Orthogonal diagonalization

Question

Is there going to be a similarity in orthogonal diagonalization of these matrixes??

$$ \begin{matrix} 3 & 1 & \\ 1 & 3 & \\ \\ \end{matrix} $$

$$ \begin{matrix} 3 & 1 & 1 & 1 \\ 1 & 3 & 1 & 1\\ 1 & 1 & 3 & 1\\ 1 & 1 & 1 & 3\\ \end{matrix} $$

I know that they both have one eigenvalue that that is the sum of the numbers in the row, and then eigenvalue 2 which has multiplicity (??? not sure how to translate it) 1 and 3.

And I also know that the eigenvectors for eigenvalue 4, 6 is going to be the same.

How can the eigenvectors be the same when they don’t even have the same dimension? — amd, Mar 27 '18 at 22:52
I meant similiar. It's going to be (1,1) for value 4 and (1,1,1,1) for value 6. — Zuzana Mitterová, Mar 28 '18 at 22:09

amd · Accepted Answer · 2018-03-28T23:23:13.300

There will be two eigenspaces for such matrices. One is the span of $(1,1,\dots,1)$, which you can normalize to $\frac1{\sqrt n}(1,1,\dots,1)$, and the other is the orthogonal complement of this space. A basis for this space can be found by inspection: $$(1,-1,0,0,\dots,0) \\ (1,0,-1,0,\dots,0) \\ (1,0,0,-1,\dots,0) \\ \vdots \\ (1,0,0,0,\dots,-1).$$ If you then orthogonalize this basis via the Gram-Schmidt process, a pattern emerges: the $i$th basis vector is $${1\over\sqrt{(i+1)^2+i}}(1,\dots,1,-i,0,\dots),$$ with the initial run of $1$s having length $i$.

Applying this pattern to your examples, an orthogonal basis that diagonalizes the first matrix is $$\frac1{\sqrt2}(1,1) \\ \frac1{\sqrt2}(1,-1)$$ and for the second: $$\frac12(1,1,1,1) \\ \frac1{\sqrt2}(1,-1,0,0) \\ \frac1{\sqrt6}(1,1,-2,0) \\ \frac1{\sqrt{12}}(1,1,1,-3).$$

Of course, since one of the eigenspaces has dimension $n-1$, there are other orthonormal bases that will also diagonalize these matrices.

Orthogonal diagonalization

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