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Every real $2$ by $2$ matrix that is not diagonalizable is similar to the $2$ by $2$ jordan canonical form,

$$ J_2=\begin{bmatrix}s&1\\0&s\end{bmatrix}, $$

where $s$ is the eigenvalue (with multiplicity $2$). My question: Is $J_2$ almost diagonalizable? I mean is it similar to

$$ \begin{bmatrix}s&\epsilon\\0&s\end{bmatrix} $$

for every $\epsilon>0$ no matter how small? And what happens in higher dimensions, is every matrix similar to an arbitrarily close to diagonal matrix?

My guess is yes but I just can't find the similarity transformation. It could be something very simple.

Thanks in advance, all ideas welcome.

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  • $\begingroup$ Hey plus1. My solution is satisfactory for you or not? I don't know because there is no reaction.. for me it seems complete. $\endgroup$ – Widawensen Mar 15 at 8:04
  • $\begingroup$ @Widawensen Yes, that's what I was looking for ! $\endgroup$ – plus1 Mar 18 at 5:34
  • $\begingroup$ @Widawensen I just did it $\endgroup$ – plus1 Mar 26 at 10:24
  • $\begingroup$ Thank you plus1 :) $\endgroup$ – Widawensen Mar 26 at 10:36
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The searched similarity transformations:

  • for dimension $2$ :

    $$ \begin{bmatrix}1&0\\0&\epsilon\end{bmatrix} \begin{bmatrix}s&\epsilon\\0&s\end{bmatrix} \begin{bmatrix}1&0\\0&\epsilon\end{bmatrix}^{-1} = \begin{bmatrix}s&1\\0&s\end{bmatrix} $$

  • for dimension $3$:

    $$ \begin{bmatrix}1&0 & 0 \\0&\epsilon & 0 \\0&0 & \epsilon^2 \end{bmatrix} \begin{bmatrix}s&\epsilon & 0 \\0& s & \epsilon \\0&0 & s \end{bmatrix} \begin{bmatrix}1&0 & 0 \\0&\epsilon & 0 \\0&0 & \epsilon^2 \end{bmatrix}^{-1} = \begin{bmatrix}s&1 & 0 \\0& s & 1 \\0&0 & s \end{bmatrix} $$

The pattern for higher dimensions is visible $\dots$

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  • $\begingroup$ Formula for dim. 2 was found by me manually, starting from general form of similarity matrix {{a,b},{c,d}}.. For dimension 3 I assumed that it is diag(1,a,b) and calculated a and b. I've checked formulas at Wolphram Alpha for dimension 2,3 and additionally for 4 so I assume that they are correct. $\endgroup$ – Widawensen Mar 13 at 12:17
  • $\begingroup$ In the case of doubts one can copy below formula to Wolphram Alpha {{1,0,0,0},{0,x,0,0},{0,0,x^2,0 },{0,0,0,x^3}}.{{s,x,0,0},{0,s,x,0},{0,0,s,x},{0,0,0,s}}.inverse{{1,0,0,0},{0,x,0,0},{0,0,x^2,0 },{0,0,0,x^3}} and check.. $\endgroup$ – Widawensen Mar 18 at 14:19
  • $\begingroup$ We have general solution $J=VJ_eV^{-1}$ where $V=\text{diag}(1, \epsilon,\epsilon^2 \dots \epsilon^{n-1})$ $\endgroup$ – Widawensen Mar 18 at 14:23
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Yes, this is true and you gave the answer almost by yourself. The key is the Jordan normal form. To get $\epsilon$ instead of $1$, you just need to take the Jordan normal form of $\frac{1}{\epsilon}A$ and then multiply by $\epsilon$.

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  • $\begingroup$ Nice one! thanks! $\endgroup$ – plus1 Jan 18 at 10:34
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Not all real matrices are close to matrices diagonalizable over $\mathbb{R}$; there are open sets on which every matrix has irreducible quadratic factors.

For example, $\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ is not diagonalizable over $\mathbb{R}$, because its characteristic polynomial $x^2+1$ doesn't factor. Moreover, adding anything with coefficients less than $\frac13$ won't change that; the discriminant will still be negative.

Over $\mathbb{C}$, where every polynomial factors, the answer is yes; we can perturb the matrix slightly by adding small random elements to the diagonal so that the eigenvalues all become different.

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Yes, it is true. For every $n\times n$ matrix $A$ and for every $\varepsilon>0$, there is some diagonal matrix $D$ such that $A$ is similar to a matrix $D^\star$ such that $\lVert D-D^\star\rVert<\varepsilon$. Here$$\lVert M\rVert=\sqrt{\sum_{i,j=1}^n\lvert m_{ij}\rvert^2},$$if $M=(m_{ij})_{1\leqslant i,j\leqslant n}$.

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  • $\begingroup$ I guess it's also true for any other norm. I buy that but how do you prove it? $\endgroup$ – plus1 Jan 18 at 10:35
  • $\begingroup$ On a finite dimensional real vector space, all norms are equivalente. So, yes, if it holds for one of them, it holds for all of them. Concerning a proof, your idea is good: consider Jordan matrices with the $1$'s above the main diagonal replaced by a tiny real number. $\endgroup$ – José Carlos Santos Jan 18 at 10:45

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