According to theorem the multiplication of all eigenvalues is equal to the determinant, so if one of them equals 0 the determinant is always 0. But is it true for the opposite statement? If determinant is equal to 0 is there for sure an eigenvalue equal to 0?
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$\begingroup$ Can you post the theorem that you are talking about? I would assume that a proof by contradiction would be sufficient, but I need to see what we're working with. $\endgroup$– JacobCheverieApr 18, 2019 at 19:26
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$\begingroup$ The product of the n eigenvalues of A is the same as the determinant of A, where A is n x n matrix. $\endgroup$– dziApr 18, 2019 at 19:30
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$\begingroup$ @dzi for general matrices? Or over the reals? Over complex numbers? $\endgroup$– Clement C.Apr 18, 2019 at 19:31
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$\begingroup$ The theorem says that the two quantities are the same, therefore they are equal. Thus, if $det(A) = \lambda_1 \cdot \lambda_2 \cdots \lambda_n = 0$, then at least one $\lambda_i = 0$. I would assume that this theorem assumes the existence of eigenvalues. $\endgroup$– JacobCheverieApr 18, 2019 at 19:33
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$\begingroup$ @ClementC. Pardon me for not mentioning that as I didn't realise if it made a difference. The question is for general matrices. $\endgroup$– dziApr 18, 2019 at 19:33
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4 Answers
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If a square matrix $A$ has zero determinant, this implies that $A$ is not injective, i.e. the kernel is nonempty. So, there exists $v \neq 0$ such that $Av = 0 = 0\cdot v$. By definition, $v$ is an eigenvector for $A$ corresponding to eigenvalue $0$.
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Maybe you've heard of the characteristic polynomial. For a size$~n$ square matrix $A$ it is defined as $\det(XI_n-A)$, it is a monic polynomial of degree$~n$, and it has the property that $\lambda$ is an eigenvalue if and only if $\lambda$ is a root of the characteristic polynomial, in other words if $\det(\lambda I_n-A)=0$. Setting $\lambda=0$ in this statement, it says that $0$ is an eigenvalue if and only if $\det(-A)=0$. It is not very hard to see that this is equivalent to $\det(A)=0$.
answered Apr 19, 2019 at 10:25
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Yes, the determinant of a matrix with real/complex entries is the product of its (complex) eigenvalues, so it has a matrix has a $0$ eigenvalue if and only if its determinant is $0$.
If you care about matrices with entries in a general field $F$, then as Clement points out the determinant will be the product of the eigenvalues which lie in the algebraic closure of $F$, and so once again a matrix has a $0$ eigenvalue if and only if its determinant is $0$.
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$\begingroup$ What if the matrix is not diagonalizable? (Or more general statements about eigenvalues not existing) I.e., what about the general question from the OP: "can the determinant of a matrix be zero if 0 is not an eigenvalue" (without further assumptions) $\endgroup$ Apr 18, 2019 at 19:27
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$\begingroup$ Complex eigenvalues always exist, and their product is the determinant. $\endgroup$– kccuApr 18, 2019 at 19:33
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$\begingroup$ and all matrices are over real or complex numbers, right? $\endgroup$ Apr 18, 2019 at 19:35
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$\begingroup$ @ClementC. It doesn't matter whether the matrix is over the real or complex numbers, in either case the determinant of $M$ is the product of all the complex eigenvalues of $M$. $\endgroup$– kccuApr 18, 2019 at 19:36
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2$\begingroup$ This is what I like about James Yang's answer, it lets you bypass all these annoying details about the statement "the determinant is the product of eigenvalues". $\endgroup$ Apr 18, 2019 at 23:57
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Fun, but overkill solution:
Zero determinant implies one of the singular values must be zero. By Weyl's Inequality, one of the eigenvalues must be zero.
answered Apr 22, 2019 at 6:13