0

Sorry to ask an elementary (even trivial) linear algebra question here but I have almost forgot every trick in linear algebra and I cannot work it out.

I know the identity matrix $I$ is positive definite.

Suppose we have a matrix $P$ whose entries are all in $[0,1]$. (Entries of $P$ are just the probabilities of something).

Now we have a $\lambda\in(-1,1)$.

Do we have $I-\lambda P$ is positive definite? (Actually invertibility is enough for my study)

Edit. What if we impose an additional condition that all row sums of $P$ are equal to $1$?

Thanks for help.

Rodrigo de Azevedo
  • 23,223
Sam Wong
  • 2,481
  • 1
    It is pos def for lambda sufficiently small (if $P$ is symmetric), see https://en.wikipedia.org/wiki/Diagonally_dominant_matrix for invertibility and https://en.wikipedia.org/wiki/Gershgorin_circle_theorem for control on eigenvalues – Calvin Khor Nov 23 '22 at 06:13
  • @CalvinKhor Hi Calvin, thanks for the reference. Unfortunately $P$ is not symmetric in my case. Do we have similar result/control for non-symmetric $P$? Invertibility results are enough. – Sam Wong Nov 23 '22 at 06:20
  • Please don't edit your question in a way that invalidates existing answers. – user1551 Nov 23 '22 at 06:37
  • @user1551 Alright. We can put the important extra conditions into the edit section but do not just delete it. – Sam Wong Nov 23 '22 at 06:42
  • 3
    none of the two results I said care if the matrix is pos def, but it is impossible for $I-\lambda P$ to be symmetric (a requirement for pos def) if $P$ is not symmetric – Calvin Khor Nov 23 '22 at 07:03
  • 1
    I would call it $\gamma$ instead, as $\lambda$ is kind of reserved for eigenvalues. Also, think of the symmetry (or lack thereof). – Rodrigo de Azevedo Nov 23 '22 at 10:24
  • @RodrigodeAzevedo Thanks for the advice. I was mainly working with probability stuff and didn't realize the notation may cause confusion. – Sam Wong Nov 23 '22 at 11:56

2 Answers2

5

The set of invertible matrices is open (preimage of $\Bbb R\backslash \{0\}$ by determinant function). Therefore, for small enough $\lambda$, $I-\lambda P$ is invertible.

If you add the extra condition that $P$ is stochastic, then $I-\lambda P$ is invertible for $\lambda\in(-1,1)$. That's because the largest eigenvalue (in absolute value) of a stochastic matrix is $1$. (see this).

And there is a little notational trap in your question: $I-\lambda P$ is invertible iff $\frac1\lambda$ is not an eigenvalue of $P$.

Therefore, if $|\lambda|<1$, $\frac1\lambda$ can't be an eigenvalue, and $I-\lambda P$ is invertible.

Jean-Claude Arbaut
  • 23,601
  • 7
  • 53
  • 88
  • You probably meant to say "it's not an eigenvalue of $A$" – lcv Nov 23 '22 at 09:26
  • 1
    @lcv Right, thanks! ${}{}{}$ – Jean-Claude Arbaut Nov 23 '22 at 09:30
  • Thanks Jean. This is exactly what I am looking for. – Sam Wong Nov 23 '22 at 11:57
  • Hi Jean, sorry to comment on your old post. I was reviewing some materials and came back to this post. I suddenly couldn't figure out why the condition that, the largest eigenvalue (in absolute value) of a stochastic matrix $P$ is $1$, can lead to the invertibility of $P$. I searched a bit and found an answer using the spectral theorem which I didn't know. Can you give me an alternative proof of this implication without involving the spectral theorem? Many thanks. – Sam Wong Jul 19 '23 at 09:49
  • 1
    @SamWong A matrix $A$ has eigenvalue $\lambda$ iff $\det(A-\lambda I)=0$. If $A$ is stochastic, its eigenvalues are $\le1$ in absolute value. Therefore $\det(A-\lambda I)\ne0$ if $|\lambda|>1$. That's just from the definition of an eigenvalue. Therefore, $\det(\frac1\lambda A-I)\ne0$ if $|\lambda|>1$, or equivalently $\det(\lambda A-I)\ne0$ if $|\lambda|<1$. Therefore if $|\lambda|<1$ the matrix $\lambda A-I$ is regular, and so is $I-\lambda A$. – Jean-Claude Arbaut Jul 19 '23 at 10:04
3

No, it is not necessarily positive definite or invertible. For example, try $\lambda = 1/3$ with $$ P = \pmatrix{1 & 1 & 1\cr 1 & 1 & 1\cr 1 & 1 & 1\cr}$$

Robert Israel
  • 470,583
  • Yea, you are right. What if the entries of $P$ can not be all the same? (Because in my study, such $P$ can not happen). I will put this extra condition into the edit. – Sam Wong Nov 23 '22 at 06:06
  • 1
    If you want an example where the entries of $P$ are not all the same, subtract different sufficiently small positive numbers from each of the $1$'s, and then adjust $\lambda$ as necessary. – Robert Israel Nov 23 '22 at 06:12
  • Make sense. I will think of my model again. Thanks. – Sam Wong Nov 23 '22 at 06:17
  • Hi Robert, I just reviewed my probability model and found one more missing condition in my case. If we require the row sum of $P$ to be $1$ for all rows, do we still have the non-invertibility result for $P$? – Sam Wong Nov 23 '22 at 06:35