1
$\begingroup$

I'm having some trouble diagonalizing this nxn matrix with ones along both diagonals:

$\begin{bmatrix} 1&0&0&\cdots&0&0&1\\0&1&0&\cdots&0&1&0\\ 0&0&1&\cdots&1&0&0\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ 0&0&1&\cdots&1&0&0\\0&1&0&\cdots&0&1&0\\1&0&0&\cdots&0&0&1 \end{bmatrix}$

How should I approach this problem? At first, I tried cofactor expansion to find the eigenvalues and eigenspaces, but it quickly became really messy, so I'm wondering if there's any simpler way to diagonalize such a matrix. Any help would be appreciated!

$\endgroup$
1
$\begingroup$

Let's look at $n$ even:

If you let an eigenvector be $(x_0, x_1, \ldots x_{n-2})^T$ and just write out the eigen equations, you get $$ x_0 + x_{n-1} = \lambda x_0 \\ x_1 + x_{n-2} = \lambda x_1 \\ \cdots + x_{n-2}+x_1 = \lambda x_{n-2} \\ x_{n-1}+x_0 = \lambda x_{n-1} $$ From this it is easy to see that either $\lambda = 0$ (in which case pairs $x_0 = - x_{n-1}$ and so forth), or $\lambda = 2$ (in which case pairs $x_0 = x_{n-1}$ and so forth). When $n$ is odd, ther is another case available for $\lambda = 0$, in whihc only the middle entry is non-zero.

So the eigenvalues are $2$ (with multiplicity $\left\lfloor \frac{n}{2}\right\rfloor$) and 0 (with multiplicity $\left\lfloor \frac{n+1}{2}\right\rfloor$). An orthonormal set of corresponding eigenvectors would be $$ \left( \begin{array}{c} 1/\sqrt{2} \\ 0 \\ 0\\0 \\ \cdots \\ 0 \\ 0\\0\\1/\sqrt{2}\end{array} \right) \left( \begin{array}{c} 0\\ 1/\sqrt{2} \\ 0 \\ \\ \cdots \\ 0\\ \\ 1/\sqrt{2} \\ 0\end{array} \right) \cdots \left( \begin{array}{c} 0 \\ 0 \\0 \\ \cdots \\ 1/\sqrt{2} \\ 1/\sqrt{2} \\ \cdots \\ 0 \\ 0\end{array} \right) \left( \begin{array}{c} 0 \\ 0 \\0 \\ \cdots \\ 1/\sqrt{2} \\ -1/\sqrt{2} \\ \cdots \\ 0 \\ 0\end{array} \right) \cdots \left( \begin{array}{c} 0\\ 1/\sqrt{2} \\ 0 \\ \\ \cdots \\ 0\\ \\ -1/\sqrt{2} \\ 0\end{array} \right) \cdots \left( \begin{array}{c} 1/\sqrt{2} \\ 0 \\ 0\\0 \\ \cdots \\ 0 \\ 0\\0\\1/\sqrt{2}\end{array} \right) $$ where the first half of those have eigenvalue $2$ and the second half $0$.

If $n$ is odd, there is another possibility: All of the $x_i$ except the middle one are zero, and the eigenvalue is 1. So our set looks like $$ \left( \begin{array}{c} 1/\sqrt{2} \\ 0 \\ 0\\0 \\ \cdots \\ \cdots \\ 0 \\ 0\\0\\1/\sqrt{2}\end{array} \right) \cdots \left( \begin{array}{c} 0 \\ 0 \\0 \\ \cdots \\ 1/\sqrt{2} \\ 0 \\ -1/\sqrt{2} \\ \cdots \\ 0 \\ 0\end{array} \right) \left( \begin{array}{c} 0 \\ 0 \\0 \\ \cdots \\ 0 \\ 1 \\ 0 \\ \cdots \\ 0 \\ 0\end{array} \right) \left( \begin{array}{c} 0 \\ 0 \\0 \\ \cdots \\ 1/\sqrt{2} \\ 0 \\ -1/\sqrt{2} \\ \cdots \\ 0 \\ 0\end{array} \right) \cdots \left( \begin{array}{c} 1/\sqrt{2} \\ 0 \\ 0\\0 \\ \cdots \\ \cdots \\ 0 \\ 0\\0\\1/\sqrt{2}\end{array} \right) $$ where the first almost-half of those have eigenvalue $2$, the middle one has eigenvalue $1$, and the second almost-half have eigenvalues $0$.

For example, $$ \left( \begin{array}{ccccc} \frac{1}{\sqrt{2}} &0 & 0 & 0 & \frac{1}{\sqrt{2}} \\ 0 & \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} & 0 \\0 &0 & 1 & 0 & 1 \\ 0 & \frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} & 0\\ \frac{1}{\sqrt{2}} & 0 &0 & 0 & -\frac{1}{\sqrt{2}} \end{array} \right) \left( \begin{array}{ccccc} 1 &0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0\\ 0 &0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0\\ 1 & 0 & 0 &0 & 1 \end{array} \right) \left( \begin{array}{ccccc} \frac{1}{\sqrt{2}} &0 & 0 & 0 & \frac{1}{\sqrt{2}} \\ 0 & \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} & 0 \\0 &0 & 1 & 0 & 1 \\ 0 & \frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} & 0\\ \frac{1}{\sqrt{2}} & 0 &0 & 0 & -\frac{1}{\sqrt{2}} \end{array} \right) = \left( \begin{array}{ccccc} 2 &0 & 0 & 0 & 0 \\ 0 & 2 & 0 & -0 & 0 \\0 &0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \end{array} \right) $$

$\endgroup$
0
$\begingroup$

In the case that the matrix has an even size, we can find the eigenvalues of this matrix by adding $1$ to the eigenvalues of the matrix $$ A = \begin{bmatrix} 0&0&0&\cdots&0&0&1\\ 0&0&0&\cdots&0&1&0\\ 0&0&0&\cdots&1&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ 0&0&1&\cdots&0&0&0\\ 0&1&0&\cdots&0&0&0\\ 1&0&0&\cdots&0&0&0 \end{bmatrix} $$ Because $A^2 = I$ (check that this is the case), we can prove that the only eigenvalues $A$ can have are $\pm 1$.

Check that $A$ has an eigenvector for each eigenvalue (for example, try $e_1 + e_n$ and $e_1 - e_n$).

It follows that the original matrix has eigenvalues $0,2$.

In the case that the original matrix has an odd size, we will also have an eigenvalue $1$.


In fact, in order to diagonalize the matrix, we need to find all eigenvalues. In order to do so, verify that for each $j$, the vector $e_j + e_{n-j}$ is an eigenvector of the matrix, as is $e_j - e_{n-j}$. Here, $e_1,\dots,e_n$ denote the standard basis vectors.

Whether $n$ is even or odd, this will give you $n$ linearly independent eigenvectors, allowing you to diagonalize the matrix.

In general, a nice approach to diagonalizing matrices following a certain pattern of arbitrary size is to "guess" the eigenvectors.

$\endgroup$
  • 1
    $\begingroup$ See also the answer that I gave here $\endgroup$ – Omnomnomnom Apr 28 '15 at 18:47
  • $\begingroup$ Why would it have an eigenvalue of 1 if it has an odd size? I'm still new to eigenvalues and eigenvectors, so my understanding of them hasn't developed much yet. $\endgroup$ – Gary Chien Apr 28 '15 at 18:55
  • 1
    $\begingroup$ Sorry, glossed over that one. The quickest explanation is this: try multiplying the vector $e_{(n+1)/2}$ and see what happens. $\endgroup$ – Omnomnomnom Apr 28 '15 at 18:56

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.