# For even dimensional vector space, any linear map has to have a positive and a negative eigenvalue provided that its determinant is negative

In the book of Linear Algebra by Werner Greub, at page 123, it is given that

Let $\phi: E \to E$ and $\dim E=n$, where E is vector space of the real numbers. Then define $$f(\lambda) = \sum_{v=0}^n \alpha_v \lambda^{n-v} = \det(\phi - \lambda i),$$ i.e., $f(\lambda)$ represents the characteristic polynomial of $\phi$

[...]

If $n$ is odd, then $\det(\phi) > 0$, $\phi$ has at least one positive e eigenvalue, and if $\det(\phi) < 0$, $\phi$ has at least one negative eigenvalue provided that $n$ is odd.

[...]

If $n$ is even, and $\det(\phi) < 0$, there exists at least one positive and one positive eigenvalues of $\phi$.

However, I can't see why is that the case ? I mean as far as I know the constant term in the characteristic polynomial is the determinant of the $\phi$ but how is that related with the sign of the zero of that polynomial ? Moreover, it $n$ is even, why does the characteristic polynomial have to have 2 zeros, one positive and one negative ?

Edit:

The determinant of a linear map is the determinant of the representing matrix of that linear map.

Edit 2:

We are working on a real inner product space.

• I find your question difficult to read and asks the reader to assume too much information because it removes too many details from your original source. For example, it isn't stated how $\phi$ is related to $f$. Similarly, an $f$ is defined, but the question doesn't provide any information about what $\alpha$ is or why this is defined if nowhere in the question does it make use of an $f$. – JessicaK Dec 3 '17 at 12:49
• @JessicaK You are right. I have edited the question, see my edit. – onurcanbektas Dec 3 '17 at 12:51
• Okay, please post more background of this question, such as the base field, the objective to introduce the polynomial, and the definition of the determinant of linear mapping in your book. – xbh Dec 18 '17 at 9:09
• @user514490 That polynomial is the characteristic polynomial, i.e the solutions of that polynomial are the eigenvalues of $\phi$, and for the rest, see my edit. – onurcanbektas Dec 18 '17 at 10:04
• Something is missing in your question. It is certainly not true the determinant of an endomorphism of a real vector space of finite, odd dimension is positive, for example. In the first «[…]» you hid some hypothesis. – Mariano Suárez-Álvarez Jan 6 '18 at 8:44

## 1 Answer

This is easy. The constant term of the polynomial is the determinant, which is also the product of all eiegnvalues.

If $n$ is odd, and $\det (\varphi) >0$, then there is at least one positive eigenvalue. Otherwise all eigenvalues are negative, and the product of them must be negative since $n$ is odd. If $\det (\varphi) < 0$, then it also cannot have all positive eigenvalues.

If $n$ is even, an assume that there is some $\varphi$ whose eigenvalues are either all positive or all negative. Then $\det (\varphi) > 0$ as the product of all eigenvalues, contradicting $\det(\varphi) < 0$.

UPDATE: The reasoning above has some flaws. Assume that the determinant is defined by the sum of the alternating products of the entries in the matrix after fixing a basis in $E$, and the base field is $\mathbb R$ [the statement on $\mathbb C$-vector spaces clearly fails]. Then the characteristic polynomial with real coefficients always has a decomposition $$f(z) = (-1)^n\prod_1^r (z + a_j) \prod_1^s (z^2 +p_k z + q_k)[a_j, p_k, q_k \in \mathbb R, p_k^2 - 4q_k^2 < 0, r , s\in \Bbb N].$$

Since $p_k^2 - 4q_k < 0$, $q_k > 0$ for all $k$. Then the constant term of $f$ is $(-1)^{n}\prod_1^r a_j \prod_1^s q_k$.

If $n$ is odd, then the degree of $f$ is $n = r + 2s$, hence $r \geqslant 1$, which means $\varphi$ has at least one eigenvalue. If $\det (\varphi) \gtrless 0$, then$\prod a_j \lessgtr 0$, and by the argument before the UPDATE, there is some $a_j \lessgtr 0$ hence $-a_j \gtrless 0$ is an eigenvalue of $\varphi$.

If $n$ is even and $\det(\varphi) < 0$, then there exists at least one negative $a_j$. Also not all $a_j$ are negative, otherwise the product $\prod a_j \prod q_k > 0$, contradiction. Thus there is another $a_\ell > 0$, and $-a_j >0, -a_\ell < 0$ are two eigenvalues of $\varphi$.

UPDATE 2:

In the book, the author applies the intermediate value theorem of continuous function to the characteristic polynomial, which is a faster method. My "UPDATE" above acquires nothing about the existence of eigenvalues. It just uses the structure of real polynomials.

ADDENDUM

First we assume that the fundamental of algebra [FTA] is proved, i.e. Every $\mathbb C$-polynomial has at least one root in $\mathbb C$. Next if a polynomial $f (z)$ has a root $z_1 \in \mathbb C$, then there exists a polynomial $g (z) \colon f(z) = g(z) (z - z_1)$ and $\deg(f) = \deg (g) + 1$. Apply the FTA to $g(z)$ and we can find another factor $(z-z_2)$. By induction we could deduce that every $\mathbb C$-polynomial $f(z)$ has a decomposition $$f (z) = c \prod_1^n (z - z_j), c \in \mathbb C.$$

Now especially for a real polynomial $f(x)$, it is easy to see that if $z \in \mathbb C$ is a root of $f$ then so is $\overline z$ [complex conjugate of $z$]. Then suppose $r_j$ are real roots of $f$, then the remained roots in $\mathbb C$ appear in pairs, i.e. two roots that are conjugate. If $z, \overline z$ are both roots of $f$, then $f$ has factors $(x - z), (x - \overline z)$ in $\mathbb C$, which means the polynomial $(x-z)(x- \overline z) = x^2 - 2\mathfrak R\{ z\}+ |z|^2$ is a factor of $f$ [here $\mathfrak R \{z\} ^2 < |z|^2$]. Thus $f(x)$ has a decomposition $$f(x) =c \prod_1^s (x - r_j) \prod_1^t (x^2 - p_k x + q_k), r_j,p_k, q_k \in \mathbb R, p_k^2 < 4q_k.$$

• Why the determinant of $\phi$ is the product of the eigenvalues of $\phi$ ? How can we show it ? I mean in general $\phi$ does not have to have eigenvalues at all. – onurcanbektas Dec 18 '17 at 7:12
• @onurcanbektas Well, could you please show the definition of determinant in your book? It maybe different if the definition is distinct. – xbh Dec 18 '17 at 8:51
• Ok, I have checked the proof of that thing, but, here is the thing: We are working on $\mathbb{C}$, so you cannot directly talk about positive & negative eigenvalues. First you need to show that at least one real eigenvalue. – onurcanbektas Dec 18 '17 at 15:33
• @onurcanbektas Well the complex case clearly fails: consider a linear mapping on a 3-dimensional space with eigenvalues -1, i, i. Thus I assume that we are talking about real vector spaces. – xbh Dec 18 '17 at 16:00
• @onurcanbektas See my ADDENDUM. This is a brief explanation of the decomposition. Details could be found in any text related to polynomial algebra and abstract algebra. – xbh Jan 6 '18 at 10:58