Are all matrices almost diagonalizable?

Question

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.

Answer 1

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 $\dots$

Answer 2

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 $\epsilon$.

Answer 3

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

Answer 4

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.