Let A be an infinite matrix with all its first column elements equal to 1 and the rest of them equal to 0.

A=\begin{pmatrix} 1 & 0 & 0 & 0 & \cdots\\ 1 & 0 & 0 & 0 &\cdots\\ 1 & 0 & 0 & 0 & \cdots\\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}

Can A be diagonalized?


Since you are in infinite dimensions, you would first need to specify in which space the operator $A$ is supposed to act, then you can try to prove that it fulfils the assumptions for the spectral theorem.

If we first look at the action of $A$ on an arbitrary sequence of real (I'm assuming that you are working in $\mathbb{R}$) numbers $a=(a_1, a_2,...)$ we see that $A(a) = (a_1, a_1, ...)$ which won't be e.g. in $l^2$, the natural Hilbert space of sequences.

Actually, $A$ might seem to only make sense in $l^\infty$ (but it doesn't, as @MartinArgerami points out, because we don't have a countable basis with which to interpret what the action of $A$ on an arbitrary vector $u \in l^\infty$ is), and this is definitely not Hilbert. Because we then lack the notion of a scalar product, we cannot define what orthogonal eigenspaces would be, hence no orthogonal diagonalisation.

Note however that we can formally find another "infinite matrix" $P$ such that $P^{-1}$ "exists" in some sense and $D = P A P^{-1}$ is a diagonal infinite matrix, namely

$$D = \left(\begin{array}{ccccc} 1 & & & & \\ 0 & 0 & & & \\ 0 & 0 & 0 & & \\ 0 & 0 & 0 & 0 & \\ \vdots & & & & \ddots \end{array}\right)$$


$$P^{- 1} = \left(\begin{array}{ccccc} 1 & & & & \\ 1 & 1 & & & \\ 1 & 0 & 1 & & \\ 1 & 0 & 0 & 1 & \\ \vdots & & & & \ddots \end{array}\right),\ \ P = \left(\begin{array}{ccccc} 1 & & & & \\ - 1 & 1 & & & \\ - 1 & 0 & 1 & & \\ - 1 & 0 & 0 & 1 & \\ \vdots & & & & \ddots \end{array}\right).$$

Edit: If you are wondering where those matrices came from, it was basically this: it is natural to see how $A$ acts on the canonical basis, and one immediately sees that $A(e_1)=u=(1,1,1,1,...)$ is an eigenvector with eigenvalue 1 and that $A(e_i)=0$ for all $i>1$, so $e_i$ are eigenvectors with eigenvalue 0. You want $P$, $P^{-1}$ such that $D=P A P^{-1}$, where $P^{-1}$ is a change from the "new" basis of eigenvectors into the "old", i.e. the matrix with columns $u, e_2, e_3, ...$ Compute its "inverse" $P$, see if $D$ is all zeros except in the first entry, and you are done. But again, this is all formal and quite wrong, since we don't have a basis to begin with. See Martin's answer for more.

  • $\begingroup$ Why is the orthogonality necessary for diagonalization? It seems to me that you already did it. $\endgroup$ – Ian Apr 15 '17 at 19:49
  • $\begingroup$ Well, you are right that it is not necessary if you don't define it to be! What I meant is that we have no orthogonal diagonalisation (with an orthogonal change of basis), as in the spectral theorem, which is what I always think of when I hear diagonalisation. I'll edit the answer to fix it, thanks. $\endgroup$ – Miguel Apr 15 '17 at 20:09
  • $\begingroup$ @Miguel: note that your $P$ does not define an operator on $\ell^\infty$ (you cannot apply it to $(1,0,0,\ldots)^T$, for instance). $\endgroup$ – Martin Argerami Apr 15 '17 at 20:33
  • $\begingroup$ @MartinArgerami: I'd say that the matrix form precisely means $P(e_1) = (1,-1,-1,...) \in l^\infty$, $P(e_i) = e_i \in l^\infty$ for all $i>1$. But I think that I understand what you mean: there is a problem with the whole idea of extending $P$ beyond the ${e_i}$ by linearity. Indeed since the ${e_i}$ are not a Schauder basis for $l\infty$, i.e. $\sum u_i e_i$ won't converge in norm to $u$, the whole thing is flawed. So in this sense, taking $l^\infty$ doesn't make any sense at all, you are right. I'll edit the question again, thanks! $\endgroup$ – Miguel Apr 15 '17 at 21:20

The question is a big vague: how do you multiply arbitrary infinite matrices? You need some restrictions, that will affect when such a "matrix" is invertible (and you need that notion to talk about diagonalization). Or, if you express diagonalization as the existence of a basis of eigenvectors, you need to tells on which space does $A$ act.

Think of it this way: it is clear that one expects the spectrum of $A$ to be $\{0,1\}$. But, note that $A$ has no eigenvalue for $1$: to have $Ax=x$ for nonzero $x$, you would need $$ x=\begin{bmatrix}1\\1\\1\\ \vdots\end{bmatrix} $$ and then $Ax$ is not defined. And that's the problem: your "matrix" $A$ does not define an operator if you want to generalize the usual action of matrices on vectors.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.