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I define the function $D(\mathbf M)$ for square matrices by taking a maximal set of all triples $(\mathbf v_i,k_i,n_i)_{i \in I}$ that satisfy $(\mathbf{M}-k\mathbf{I})^n\mathbf{v}_i = \mathbf{0}$ with the $\mathbf{v}_i$ linearly independent, and letting $$D(\mathbf M) = \prod_{i \in I} k_i.$$

For example the matrix

$$\mathbf X=\begin{pmatrix}0&1&1\\1&0&1\\1&1&0\end{pmatrix}$$

has the set:

  • $((-1\quad 0\quad 1)^T,-1,1)$
  • $((-1\quad 1\quad 0)^T,-1,1)$
  • $((1\quad 1\quad 1)^T,2,1)$

so $D(\mathbf X) = -1 \cdot -1 \cdot 2 = 2$. It is equal to the determinant but I don't want to use that.


How can it be shown that $D(\mathbf A\mathbf B) = D(\mathbf A)D(\mathbf B)$?

  • 0
    So, you want to prove that the product of the eigenvalues is the determinant first?2010-12-30
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    You don't really take *all* pairs. It seems that you're assuming that $v_1,\ldots,v_n$ is a basis, which means that $M$ is diagonalizable.2010-12-30
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    You need the $v_i$ to be linearly independent, right? Are you sure the way you are defining things this is the determinant?2010-12-30
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    It's equal to the determinant only if the matrix is diagonalizable. In general you need to look at the Jordan form.2010-12-30
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    I checked http://en.wikipedia.org/wiki/Jordan_normal_form and the eigenvector/eigenvalue pair ([1;0;-1;1],4) must be counted twice, I am not sure how to fix my definition to account for that..2010-12-30
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    @J.M. I would like to prove it from scratch if possible, not using properties of determinants.2010-12-30
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    @quanta: that's perfectly alright. What you ought to be worrying about are matrices like $$\bigl(\begin{smallmatrix}1&1&4\\3&0&3\\2&5&-1\end{smallmatrix}\bigr)$$2010-12-30
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    J.M. thanks for the counterexample - It would work if we count the [-1;0;1] eigenvector twice, but I don't yet know the criteria for how many times to count an eigenvalue. I'll have to figure that out so I can fix the definition.2010-12-30
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    As Jonas and Yuval have pointed out, this only works if the matrix is diagonalizable. Perhaps an explicit example will help. How do you propose to handle the matrix $\mathbf M = \begin{bmatrix}0 & -2 \\ 2 & 0\end{bmatrix}$? There are no nonzero $(\mathbf v, k)$ such that $\mathbf{Mv} = k\mathbf v$, but the determinant is 4.2010-12-30
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    I get the eigenvectors/eigenvalue pairs ([i;1],-2i),([-i;1],2i) and the product gives 4.2010-12-30
  • 2
    First, change your definition from $\mathbf{Mv}_i = k\mathbf{v}_i$ to $(\mathbf{M}-k\mathbf{I})\mathbf{v}_i = \mathbf{0}$. Then consider $(\mathbf{v}_i,k,n)$, where $(\mathbf{M}-k\mathbf{I})^n\mathbf{v}_i = \mathbf{0}$ with $n$ minimal. If you are working over $\mathbb{C}$, this will help you get the necessary multiplicity.2010-12-30
  • 0
    Artuno Magidin, Great! Thank you!2010-12-31
  • 0
    @quanta: If you use the `@` symbol to begin the comment, the person you are addressing will be notified. See http://meta.stackexchange.com/questions/1093/make-recent-activity-and-responses-show-new-comments-on-questions-answers-i-have/35913#359132010-12-31
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    @quanta: You misunderstood how my suggestion will work. You don't take $k^n$; you just take $k$. But, for example, with $\left(\begin{array}{cc}2&1\\0&2\end{array}\right)$, you will have one vector with $k=2$ and $n=1$, and one vector with $k=2$ and $n=2$. By taking the two instances of $k=2$ you capture the necessary multiplicity.2010-12-31
  • 0
    Rather than "a set of all triples" as it says currently, you may want to say "a maximal set of triples". Also, your $n$ should be an $n_i$.2010-12-31
  • 0
    I've updated it now. I think maximal set just means that it contains everything it should (as opposed to subsets).2010-12-31

1 Answers 1

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If I understand you correctly, you want to prove that, for a square matrix with complex coefficients (or more generally, any algebraically closed field) the product of the eigenvalues (with appropriate multiplicity) is a multiplicative function on the matrices, but without using that the product of the eigenvalues is the determinant.

(It is pretty easy to show that the product of the Eigenvalues, with appropriate multiplicity, is the determinant, by using the Jordan canonical form. The determinant is invariant under conjugation, so the determinant of a matrix is the same as the determinant of its Jordan canonical form; this form is upper triangular, and the diagonal contains the eigenvalues, each with the appropriate algebraic multiplicity, yielding that the determinant is indeed the product of the eigenvalues; a Jordan canonical basis yields the triples you have. Once you have that the product is indeed the determinant, the conclusion follows).

There are a number of obstacles to the project. First, we need to show that $D(\mathbf{M})$ is well-defined; that is, that it does not depend on the choice of maximal family of vectors. This can be achieved pretty much the way one shows that the generalized eigenspaces yield a decomposition of the vector space (see any book that covers the Jordan canonical form), but the simplest proofs I know will use the Cayley-Hamilton Theorem at some point, which involves the determinant. (The more complicated ones consider the vector space as a module over $\mathbb{C}[x]$ and use the structure theorem for modules over PIDs... don't ask.)

But assuming this is taken for granted (that you can find a basis of generalized eigenvectors, so that your maximal family of triples will have $n$ elements, and that the number of times that each $k_i$ occurs is independent of the choice), then we can proceed along similar lines as the way we prove that the determinant is multiplicative.

Lemma. If there is vector $\mathbf{v}\neq 0$ such that $\mathbf{M}\mathbf{v}=\mathbf{0}$, then $D(\mathbf{M})=0$. Conversely, if $D(\mathbf{M})=0$, then there exists a vector $\mathbf{v}\neq 0$ such that $\mathbf{M}\mathbf{v}=\mathbf{0}$.

Proof. Immediate. QED

Proposition. $D(\mathbf{M})=0$ if and only if $\mathbf{M}$ is not invertible.

Proof. $\mathbf{M}$ is invertible if and only if it is of full rank. By the Rank-Nullity Theorem, this occurs if and only if the nullity of $\mathbf{M}$ is zero, which happens if and only if the nullspace is trivial, if and only if for all $\mathbf{v}$, $\mathbf{M}\mathbf{v}=\mathbf{0}\Rightarrow \mathbf{v}=\mathbf{0}$. The previous lemma then completes the proof. QED

Proposition. If $D(\mathbf{B})=0$, then $D(\mathbf{AB}) = 0 = D(\mathbf{A})D(\mathbf{B})$.

Proof. Let $\mathbf{v}\neq\mathbf{0}$ be such that $\mathbf{B}\mathbf{v}=\mathbf{0}$. Then $(\mathbf{AB})\mathbf{v}=\mathbf{0}$, so by previous result, $D(\mathbf{AB})=0$. QED

Proposition. If $D(\mathbf{A}) = 0$, then $D(\mathbf{AB}) = 0 = D(\mathbf{A})D(\mathbf{B})$.

Proof. If $D(\mathbf{B})=0$, then the result follows by the previous proposition. If $D(\mathbf{B})\neq 0$, then $B$ is invertible (as above). Since $D(\mathbf{A})=0$, there exists $\mathbf{v}\neq \mathbf{0}$ such that $\mathbf{A}\mathbf{v}=\mathbf{0}$. Then $$(\mathbf{AB})(\mathbf{B}^{-1}\mathbf{v}) = \mathbf{A}\mathbf{v}=\mathbf{0} = 0(\mathbf{B}^{-1}\mathbf{v}),$$ and since $\mathbf{B}^{-1}\mathbf{v}\neq\mathbf{0}$, then $D(\mathbf{AB}) = 0$, as desired. QED

Lemma. If $\mathbf{B}$ is an elementary matrix, then

  1. If $\mathbf{B}$ is obtained from the identity by transposing two rows, then $D(\mathbf{B}) = -1$.
  2. If $\mathbf{B}$ is obtained from the identity by adding a multiple of one row to another row, then $D(\mathbf{B}) = 1$.
  3. If $\mathbf{B}$ is obtained from the identity by multiplyig a row by $\lambda\neq 0 $, then $D(\mathbf{B}) = \lambda$.

Proof.

  1. Suppose we exchange rows $i$ and $j$. That means that $\mathbf{B}(e_i) = e_j$ and $\mathbf{B}(e_j) = e_i$, while $\mathbf{B}(e_k) = e_k$ if $k\neq i,j$. We get $n-2$ triples with $(e_k,1,1)$, $k\neq i,j$, and also the triples $(e_i+e_j,1,1)$ and $(e_i-e_j,-1,1)$. Then $D(\mathbf{B}) = 1^{n-1}(-1) = -1$, as claimed.

  2. Suppose we add $r\neq 0$ times the $i$th row to the $j$th row. Then $\mathbf{B}(e_k) = e_k$ if $k\neq i$, and $\mathbf{B}(e_i) = e_i+re_j$. Take the $n-1$ triples $(e_k,1,1)$ with $k\neq i$, and the triple $(e_i,1,2)$. Note that this works, since $(\mathbf{B}-\mathbf{I})e_i = (e_i+re_j) - e_i = re_j$, and $(\mathbf{B}-\mathbf{I})(re_j) = re_j - re_j = \mathbf{0}$, so that $(\mathbf{B}-\mathbf{I})^2e_i = \mathbf{0}$, with $2$ minimal. Thus, $D(\mathbf{B}) = 1^n = 1$.

  3. If we multiply the $i$th row by $\lambda$, then the $n-1$ triples $(e_j,1,1)$ with $j\neq i$, together with $(e_i,\lambda,1)$, make a maximal family and we have $D(\mathbf{B}) = 1^{n-1}\lambda = \lambda$. QED

Lemma. If $\mathbf{A}$ and $\mathbf{B}$ are elementary matrices, then $D(\mathbf{AB}) = D(\mathbf{A})D(\mathbf{B})$.

Sketch. There are several cases to consider, depending on which kind of elementary matrices they are and how they interact. For example, if $A$ is obtained from the identity by adding $r$ times the $i$th row to the $j$th row, and $B$ is obtained by multiplying the $\ell$th row by $\lambda$, then we must consider:

  1. If $\ell\neq i,j$, then we can take the $n-3$ triples $(e_k,1,1)$ with $k\neq i,j,\ell$, the triple $(e_{\ell},\lambda,1)$, the triple $(e_j,1,1)$, and the triple $(e_i,1,2)$. Then $D(\mathbf{AB}) = 1^{n-1}\lambda = \lambda = D(\mathbf{A})D(\mathbf{B})$.

  2. If $\ell = j$, then $\mathbf{AB}(e_k) = e_k$ if $k\neq i,j$; $\mathbf{AB}e_j = \lambda e_j$, and $\mathbf{AB}(e_i) = e_i+re_j$, so we can take the $n-2$ triples $(e_k,1,1)$ with $k\neq i,j$, together with $(e_j,\lambda,1)$ and $(e_i,1,2)$. Then $D(\mathbf{AB}) = 1^{n-1}\lambda = \lambda = D(\mathbf{A})D(\mathbf{B})$.

  3. If $\ell=i$, then $\mathbf{AB}(e_k) = e_k$ if $k\neq i$, and $\mathbf{AB}(e_i) = \lambda e_i + \lambda re_j$. We can then take the $n-1$ triples $(e_k,1,1)$, $k\neq i$, and the triple $(e_i,\lambda,2)$, to get $D(\mathbf{AB})=1^{n-1}\lambda = \lambda = D(\mathbf{A})D(\mathbf{B})$.

Similar arguments hold for the other combinations of products of elementary matrices. QED

Corollary. If $E_1,\ldots,E_k$ are elementary $n\times n$ matrices with coefficients in $\mathbb{C}$, then $D(E_1\cdots E_k) = D(E_1)\cdots D(E_k)$.

Proof. Induction on $k$. QED

THEOREM. If $\mathbf{A}$ and $\mathbf{B}$ are $n\times n$ matrices with coefficients in $\mathbb{C}$, then $D(\mathbf{AB}) = D(\mathbf{A})D(\mathbf{B})$.

Proof. If either $D(\mathbf{A}) = 0$ or $D(\mathbf{B})=0$, then the result follows from previous proposition. If $D(\mathbf{A})\neq 0 \neq D(\mathbf{B})$, then each of $\mathbf{A}$ and $\mathbf{B}$ is invertible, hence a product of elementary matrices. The result now follows from the Corollary above. QED