Can the following matrix be diagonalized?

JmsNxn92 · **Joined:** Wed, 28 Sep 2011 23:37:54 UTC **Posts:** 23

Hello, I'm fairly new to matrix algebra, and I'm iffy on how to properly find diagonalizations, and determining whether a matrix can be diagonalized. I'm vague on finding eigenvectors and eigenvalues. I thought I'd just ask the question because this specific matrix is of interest to me.

It is an infinite square matrix:

$\mathbf{M} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 &...\\ 1 & 0 & 0 & 0 & 0 &...\\ 0 & 1 & 0 & 0 & 0 &...\\ 0 & 0 & 1 & 0 & 0 &...\\ 0 & 0 & 0 & 1 & 0 & ...\\ \vdots & \vdots & \vdots & \vdots & \vdots \end{pmatrix}$

The ones continue on as expected:

$\mathbf{M}_{j+1;j} = 1$

other wise:
$\mathbf{M}_{i;j} = 0$

Essentially I'm trying to understand fractional powers of the matrix, and the only method I am aware of is diagonalization. If that's not possible, does anybody know of any alternative methods? I'm fearful of the matrix logarithm and exponential, but if necessary I could try.

Shadow · **Posted:** Thu, 19 Apr 2012 01:07:59 UTC

JmsNxn92 wrote:

Hello, I'm fairly new to matrix algebra, and I'm iffy on how to properly find diagonalizations, and determining whether a matrix can be diagonalized. I'm vague on finding eigenvectors and eigenvalues. I thought I'd just ask the question because this specific matrix is of interest to me.

It is an infinite square matrix:

$\mathbf{M} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 &...\\ 1 & 0 & 0 & 0 & 0 &...\\ 0 & 1 & 0 & 0 & 0 &...\\ 0 & 0 & 1 & 0 & 0 &...\\ 0 & 0 & 0 & 1 & 0 & ...\\ \vdots & \vdots & \vdots & \vdots & \vdots \end{pmatrix}$

The ones continue on as expected:

$\mathbf{M}_{j+1;j} = 1$

other wise:
$\mathbf{M}_{i;j} = 0$

Essentially I'm trying to understand fractional powers of the matrix, and the only method I am aware of is diagonalization. If that's not possible, does anybody know of any alternative methods? I'm fearful of the matrix logarithm and exponential, but if necessary I could try.

Your matrix cannot be diagonalized. As a bit of an important technicality, it should be mentioned that what you have is not literally a matrix at all.

outermeasure · **Posted:** Thu, 19 Apr 2012 07:13:34 UTC

Shadow wrote:

JmsNxn92 wrote:

Hello, I'm fairly new to matrix algebra, and I'm iffy on how to properly find diagonalizations, and determining whether a matrix can be diagonalized. I'm vague on finding eigenvectors and eigenvalues. I thought I'd just ask the question because this specific matrix is of interest to me.

It is an infinite square matrix:

$\mathbf{M} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 &...\\ 1 & 0 & 0 & 0 & 0 &...\\ 0 & 1 & 0 & 0 & 0 &...\\ 0 & 0 & 1 & 0 & 0 &...\\ 0 & 0 & 0 & 1 & 0 & ...\\ \vdots & \vdots & \vdots & \vdots & \vdots \end{pmatrix}$

The ones continue on as expected:

$\mathbf{M}_{j+1;j} = 1$

other wise:
$\mathbf{M}_{i;j} = 0$

Essentially I'm trying to understand fractional powers of the matrix, and the only method I am aware of is diagonalization. If that's not possible, does anybody know of any alternative methods? I'm fearful of the matrix logarithm and exponential, but if necessary I could try.

Your matrix cannot be diagonalized. As a bit of an important technicality, it should be mentioned that what you have is not literally a matrix at all.

... and, interpreting as $M\in B(\ell^2)$ , there doesn't exist $N\in B(\ell^2)$ such that N^2=M

.

JmsNxn92 · **Joined:** Wed, 28 Sep 2011 23:37:54 UTC **Posts:** 23

Alright, doesn't surprise me. Thanks for the help.

outermeasure · **Posted:** Fri, 20 Apr 2012 06:53:38 UTC

JmsNxn92 wrote:

Alright, doesn't surprise me. Thanks for the help.

Indeed it shouldn't surprise you. There are no square-root of the matrix $\begin{pmatrix}0&0\\1&0\end{pmatrix}$ , for example. The theory doesn't want non-invertibles for a reason.

On the other hand, diagonalisability is something we can get around, at least in finite dimensions, by using the Jordan normal form (infinite-dimensional Jordan normal form is highly sensitive to small perturbations, plus other technical problems like the presence of pseudospectrum, which makes them rather useless in general). For example, $\begin{pmatrix}1&0\\1&1\end{pmatrix}$ has a square-root, namely $\begin{pmatrix}1&0\\\frac{1}{2}&1\end{pmatrix}$ , which you can obtain from the series $(I+N)^r=I+rN+\binom{r}{2}N^2+\binom{r}{3}N^3+\cdots$ , N nilpotent.

S.O.S. Mathematics CyberBoard

Can the following matrix be diagonalized?

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