Why is the matrix $(I - A)$ theoretically singular?

Question

I've the following Matlab code to compute the eigenvector using the inverse iteration (or power) method:

A = p * G * D + delta;
x = (I − A) \ e;
x = x / sum(x);

taken from the 4th page of this chapter about the pagerank algorithm by Cleve Moler.

First, I don't see how this code relates to the "typical" inverse iteration method. If you know how, I would appreciate an explanation.

But my main question in this post is: why $(I - A)$ is theoretically singular?

A should be sparse. Would this help?

The author says:

Because I − A is theoretically singular, with exact computation some diagonal element of the upper triangular factor of $I − A$ should be zero and this computation should fail.

Why some of the diagonals should be $0$?

The author seems to be using the structure of the matrix $A$ in some way... can you say what we know about $A$? — Ben Grossmann, Oct 06 '16 at 11:53

score 2 · Answer 1 · answered Oct 06 '16 at 11:54

First, a square matrix $A$ is singular iff its transpose is singular.

Now, as stated on page 2, $A$ is specifically set up so that all the columns sum to one. As a result, $\lambda=1$ is an eigenvalue of $A^T$ with any constant eigenvector, say $\vec{v}=\langle 1,\ldots,1\rangle$. Thus, the same vector $\vec{v}$ maps to the zero vector under $I-A^T$, since $$(I-A^T)\vec{v} = I\vec{v}-A^T\vec{v} = \vec{v}-\vec{v}=\vec{0}.$$ Thus, $A^T$ is singular.

More generally, the pagerank matrix is an example of a stochastic matrix.

Why is the matrix $(I - A)$ theoretically singular?

1 Answers1