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I've just realized that I'm not sure about the answer to this. Are determinants always real-valued? Determinants can be calculated as the product of eigenvalues. Eigenvalues can be complex-valued. Thus surely determinants can be complex-valued? Yet, I've never calculated a complex-valued determinant. So have I just not encountered one yet, or are determinants always real for some reason?

Note: I'm inclined to think they must be real because you can say that a determinant is just a generalized volume of a parallelotope made by the column vectors as the sides. But then again, how do you construct a parallelotope out of complex vectors?

user195150
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    If a matrix has real entries, its determinant is real valued although eigenvalues may be complex. In this case the nonreal eigenvalues come as pairs of complex conjugates. If the matrix has complex entries, then the determinant can well be complex. There is an explicit expression for the determinant in terms of the entries. – Joonas Ilmavirta Nov 24 '14 at 17:03
  • It's simple to just try a counter-example. What is the determinant of [[1,0],[0,i]]? If the matrix has real entries, then the determinant, consisting only of products and sums of those entries, cannot be non-real, as the field of real numbers is closed for addition and multiplication (i.e. you cannot get a non-real number as the result of the sum or product of two reals). – andrepd Nov 24 '14 at 23:34
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    What is the determinant of [i]? – user541686 Nov 25 '14 at 07:59

4 Answers4

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A determinant is always a member of the field (or ring) that the matrix entries comes from -- for any given size of the matrix the determinant is a particular polynomial in the entries. Thus, if the matrix entries are all real, then so is the determinant.

But with, say, complex entries in the matrix it is easy to find a matrix with a non-real determinant:

$$\left|\begin{matrix}i & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right|=i $$

By the way, it doesn't really work well to define the determinant as the product of the eigenvalues, because some matrices have fewer different eigenvalues than the size of the matrix, and then you need to include some eigenvalues in the product several times, according to their (algebraic) multiplicity. But the way to define the multiplicity of eigenvalues is through the characteristic polynomial which in itself is defined using determinants!

hmakholm left over Monica
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    Does the determinant of a complex-valued matrix have a geometric interpretation like a real-values one does (as volume)? – user195150 Nov 24 '14 at 17:08
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    @user195150: I don't know of any. – hmakholm left over Monica Nov 24 '14 at 17:12
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    It does! The norm of the determinant is how much it expands volume, and the phase of the determinant is how much it rotates $\mathbb{R}^n$ inside $\mathbb{C}^n$. What does that mean? It's a bit much for a comment, but if you consider the set of all totally real (i.e. containing no complex linear subspace) $n$-dimensional real subspaces of $\mathbb{C}^n$, there's precisely one phase parameter describing these, and it's this one. – aes Nov 25 '14 at 00:21
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Given a real $n\times n$ matrix $A$, its determinant is equal to the product of its eigenvalues. As you point out, even though $A$ is real, its eigenvalues may not be. However, $\det A$, the product of these eigenvalues, will be real. To see this, note that the eigenvalues of $A$ are precisely the zeroes of $p(\lambda) = \det(\lambda I - A)$ which is a polynomial with real coefficients (the coefficients are sums and products of entries of $A$). Therefore, if $\lambda$ is a zero of $p$ of algebraic multiplicity $k$, then so is $\bar{\lambda}$. As $\lambda\bar{\lambda} = |\lambda|^2 \in \mathbb{R}$, the determinant will always be real.

Michael Albanese
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By its definition (sum of products), the determinant of a real-valued matrix is a real number.

If you diagonalize the matrix, complex Eigen values will come in conjugate pairs.

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The determinant of an $n \times n$ matrix with real entries is always real when you use the cofactor expansion, you are essentially adding and multiplying real numbers, and since $\mathbb{R}$ is a field, $\mathbb{R}$ is closed under addition and multiplication. However, the eigenvalues of a matrix with real entries may not all necessarily be real. This depends on the characteristic polynomial, in which you take $\det(A-\lambda I)$, and check its roots. Verify that the matrix $$\begin{bmatrix} 3&&2 \\ -5 &&-3\\ \end{bmatrix}$$ has a real determinant but imaginary eigenvalues.

However, the determinant of an $n \times n$ matrix with complex entries may not be real. But since $\mathbb{C}$ is a field, the determinant will always be complex.

Vivek Kaushik
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