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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?

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    $\begingroup$ 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. $\endgroup$ – Will Jagy Mar 24 '17 at 18:00
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    $\begingroup$ What does a change of basis do to the determinant of a matrix? $\endgroup$ – amd Mar 24 '17 at 18:02
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One way to define a determinant of a linear transformation that is independent of a basis is by

$$\text{det}(T) = \Pi_i \lambda_i = \lambda_1 \lambda_2 \cdots \lambda_n$$

where $\lambda_i$ are the eigenvalues of $T$.

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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.

Bottom line: the algebraic multiplicity (dimension of the space of generalized eigenvectors associated to the eigenvalue) is so geometric as the geometric multiplicity (dimension of the space of eigenvectors associated the the eigenvalue).

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