A problem on diagonalisation

Give examples of the following types of operators defined by a $2\times 2$ matrix or explain why such an operator can't exist:
i) diagonalisable, invertible,
ii) diagonalisable, not invertible,
iii) not diagonalisable, invertible,
iv) not diagonalisable, not invertible.

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Let's just consider the case in $\Bbb R$.

i)  $$\boxed{\begin{pmatrix}a&0\\0&b\end{pmatrix},\quad \text{a and b are not necessarily different},\quad a,b\neq 0}$$ Any diagonal matrix $D$ is diagonalisable: $$I_n DI_n=D.\\ \boxed{\begin{pmatrix}a&b\\b&a\end{pmatrix},\quad a\neq b,\quad a\neq -b,\quad\text{a can be zero.}}$$ Calculations: $(a-\lambda)^2-b^2=0\Rightarrow \lambda=a-b,a+b$. $$\ker\begin{pmatrix}a-(a+b)&b\\b&a-(a+b)\end{pmatrix}=\text{span}\bigg\{\begin{pmatrix}1\\1\end{pmatrix}\bigg\}\\ \ker\begin{pmatrix}a-(a-b)&b\\b&a-(a-b)\end{pmatrix}=\text{span}\bigg\{\begin{pmatrix}1\\-1\end{pmatrix}\bigg\}$$ The two eigenvectors form an eigenbasis. Alternatively, one can use the result: if $A\in M_{n\times n}(\Bbb F)$ has $n$ distinct eigenvalues in $\Bbb F$, then $A$ is diagonalisable over $\Bbb F$. $$\boxed{\begin{pmatrix}a&b\\b&c\end{pmatrix},\quad a\neq c,\quad b\neq 0,\quad\text{a can be zero.}}$$ The calculation is similar, but involves the use of quadratic formula. Since $a\neq c$ and $b\neq 0$, we have two distinct eigenvalues. Symmetric matrices are thus diagonalisable. $$\boxed{\begin{pmatrix}a&b\\0&c\end{pmatrix},\quad a\neq c,\quad b\neq 0}$$ Upper triangular matrices with distinct diagonal entries are diagonalisable because of the same reason: they have distinct eigenvalues. The above matrices are invertible since their determinants are non-zero.

ii) $$\boxed{\begin{pmatrix}a&0\\0&0\end{pmatrix},\quad a\neq 0}$$ Again, any diagonal matrix is diagonalisable. $$\boxed{\begin{pmatrix}a&a\\a&a\end{pmatrix},\quad a\neq 0}$$ Calculation: $(a-\lambda)^2-a^2=0\Rightarrow \lambda=0,2a$. When $a=1$, the matrix is called matrix of ones. $$\boxed{\begin{pmatrix}a&a\\b&b\end{pmatrix}, \begin{pmatrix}a&b\\a&b\end{pmatrix},\quad a+b\neq 0,\;\;\text{a,b not both zero.}}$$ Calculation: $(a-\lambda)(b-\lambda)-ab=0\Rightarrow \lambda=0,a+b$.

iii) $$\boxed{\begin{pmatrix}a&b\\0&a\end{pmatrix},\begin{pmatrix}a&0\\b&a\end{pmatrix},\quad \text{a and b are not necessarily different},\quad a,b\neq 0}$$ These are known as the shear transformations, which are not diagonalisable. Proof by contradiction: Suppose $\begin{pmatrix}a&b\\0&a\end{pmatrix}$ is diagonalisable. Then there exist a diagonal matrix $D$ and an invertible matrix $V$ s.t. $$\begin{pmatrix}a&b\\0&a\end{pmatrix}=V^{-1}DV.$$ We know just by looking at the matrix that $a$ is an eigenvalue with multiplicity $2$. So $D=\begin{pmatrix}a&0\\0&a\end{pmatrix}=aI_2$. We then have $\begin{pmatrix}a&b\\0&a\end{pmatrix}=aV^{-1}I_2V=aI_2=\begin{pmatrix}a&0\\0&a\end{pmatrix}$ which is a contradiction.

iv) $$\boxed{\begin{pmatrix}0&a\\0&0\end{pmatrix},\quad a\neq 0}$$ Its only eigenvalue is $0$ of multiplicity $2$.

From this problem, we see that invertibility does not necessarily nor sufficiently imply diagonalisability! Another important fact that one can observe from (ii) and (iv) is that if a square matrix $A$ is not invertible, then $\lambda=0$ is an eigenvalue of $A$. The converse also holds: if $\lambda=0$ is an eigenvalue of $A$, then $A$ is not invertible.

Claim: $A\in M_{n\times n}(\Bbb F)$ is singular iff $\lambda=0$ is an eigenvalue of $A$.
Proof: $\lambda=0$ is a solution of the characteristic equation $\lambda^n+c_1\lambda^{n-1}+\cdots+c_n=0$ iff $c_n=0$. By definition, $\det(\lambda I-A)=\lambda^n+c_1\lambda^{n-1}+\cdots+c_n$. When $\lambda=0$, we have $\det(\lambda I-A)=\det(-A)=(-1)^n\det(A)=c_n$. We thus have $\det(A)=0$ iff $c_n=0$ iff $\lambda=0.\;\Box$

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Here's a summary:

i) diagonalisable, invertible: symmetric, upper triangular, diagonal
ii) diagonalisable, not invertible: nilpotent, scalar multiples of matrix of ones
iii) not diagonalisable, invertible: shear
iv) not diagonalisable, not invertible: nilpotent.

I tried to enumerate all possibilities but failed. I can't imagine the amount of work it takes to characterise diagonalisation for $3\times 3$ matrices...

Questions:
$$\boxed{\begin{pmatrix}a&b\\c&d\end{pmatrix},\quad (a-d)^2=-4bc,\quad ad-bc\neq 0}$$ has repeated eigenvalues, so I consider it fitting (iii), but can the kernel corresponding to that one eigenvalue be spanned by two eigenvectors? Similar questions for $$\boxed{\begin{pmatrix}a&a\\a&b\end{pmatrix},\quad 5a^2+b^2=2ab,\quad a(b-a)\neq 0}\\ \boxed{\begin{pmatrix}a&a\\b&c\end{pmatrix},\quad a^2+c^2=2ac-4ab,\quad a(c-b)\neq 0}\\ \boxed{\begin{pmatrix}0&a\\b&c\end{pmatrix},\quad c^2=-4ab,\quad ab\neq 0}$$ I'm not even sure if there are other possibilities for (iii) and (iv).