# Is there a eigenvalue equal to 0 if determinant is equal to 0?

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?

• 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. Apr 18, 2019 at 19:26
• The product of the n eigenvalues of A is the same as the determinant of A, where A is n x n matrix.
– dzi
Apr 18, 2019 at 19:30
• @dzi for general matrices? Or over the reals? Over complex numbers? Apr 18, 2019 at 19:31
• 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. Apr 18, 2019 at 19:33
• @ClementC. Pardon me for not mentioning that as I didn't realise if it made a difference. The question is for general matrices.
– dzi
Apr 18, 2019 at 19:33

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 $$0$$.

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 $$\det(A)=0$$.

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 $$0$$.

• 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) Apr 18, 2019 at 19:27
• Complex eigenvalues always exist, and their product is the determinant.
– kccu
Apr 18, 2019 at 19:33
• and all matrices are over real or complex numbers, right? Apr 18, 2019 at 19:35
• @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 $M$.
– kccu
Apr 18, 2019 at 19:36
• 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". Apr 18, 2019 at 23:57

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.