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For people on this board I have a probably pretty modest question, but since I'm not a mathematician (just an economist), I'm having trouble. The full pdf can be found here: http://www.pnas.org/content/early/2012/05/16/1206569109.full.pdf+html

The question is regarding the following passage and has to do with linear algebra. They write: "where Adj(M′) is the adjugate matrix (also known as the classical adjoint or, as in high-school algebra, the “matrix of minors”). Eq. 2 implies that every row of Adj(M′) is proportional to v. Choosing the fourth row, we see that the components of v are (up to a sign) the determinants of the 3 × 3 matrices formed from the first three columns of M′, leaving out each one of the four rows in turn. These determinants are unchanged if we add the first column of M′ into the second and third columns. The result of these manipulations is a formula for the dot product of an arbitrary four-vector f with the stationary vector v of the Markov matrix, v · f ≡ D(p; q; f), where D is the 4 × 4 determinant shown explicitly in Fig. 2B. This result follows from expanding the determinant by minors on its fourth column and noting that the 3 × 3 determinants multiplying each fi are just the ones described above."

To understand the full context you will probably have to read the beginning of the passage, which is also very short. Yet my question is specifically regarding the formulated relationship between the stationary vector v of the Markov transition-matrix M and the Adj(M'), which is Adj.(M-I). As they say: Every row of Adj.(M') is proportional to v, which is sort of intuitive looking at Eq. 2, but I simply do not understand how they got that. Also the immediately following conclusion that the elements of v are the 3x3 column determinants of M' if you were to eliminate from the bottom of the fourth column.

Also to point out a petty mistake but v · f ≡ D(p; q; f) can't be correct as the dimensions do not link up correctly. v' · f ≡ D(p; q; f) is correct. But yet again I grasp that this formulation makes sense, but fail to understand how this can be arrived at.

If you can point me in the direction of a book or can flat-out explain this to me, I would be very obliged.

Thanks in advance o_s

2 Answers 2

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A very interesting paper!

On every row of $\operatorname{Adj}(\mathbf M')$ being proportional to $\mathbf v$: This isn't fully justified in the paper. In speaking of "the stationary vector $\mathbf v$ of the Markov matrix", they're implicitly using the fact that the matrix has a unique stationary vector. That's not trivial; an irreducible Markov chain has a unique stationary distribution if and only if all its states are positive recurrent. (For that statement and an explanation of the terminology it uses see Wikipedia.) If it does have a unique stationary vector (i.e. a unique left eigenvector with eigenvalue $1$), then all vectors with $\mathbf x^\top M=\mathbf x^\top$, or equivalently $\mathbf x^\top(\mathbf M-\mathbf I)=0$, must be proportional to that vector, and equation [2] says precisely that every row of $\operatorname{Adj}(\mathbf M')$ satisfies $\mathbf x^\top(\mathbf M-\mathbf I)=0$.

On the components of $\mathbf v$ being given by these $3\times3$ determinants: This is a bit imprecise; $\mathbf v$ was introduced as the stationary vector of the Markov matrix, which should be normalized to sum to $1$; equation [2] only says that $\mathbf v$ is proportional to the vector formed from these determinants. (This is because the fourth row of $\operatorname{Adj}(\mathbf M')$ is defined by these determinants, and as discussed above each row of $\operatorname{Adj}(\mathbf M')$ is proportional to $\mathbf v$.) This imprecision is corrected on the next page, where the payoffs are noramlized using $\mathbf v\cdot\mathbf 1$.

On replacing $\mathbf v\cdot\mathbf f$ by $\mathbf v'\cdot\mathbf f$: I'm not sure what you mean by $\mathbf v'$, but if you mean $\mathbf v^\top$, the transpose of $\mathbf v$, note that the dot notation $\mathbf v\cdot\mathbf f$ for the dot product of two column vectors is quite usual; to write it in matrix notation as you seem to have in mind, you should simply write $\mathbf v^\top\mathbf f$ without a dot, since it's unusual to use dots for matrix multiplication. You can see that the dot product is given by this determinant by applying the Laplace expansion to the last column (the one containing $\mathbf f$).

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    Thank you for your detailed answer, but due to my lackluster math education I will need to do some further reading. On the last paragraph: I did not know that this notation usual and i did mean the transpose v. That the Laplace expansion on the fourth row was relevant, I was able to grasp and also that this syncs up with the determinants.2012-08-18
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    @option_select: OK, feel free to ask again if you have further questions.2012-08-18
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    Ok one more thing on your first paragraph. You state: "and equation \[2\] says precisely that every row of $\operatorname{Adj}(\mathbf M')$ satisfies $\mathbf x^\top(\mathbf M-\mathbf I)=0$." Which kind of makes intuitive sense as they both set $\mathbf M'$ to zero. But is there a higher order reason (textbook explanation) why this relationship even exists?2012-08-18
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    @option_select: The reason for $\operatorname{Adj}(\mathbf M')\mathbf M'=\det(\mathbf M')\mathbf I$ quite generally is that you can view each product of a row of $\operatorname{Adj}(\mathbf M')$ with a column of $\mathbf M'$ as the Laplace expansion of a determinant, namely the determinant of $\mathbf M'$ itself if the row and column indices coincide (i.e. for the diagonal elements of the product), and the determinant of a matrix with two equal columns (which is zero) if they don't (i.e. for the off-diagonal elements of the product).2012-08-18
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    @option_select: In the special case where $\det(\mathbf M')=0$, the diagonal elements are also zero, so all the products are zero, so $\mathbf v$ is a left eigenvector of $\mathbf M'$ with eigenvalue $0$, and thus a left eigenvector of $\mathbf M$ with eigenvalue $1$.2012-08-18
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    Ok so i spent some time with computation and one thing is actually crucial. The paper mentions that the relationship holds up to a sign (which is rather ominous). Yet the relationship $\operatorname{Adj}(\mathbf M')\mathbf M' = \mathbf 0 = \operatorname{Det}(\mathbf M')$ cannot be simply said to be correct. I will submit an example I undertook using mathematica:2012-08-18
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    $b=\left( \begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array} \right)$2012-08-18
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    So that the adjungate is: $\left( \begin{array}{ccc} -3 & -6 & -3 \\ -6 & -12 & -6 \\ -3 & -6 & -3 \end{array} \right)$ , which implies the following dot-product: $\text{Dot}\left[\left( \begin{array}{ccc} -3 & -6 & -3 \\ -6 & -12 & -6 \\ -3 & -6 & -3 \end{array} \right).b\right]=\left( \begin{array}{ccc} -48 & -60 & -72 \\ -96 & -120 & -144 \\ -48 & -60 & -72 \end{array} \right)$2012-08-18
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    Obviously not the picture one would hope for. And here is the actual desired result: $\text{Dot}\left[\left( \begin{array}{ccc} -3 & 6 & -3 \\ 6 & -12 & 6 \\ -3 & 6 & -3 \end{array} \right).b\right]=\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right)$ So the crucial point is that one needs to leave the sign permutation of the adjungate formula out. So the statement of $\operatorname{Adj}(\mathbf M') \mathbf M'= 0$ is false.2012-08-18
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    Also I have checked the validity of your last post and can say that to be correct. $ \mathbf v$ is a left eigenvector of $\mathbf M'$ with eigenvalue 0 and one can identify it from the non-permutated adjugate.2012-08-18
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    Here my results from my example in mathematica: $\text{Dot}\left[\left( \begin{array}{ccc} -3 & 6 & -3 \end{array} \right).\left( \begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array} \right)\right]=\left( \begin{array}{ccc} 0 & 0 & 0 \end{array} \right)$ And therefore: $\text{Dot}\left[\left( \begin{array}{ccc} -3 & 6 & -3 \end{array} \right).\left( \begin{array}{ccc} 2 & 2 & 3 \\ 4 & 6 & 6 \\ 7 & 8 & 10 \end{array} \right)\right]=\left( \begin{array}{ccc} -3 & 6 & -3 \end{array} \right)$2012-08-18
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    @option_select: I'm not sure what you mean by "leave the sign permutation of the adjugate formula out". The adjugate of your matrix is $$ \pmatrix{ -3 & 6 & -3 \\ 6 & -12 & 6 \\ -3 & 6 & -3 } $$ *including* the correct signs, so as far as I can tell it's not that one has to leave the signs out but that you left them out at first and then included them to get the correct result.2012-08-18
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    In this example $\mathbf b$ is supposed to be $\mathbf M'$. The example is valid up to the point that $\left( \begin{array}{ccc} 2 & 2 & 3 \\ 4 & 6 & 6 \\ 7 & 8 & 10 \end{array} \right)$ is obviously not a Markov matrix.2012-08-18
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    You are correct. I looked up the specification in mathematica and it does not return the adjugate but rather just the matrix of minors without the sign. My bad...2012-08-18
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    Ok, barring my misunderstanding of what mathematica does, I think i now get how you can extrapolate $\mathbf v$ from the adjugate of $\mathbf M'$. To set the elements of $\mathbf f$ to one is then the norming of the vector to be a stationary vector. I thank you both tremendously for your help.2012-08-18
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    @option_select: You're welcome!2012-08-18
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Here $M$ is a matrix with $1$ as an eigenvalue, $M'=M-I$ and $v$ is a non-zero vector such that $M'v=0$. We have $\operatorname{Adj}(M') M' = 0 = M' \operatorname{Adj}(M')$. If $v_1, \ldots $ are the columns of $M'$ then $M' \operatorname{Adj}(M')$ is the matrix whose columns are $M'v_1, M'v_2,\ldots$, hence all these are zero, i.e. $Mv_i = v_i$ for all $i$.

If the space of 1-eigenvectors of $M$ is one-dimensional, it follows each $v_i$ must be a scalar multiple of $v$. Presumably this eigenspace condition somehow follows from the set-up in this paper. Otherwise the statement is false.

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    Thank you for your answer. Yet I will need some time to understand the statements made.2012-08-18