# Are all matrices almost diagonalizable?

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.

• Hey plus1. My solution is satisfactory for you or not? I don't know because there is no reaction.. for me it seems complete. – Widawensen Mar 15 '19 at 8:04
• @Widawensen Yes, that's what I was looking for ! – plus1 Mar 18 '19 at 5:34
• @Widawensen I just did it – plus1 Mar 26 '19 at 10:24
• Thank you plus1 :) – Widawensen Mar 26 '19 at 10:36

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 $$\dots$$

• 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. – Widawensen Mar 13 '19 at 12:17
• 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.. – Widawensen Mar 18 '19 at 14:19
• We have general solution $J=VJ_eV^{-1}$ where $V=\text{diag}(1, \epsilon,\epsilon^2 \dots \epsilon^{n-1})$ – Widawensen Mar 18 '19 at 14:23

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 $$\epsilon$$.

• Nice one! thanks! – plus1 Jan 18 '19 at 10:34

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.

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 $$M=(m_{ij})_{1\leqslant i,j\leqslant n}$$.

• I guess it's also true for any other norm. I buy that but how do you prove it? – plus1 Jan 18 '19 at 10:35
• 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. – José Carlos Santos Jan 18 '19 at 10:45