial is 1? Or is that obvious? – DanielV Sep 28 '13 at 9:05
• $\det(A-\lambda I)=(\lambda_1 - \lambda)^{m_1}(\lambda_2 - \lambda)^{m_2}\cdots (\lambda_n - \lambda)^{m_n}.$ so $\det(A) = \lambda_1^{m_1} \lambda_2^{m_2}\cdots\lambda_n^{m_n}.$ with $m_i$ is the multiplicity of $\lambda_i$ – Mohamez Jan 14 '14 at 7:39
• @Muno: I believe that "Expanding the determinant along the diagonal" refers to the permuatation method of computing the determinant. This gives an polynomial with $n!$ terms, but only one will contribute $n$th power of $\lambda$: when the factors come from the diagonal. – Jason DeVito Jan 29 '18 at 20:18
• I was confused how to justify the coefficient $(-1)^n$ and understood it like this: We know $p(\lambda)$ factors in $(\lambda-\lambda_1) \ldots (\lambda-\lambda_n)$. But if we multiply it out we get that the coefficient of $\lambda^n$ is $1$. This is not what we get when we compute $\det(A-\lambda I)$, because there we get an $(-1)^n \lambda^n$. So we need to add $(-1)^n$ manually so that our factorization is correct. – philmcole Apr 10 '18 at 12:29
• @philmcole would you please explain why the coefficient of $\lambda^n$ is $(-1)^n$? I did not get it... – Peng Zhao Oct 16 '18 at 14:42

I am a beginning Linear Algebra learner and this is just my humble opinion.

One idea presented above is that

Suppose that $\lambda_1,\ldots \lambda_2$ are eigenvalues of $A$.

Then the $\lambda$s are also the roots of the characteristic polynomial, i.e.

$$\det(A−\lambda I)=(\lambda_1-\lambda)(\lambda_2−\lambda)\cdots(\lambda_n−\lambda)$$.

Now, by setting $\lambda$ to zero (simply because it is a variable) we get on the left side $\det(A)$, and on the right side $\lambda_1\lambda_2\ldots \lambda_n$, that is, we indeed obtain the desired result

$$\det(A)=\lambda_1\lambda_2\ldots \lambda_n$$.

I dont think that this works generally but only for the case when $\det(A) = 0$.

Because, when we write down the characteristic equation, we use the relation $\det(A - \lambda I) = 0$ Following the same logic, the only case where $\det(A - \lambda I) = \det(A) = 0$ is that $\lambda = 0$. The relationship $\det(A - \lambda I) = 0$ must be obeyed even for the special case $\lambda = 0$, which implies, $\det(A) = 0$

UPDATED POST

Here i propose a way to prove the theorem for a 2 by 2 case. Let $A$ be a 2 by 2 matrix.

$$A = \begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\\\end{pmatrix}$$

The idea is to use a certain property of determinants,

$$\begin{vmatrix} a_{11} + b_{11} & a_{12} \\ a_{21} + b_{21} & a_{22}\\\end{vmatrix} = \begin{vmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\\\end{vmatrix} + \begin{vmatrix} b_{11} & a_{12}\\b_{21} & a_{22}\\\end{vmatrix}$$

Let $\lambda_1$ and $\lambda_2$ be the 2 eigenvalues of the matrix $A$. (The eigenvalues can be distinct, or repeated, real or complex it doesn't matter.)

The two eigenvalues $\lambda_1$ and $\lambda_2$ must satisfy the following condition :

$$\det (A -I\lambda) = 0$$ Where $\lambda$ is the eigenvalue of $A$.

Therefore, $$\begin{vmatrix} a_{11} - \lambda & a_{12} \\ a_{21} & a_{22} - \lambda\\\end{vmatrix} = 0$$

Therefore, using the property of determinants provided above, I will try to decompose the determinant into parts.

$$\begin{vmatrix} a_{11} - \lambda & a_{12} \\ a_{21} & a_{22} - \lambda\\\end{vmatrix} = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} - \lambda\\\end{vmatrix} - \begin{vmatrix} \lambda & 0 \\ a_{21} & a_{22} - \lambda\\\end{vmatrix}= \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\\\end{vmatrix} - \begin{vmatrix} a_{11} & a_{12} \\ 0 & \lambda \\\end{vmatrix}-\begin{vmatrix} \lambda & 0 \\ a_{21} & a_{22} - \lambda\\\end{vmatrix}$$

The final determinant can be further reduced.

$$\begin{vmatrix} \lambda & 0 \\ a_{21} & a_{22} - \lambda\\\end{vmatrix} = \begin{vmatrix} \lambda & 0 \\ a_{21} & a_{22} \\\end{vmatrix} - \begin{vmatrix} \lambda & 0\\ 0 & \lambda\\\end{vmatrix}$$

Substituting the final determinant, we will have

$$\begin{vmatrix} a_{11} - \lambda & a_{12} \\ a_{21} & a_{22} - \lambda\\\end{vmatrix} = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\\\end{vmatrix} - \begin{vmatrix} a_{11} & a_{12} \\ 0 & \lambda \\\end{vmatrix} - \begin{vmatrix} \lambda & 0 \\ a_{21} & a_{22} \\\end{vmatrix} + \begin{vmatrix} \lambda & 0\\ 0 & \lambda\\\end{vmatrix} = 0$$

In a polynomial $$a_{n}\lambda^n + a_{n-1}\lambda^{n-1} ........a_{1}\lambda + a_{0}\lambda^0 = 0$$ We have the product of root being the coefficient of the term with the 0th power, $a_{0}$.

From the decomposed determinant, the only term which doesn't involve $\lambda$ would be the first term

$$\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\\end{vmatrix} = \det (A)$$

Therefore, the product of roots aka product of eigenvalues of $A$ is equivalent to the determinant of $A$.

I am having difficulties to generalize this idea of proof to the $n$ by $$case though, as it is complex and time consuming for me. • In the previously given proof, the fact that p(λ) = 0 was not used. Hence it is valid for all A and not just det(A) = 0 – Tyg13 Apr 1 '17 at 23:57 • The first part seems accurate as long as the characteristic polynomial can be factored in linear terms as you did. This always happens over \mathbb{C} (due to the Fundamental Theorem of Algebra), but not always over \mathbb{R}. What happens if p(\lambda) may not be factored in linear terms? Does the formula still holds? – Marra Oct 4 '18 at 23:54 The approach I would use is to Decompose the matrix into 3 matrixes based on the eigenvalues. Then you know that the det(A*B) = det(A)*det(B), and that det(inv(A)) = 1/det(A). You can probably fill in the rest of the details from the article, depending on how rigorous your proof needs to be. Edit: I just realized this won't work on all matrices, but it might give you an idea of an approach. • Why not? Any matrix has some "eigendecomposition": Schur, Jordan,... – Algebraic Pavel Sep 29 '13 at 11:20 • I liked this. I did it this way, but is this a correct proof? a) If matrix A has linearly independent columns.$$A=SDS^{-1}$$now take \det of both sides$$\det(A)=\det(SDS^{-1})=\det(S)\det(D)\det(S^{-1})=\det(D)$$and \det(D) is just the product of all \lambda_i. b) If matrix A has linearly de pendent columns. Then$$\det(A)=0$$but what are it's eigenvalues? – jacob Mar 8 '14 at 7:11 • @jacob Having linearly independent columns does not imply diagonalisable... – user1537366 Jan 22 '15 at 5:47 • His Idea is good but needs more arguments the D will be a Jordan Block matrix. Such matrices are almost diagonal. – Kori Sep 25 '16 at 2:07 You must know the following: == If we take an extension of the basis field then both the determinant and the trace of a (square) matrix remain unchanged when evaluating them in the new field == Take a splitting field of the characteristic polynomial of \;A\; and calculate this matrix's Jordan Canonical form. Since this last is a triangular matrix its determinant is the product of the elements in its main diagonal, and we know that in this diagonal appear the eigenvalues of \;A\; so we're done. From eigen decomposition A = S \lambda S^{-1}, where \lambda is a matrix formed by eigen values of A. \implies det(A) = det(S)\phantom{1}det(\lambda)\phantom{1}det(S^{-1}) \implies det(A) = det(\lambda)   det(\lambda) is nothing but \lambda_1$$\lambda_2$....$\lambda_n$• Is this shown by saying$det(S) = 1/det(S^-1)$? – Hunle Nov 1 '17 at 4:27 • Indeed. determinant satisfies that$det(AB) = det(A)det(B)$for any two$n\times n$matrices A, B. Since$S S^{ -1}=Id$then$det(S) det(S^{ -1})= det(Id) = 1$. – eduard Nov 11 '17 at 14:06 • but this decomposition only exists, if A is diagonalizable – user519338 Apr 1 '18 at 21:01 • As noted above, this does not work if$A$is not diagonalizable. Instead, you could use a square matrix$T$over the algebraic closure of the given field s.t.$T^{-1}AT$has Jordan form. – qwertz Jul 9 '18 at 14:06 A few places in this thread I noticed people raised issues about 'what if A doesn't have independent columns' or 'what if the determinant is 0'. I believe the following are all equivalent: • 0 is an eigenvalue of A • A has linearly dependent columns (or rows) •$det(A)=0$•$\prod_i \lambda _i = 0$(the product of eigenvalues of A) so we can take care of this issue by saying "Suppose one of the above is true. Then$det(A)=\prod_i \lambda _i = 0$, otherwise... (note it is clear that 0 being an eigenvalue results in the product being 0) I think this is right... Write A = $$\begin{pmatrix}a_{11} & a_{12}&\cdots&a_{1n}\\\ \vdots & \vdots&&\vdots\\\ a_{n1}& a_{n2}&\cdots& a_{nn} \end{pmatrix}$$ Let the n eigenvalues of A be $$\lambda_1, \cdots , \lambda_n$$. Finally, denote the characteristic polynomial of A by $$p(\lambda) = |\lambda I − A| = \lambda_n + c_{n−1}\lambda{n−1} + \cdots + c_1λ + c_0$$. Note that since the eigenvalues of A are the zeros of $$p(\lambda)$$, this implies that $$p(\lambda)$$ can be factorised as $$p(\lambda) = (\lambda − \lambda_1)\cdots(\lambda − \lambda_n)$$. Consider the constant term of $$p(λ), c_0$$. The constant term of $$p(\lambda)$$ is given by $$p(0)$$, which can be calculated in two ways: Firstly, $$p(0) = (0 − λ_1)\cdots(0 − λ_n) = (−1)^nλ_1 \cdots λ_n$$. Secondly, $$p(0) = |0I − A| = | − A| = (−1)^n |A|$$. Therefore $$c_0 = (−1)^nλ_1 \cdots λ_n = (−1)^n |A|$$, and so $$λ_1 \cdots λ_n = |A|$$. That is, the product of the n eigenvalues of A is the determinant of A. • This question has been answered for a year now. It is fine to add an answer after a long time, but don't you think it should be different from the other answers? In particular, the accepted answer explains the same argument, and it has a math formatting. – zarathustra Nov 5 '14 at 18:41 • @zarathustra your and others' complaint was only partly valid as Joey had definitely given more algebraic details as to how to prove the theorem; whereas the accepted answer didn't use the actual$p(\lambda)\$ for any deduction and his statement of expanding the determinant along diagonal was also questioned by commenters. This answer actually at least serves me well and had I read this one first I'd have comfortably disregarded the accepted answer but not other way round. – stucash Feb 25 at 0:16
• @zarathustra I don't think other answer use characteristic polynomial as like this answer do. Perhaps I found logic in it :) – emonhossain May 18 at 12:07

protected by user99914 May 21 '18 at 8:55

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