# Determinant of a matrix corresponding to a linear mapping $T:V \to V$ is positive

Letting $V$ be a general $n$-dimensional vector space such that $T:V\to V$. There will always be a matrix $A$ corresponding to $T$ such that $T(x) = Ax$ for all $x \in V$. Of course the matrix $A$ has different components depending on the basis we use to describe it.

It is true that $\det(A)$ is independent of the basis used to describe $A$, but I am interested in how we can describe $\det(T)$ $without$ resorting to a specific basis. Is there a general definition of the determinant?

Context: I've got this problem; given the quaternions $\ \mathbb{H} = \{ a + b i + cj + dk | a,b,c,d\in\mathbb{R} \} \approx \mathbb{R}^{4}$. Fixing any $q, p \in \mathbb{H}$ such that $qq^{\star} = p p^{\star} = 1$ and defining the linear map $M : \mathbb{R}^{4} \to \mathbb{R}^{4}$ as $$M(x) = q x p^{\star} \ \ \ \ \mathrm{for\ all\ }x \in \mathbb{H} \approx \mathbb{R}^{4}$$

I already know that $M$ is an orthogonal map, in the sense that $(Mx)\cdot (My) = x \cdot y$. I'd like to show that $\det(M) > 0$ (so that $M$ is in the special orthogonal group). I really don't want to write out the whole matrix corresponding to $M$; so what properties of the determinant are there which can help here?

• multiplication on the left by fixed $q$ preserves orientation, so positive determinant. This can be argued by taking a curve among unit quaternions from $1$ to $q,$ as determinant is continuous. Parameter $t,$ should be $1 \cos t + q \sin t$ or similar. Actually, assuming $q \neq \pm 1,$ we should take $v$ as the normalized (to norm $1$) vector (nonreal) part of $q,$ then use $1 \cos t + v \sin t.$ Along the way we do get $q$ itself. – Will Jagy Mar 24 '17 at 18:00
• What does a change of basis do to the determinant of a matrix? – amd Mar 24 '17 at 18:02

$$\text{det}(T) = \Pi_i \lambda_i = \lambda_1 \lambda_2 \cdots \lambda_n$$
where $\lambda_i$ are the eigenvalues of $T$.
See Down with Determinants! The definition: $$\det = \text{product of eigenvalues of the operator}.$$ Obviously, the definition of eigenvalue is purely geometric and does not depend of determinat, but how to define the (algebraic) multiplicity of an eigenvalue?
A vector $v\in V$ is called a generalized eigenvector of $T$ if $$(T − \lambda)^kv = 0$$ for some eigenvalue $\lambda$ of $T$ and some positive integer $k$.
The multiplicity of an eigenvalue $\lambda$ of $T$ is defined to be the dimension of the set of generalized eigenvectors of $T$ corresponding to $\lambda$. We see immediately that the sum of the multiplicities of all eigenvalues of $T$ equals $n$, the dimension of $V$ (from Theorem 3.11(a)). Note that the definition of multiplicity given here has a clear connection with the geometric behavior of $T$, whereas the usual definition (as the multiplicity of a root of the polynomial $\det(zI − T))$ describes an object without obvious meaning.