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.

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    $\begingroup$ 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 $\det$. $\endgroup$ – user1537366 Jan 22 '15 at 5:45
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    $\begingroup$ @user1537366 are you saying that this is not necessarily true in cases where the eigenvalues have multiplicity > 1? $\endgroup$ – Hunle Nov 1 '17 at 4:29
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    $\begingroup$ @user1537366, is it the product of all eigenvalues, or only a product of the set of distinct eigenvalues? thanks you. $\endgroup$ – Hunle Aug 26 '18 at 22:47
  • $\begingroup$ 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. $\endgroup$ – inavda Mar 23 at 20:40

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.

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    $\begingroup$ Interesting, but don't you also have to show that the leading coefficient of the polynomial is 1? Or is that obvious? $\endgroup$ – DanielV Sep 28 '13 at 9:05
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    $\begingroup$ $\det(A-\lambda I)=(\lambda_1 - \lambda)^{m_1}(\lambda_2 - \lambda)^{m_2}\cdots (\lambda_n - \lambda)^{m_n}.$ so $\det(A) = \lambda_1^{m_1} \lambda_2^{m_2}\cdots\lambda_n^{m_n}.$ with $m_i$ is the multiplicity of $\lambda_i$ $\endgroup$ – Mohamez Jan 14 '14 at 7:39
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    $\begingroup$ @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. $\endgroup$ – Jason DeVito Jan 29 '18 at 20:18
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    $\begingroup$ 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. $\endgroup$ – philmcole Apr 10 '18 at 12:29
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    $\begingroup$ @philmcole would you please explain why the coefficient of $\lambda^n$ is $(-1)^n$? I did not get it... $\endgroup$ – 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, $\det(A) = 0$


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.

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    $\begingroup$ 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 $\endgroup$ – Tyg13 Apr 1 '17 at 23:57
  • $\begingroup$ 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? $\endgroup$ – 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.

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    $\begingroup$ Why not? Any matrix has some "eigendecomposition": Schur, Jordan,... $\endgroup$ – Algebraic Pavel Sep 29 '13 at 11:20
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    $\begingroup$ 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? $\endgroup$ – jacob Mar 8 '14 at 7:11
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    $\begingroup$ @jacob Having linearly independent columns does not imply diagonalisable... $\endgroup$ – user1537366 Jan 22 '15 at 5:47
  • $\begingroup$ His Idea is good but needs more arguments the D will be a Jordan Block matrix. Such matrices are almost diagonal. $\endgroup$ – 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$

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    $\begingroup$ Is this shown by saying $det(S) = 1/det(S^-1)$? $\endgroup$ – Hunle Nov 1 '17 at 4:27
  • $\begingroup$ 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 $det(S) det(S^{ -1})= det(Id) = 1$. $\endgroup$ – eduard Nov 11 '17 at 14:06
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    $\begingroup$ but this decomposition only exists, if A is diagonalizable $\endgroup$ – user519338 Apr 1 '18 at 21:01
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    $\begingroup$ 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. $\endgroup$ – 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 = ( a11 . . . a1n

       .           .

       an1 . . . ann)

Let the n eigenvalues of A be λ1, . . . , λn. Finally, denote the characteristic polynomial of A by p(λ) = |λI − A| = λn + cn−1λn−1 + · · · + c1λ + c0.

Note that since the eigenvalues of A are the zeros of p(λ), this implies that p(λ) can be factorised as p(λ) = (λ − λ1). . .(λ − λn). Consider the constant term of p(λ), c0. The constant term of p(λ) is given by p(0), which can be calculated in two ways:

Firstly, p(0) = (0 − λ1). . .(0 − λn) = (−1)nλ1 . . . λn.

Secondly, p(0) = |0I − A| = | − A| = (−1)n |A|.

Therefore c0 = (−1)nλ1 . . . λn = (−1)n |A|, and so λ1 . . . λn = |A|.

That is, the product of the n eigenvalues of A is the determinant of A.

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    $\begingroup$ 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. $\endgroup$ – zarathustra Nov 5 '14 at 18:41
  • $\begingroup$ @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. $\endgroup$ – stucash Feb 25 at 0:16

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