# Definition of characteristic polynomial

The question is very simple but it's giving me a hard time. I've been given the following definition for the characteristic polynomial of a linear transformation $$p_c(x)=\det(xI-T)$$ But what does it mean to take the determinant of a linear transformation? I understand that you can find the characteristic polynomial by computing $\det(xI-A)$ where $A=[T]_B$ for $B$ an ordered basis of the space but that's taking the determinant of a matrix, which I understand.

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For finite-dimensional vector spaces, there is a one-to-one correspondence between linear transformations and matrices (up to a change of basis). – Phrohlych Aug 23 '14 at 23:48
@Phrohlych the one-to-one correspondence up to change of basis is not the issue here. Firstly, it is irrelevant that the correspondence is 1-1. Secondly, you need much more than a bijective correspondence. You need to know that the determinant is multiplicative and you need to know how change of base affects the matrix. – Ittay Weiss Aug 24 '14 at 0:18
Perhaps the title of this question should really be "Definition of determinant without a basis"? The motivating application is actually peripheral. – Ryan Reich Aug 24 '14 at 3:28

Conceptually, taking $\det(xI−T)$ means that, no matter what basis B you use to obtain the matrix $A=[T]_B$, the resulting characteristic polynomial will be the same. So $\det(xI−T)$ is defined with respect to a basis, perhaps, but is ultimately basis independent. This is done a lot when mathematical entities are defined; e.g., if $H$is a normal subgroup of $G$, we can define coset multiplication in the factor group using coset representatives, then show that the product coset is the same no matter what representatives we choose. Representation theory is BIG.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

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For the purposes of computing the determinant it suffices to represent the matrix according to any basis you like since it can be shown the result is independent of the chosen basis. However, this does not explain what the determinant of a linear transformation "really" is.

For a linear transformation $T:V\to V$, where $V$ is a vector space in which it makes sense to speak of the determinant of $n$ vectors (e.g., by means of taking the determinant of the corresponding matrix, and specifically if $V=\mathbb R^n$ then this is related to volume) you can construct the multi-linear function $F:V\times V\cdots \times V \to \mathbb R$ by $F(v_1,\ldots ,v_n)={\rm det}(Tv_1,\ldots, Tv_n)$. It can then be shown that no matter which $n$ vectors you choose you will get that $F(v_1,\ldots, v_n)=\alpha \cdot {\rm det}(v_1,\ldots, v_n)$. This $\alpha$ is the determinant of the transformation T.

In particular, if you choose $v_1,\ldots, v_n$ to be the standard basis in $\mathbb R^n$, then the determinant of $T$ is the determinant of the vectors $Tv_1,\ldots, Tv_n$. Thinking of determinants as expressing volume we see that the determinant of a transformation is the factor by which the transformation distorts the volume of standard cube. By the above, its also the factor by which it distorts any other parallelpipe determined by any $n$ vectors. This is geometrically what the determinant of a transformation is.

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Given an ordered basis $\beta= (b_1,...,b_n)$ you can find a corresponding matrix $T_\beta$, and we define the characteristic polynomial as $x \mapsto \det (xI-T_\beta)$.

To see that it is well defined, suppose $\beta'= (b_1',...,b_n')$ is another basis, then for some matrix $A$, we have $T_{\beta'} = A^{-1} T_\beta A$, and then \begin{eqnarray} \det (xI-T_{\beta'}) &=& \det (xI-A^{-1}T_{\beta}A) \\ &=& \det (xA^{-1}A-A^{-1}T_{\beta}A) \\ &=& {1 \over \det A} \det (xI-T_\beta) \det A \\ &=&\det (xI-T_\beta) \end{eqnarray}

Hence the characteristic polynomial is independent of the choice of basis.

Another way of viewing it is to notice that if we choose a basis $\beta_J$ so that $T_{\beta_J}$ is the Jordan normal form, then $\det(xI - T_{\beta_J}) = (x-\lambda_1)\cdots (x -\lambda_n)$, where $\lambda_k$ are the eigenvalues of $T$. This is, perhaps, a more intrinsic viewpoint?

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Yes, I do understand how it works once I take a basis. But the question is if it makes sense to talk about the determinant on a linear transformation – Shomar Aug 24 '14 at 0:06
@Shomar: Well, it is a little unsatisfactory, but defining it via a matrix is how its done. – copper.hat Aug 24 '14 at 1:41

It seems to me the problem is that you think the determinant is a property of a matrix, and a matrix alone, but it's commonly taken that the determinant is a property of a linear operator, and as such, the computation of the determinant using a matrix is just one way to figure out the value of this property.

You can define the determinant of a linear operator without reference to some matrix representation. A common way to do this is with exterior algebra and wedge products: the wedge product (denoted $\wedge$) produces elements of the exterior algebra. If $k$ vectors are wedged together, then the object is an element of the grade-$k$ exterior algebra on $V$, denoted $\bigwedge^k V$.

Now, define the action of a linear operator on an element of $\bigwedge^2 V$ like so: if $B = a \wedge b$, then

$$T(B) = T(a \wedge b) \equiv T(a) \wedge T(b)$$

and do so recursively for any $k > 2$ to generalize this to $\bigwedge^k V$.

The vector space $\bigwedge^n V$, where $n = \dim V$, is "one-dimensional", in the sense that all the elements are scalar multiples of each other. Hence, if $I \in \bigwedge^n V$, then it must be true that

$$T(I) = \alpha I$$

for some scalar $\alpha$. $\alpha$ obeys all the usual properties of the determinant of the matrix representation for $T$, so we can say it is the determinant.

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Determinants of linear maps in real vector spaces can be understood geometrically. For example, if $u, v, w$ are three linearly independent vectors belonging to a three-dimensional vector space $\mathcal{V}$ associated with Euclidean space, then the determinant of the linear map $S$ $$\det(S) = \frac{Su.(Sv \times Sw)}{u. (v \times w)}$$

which is a measure of the ratios of the volumes formed by the triad $u,v,w$ after and before the linear map $S$ acts on them.

Note that this definition is basis independent.

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I wonder if this is the same thing as Muphrid is saying, except using other words, and restricted to 3D Euclidean space, given the relationship of cross product to exterior product. – Evgeni Sergeev Aug 24 '14 at 9:06

Every linear transformation $\mathbb{R}^n \to \mathbb{R}^n$ can be represented by a square matrix. So the determinant can be calculated using this matrix in place of the linear transformation. Its just substitution really.

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I understand but the question is conceptually what does it mean to take $det(xI-T)$ – Shomar Aug 23 '14 at 23:49
just take it man, don't ask! – A B C Aug 23 '14 at 23:51
And I understand the problem. It seems to me that it is in the notation. They are using a convention that says "don't think of $\det{(xI-T)}$ where $T$ acts on $v$ by $T(v)$ instead think of $T(v)$ as being $Av$ and so $T$ can be thought of as $A$". Again, it is a notation convention. – Paul Sundheim Aug 23 '14 at 23:56
Conceptually, taking $\det(xI - T)$ means that, no matter what basis $B$ you use to obtain the matrix $A = [T]_B$, the resulting characteristic polynomial will be the same. So $\det(xI - T)$ is defined with respect to a basis, perhaps, but is ultimately basis independent. This is done a lot when mathematical entities are defined; e.g., if $H$ is a normal subgroup of $G$, we can define coset multiplication in the factor group using coset representatives, then show that the product coset is the same no matter what representatives we choose. Representation theory is BIG. – Robert Lewis Aug 24 '14 at 7:01
Every linear operator (i.e. from a vector space into itself) on a finite-dimensional vector space can be realized as a square matrix. Say you have a linear transformation: $$f: \mathbb{R}^n \to \mathbb{R}^n$$
Then, given an ordered basis $B = ( b_1, \ldots, b_n)$ define $T_f^B$ to be
$$T_f^B = \begin{bmatrix} f(b_1) &f(b_2)& \ldots& f(b_n) \end{bmatrix}$$
where $f(b_i)$ is the column vector of $b_i$ transformed by $f$. You may now use the matrix $T_f^B$ as the 'linear transformation', since in the context of this basis, they are indistinguishable.