(i) : $\left(\begin{array}{ll} \lambda_{1} & 0\\ 0 & \lambda_{2} \end{array}\right)$, (ii) : $\left(\begin{array}{ll} \lambda & 0\\ 0 & \lambda \end{array}\right)$, (iii) : $\left(\begin{array}{ll} \lambda & 1\\ 0 & \lambda \end{array}\right)$, where $\lambda_{1}\neq\lambda_{2}$, and no two of these are similar.

$\bbox[5px,border:2px solid gray]{ \text{ Case 1 } }$ If A has distinct eigenvalues, $\lambda_{1}\neq\lambda_{2}$ then $A$ is diagonalizable which means similar.

Otherwise, $\lambda_{1}=\lambda_{2}=\lambda$, and two cases arise according as $\dim E_{\lambda}=2$ and $\dim E_{\lambda}=1$.

$\bbox[5px,border:2px solid gray]{ \text{ Case 2 } }$ If $\dim E_{\lambda} = $ dim of eigenspace wrt $\lambda =2$ then $E_{\lambda}=\mathbb{C}^{2}$. Let $\mathfrak{B}=\{u,\ v\}$ be a basis of two linearly independent vectors, which transforms $A$ transforms to a diagonal matrix, so there exists a transformatin (which can be proven to be invertible) matrix $P$ such that $P^{-1}AP = \lambda I \implies$ (Lay P326 Ch 5 Supplementary Q8) $\implies A = \lambda I$

$\bbox[5px,border:2px solid gray]{ \text{ Case 3 } }$ If $\dim E_{\lambda}=1$, take a nonzero $v\in E_{\lambda}$, then $\{v\}$ is a basis for $E_{\lambda}$. Extend this to a basis $\mathfrak{B}=\{v,\ w\}$ for $\mathbb{C}^{2}$ by choosing $w\in \mathbb{C}^{2}\backslash E_{\lambda}$. If $Aw = \alpha v+\beta w$, then there exists a transformation matrix $P$ (that transforms the original $\{v\}$ to $\mathfrak{B}$), such that $C =P^{-1}AP=\left(\begin{array}{ll} \lambda & \alpha\\ 0 & \beta \end{array}\right) \quad (♦) $

$1.$ How and why is $Aw = \alpha v+\beta w$ ? Please explain how and why "w NOT an eigenvector of A $\implies α≠0$ ", from Christiaan Hattingh's answer below?

But in this case 3, $C$ must satisfy $\dim E_{\lambda}=1$. So for $C$, and thence $A$, to have $\lambda$ of algebraic multiplicity of 2, then $ C =\left(\begin{array}{ll} \lambda & \alpha\\ 0 & \lambda \end{array}\right) $

$3.$ What does "thence A" mean? Is this related to: Similar matrices $\implies \not \Leftarrow$ same eigenvalues.

$4.$ Why is the (2,2) entry of C now $\lambda$? What happened to $\beta$?

Further, we note that $\color{orangered}{ (C-\lambda I)^{2} } =\left(\begin{array}{ll} 0 & \alpha \\ 0 & 0 \end{array}\right)\left(\begin{array}{ll} 0 & \alpha\\ 0 & 0 \end{array}\right) = \color{orangered}{ 0 \, matrix } $.
Since $\color{orangered}{ (C-\lambda I)^{2} }w = (C-\lambda I) \color{forestgreen}{ (C-\lambda I)w } = \color{orangered}{ \mathbf{0} } $, thus $\color{forestgreen}{ (C-\lambda I)w }$ is an eigenvector of $C$ (and $w$ is a generalised eigenvector of $C$).

$7.$ Why: $\color{forestgreen}{ (C-\lambda I)w }$ is an eigenvector of $C$ ? Why do we need this? Does the proof ever use this?

Moreover, since $ Cw=u+\lambda w, $ if we consider the alternative basis $B' =\{u,\ w\}$ then there is a transformation matrix, say $Q$, such that $Q^{-1}CQ =\left(\begin{array}{ll} \lambda & 1\\ 0 & \lambda \end{array}\right). $ Finally, use $(♦)$ to substitute A for $C$ : $Q^{-1}CQ = (PQ)^{-1}A(PQ)$.

  • $\begingroup$ @Shaun: Thanks, but would you mind if I minimised white space? $\endgroup$ – Greek - Area 51 Proposal May 26 '14 at 16:19
  • $\begingroup$ Not at all: it's your choice :) $\endgroup$ – Shaun May 26 '14 at 16:21
  • $\begingroup$ @Shaun: THank you for your efforts though! $\endgroup$ – Greek - Area 51 Proposal May 27 '14 at 10:20

1.$\;$Since $\{v,w\}$ is a basis, any vector $y$ in $\mathbb{C}^2$ can be expressed uniquely in terms of this basis. Let $y=Aw$, then we can write $y=Aw = \alpha v+\beta w$ for unique $\alpha$ and $\beta$ (the $y$ is just to show that $Aw\in \mathbb{C}^2$, and every such a vector can be uniquely expressed in terms of a basis for $\mathbb{C}^2$). Furthermore, since $w$ is not an eigenvector of $A$ we know that $\alpha \neq 0$, and since $w$ and $v$ are linearly independent we know that $\beta \neq 0$.

As per OP's request, just to clarify in point number 1: If $\alpha=0$, we have $Aw=0v+\beta w=\beta w$ so that $w$ is an eigenvector of $A$ with associated eigenvalue $\beta$ (in which case $A$ would be diagonalizable).

3.$\;$Since $C$ and $A$ are similar, they have the same eigenvalues.

4.$\;$Since $C$ is upper triangular its diagonal entries are also its eigenvalues, so $\beta$ must actually be $\lambda$

7.$\;$To see clearly why this is so, multiply out: $C(C-\lambda I)w-\lambda I(C-\lambda I)w=0$, then rearranging, and since $IC=C$ and $\lambda I^2=\lambda I$: $C(C-\lambda I)w=\lambda(C-\lambda I)w$, so $(C-\lambda I)w$ is an eigenvector of $C$ with associated eigenvalue $\lambda$. Now this is not used explicitly in calculating the matrices of the proof, but the reason why this is mentioned is because the form (iii) is called the Jordan canonical form, and similar to the diagonalization problem (if one can find a basis consisting of eigenvectors then the matrix is diagonalizable), if one can find a basis consisting of generalized eigenvectors then a matrix is similar to a Jordan Canonical Form matrix.

  • $\begingroup$ Thanks. Regarding 1, what legitimates "letting y = Aw" ? How and why can you do this? $\endgroup$ – Greek - Area 51 Proposal May 28 '14 at 10:41
  • $\begingroup$ Will you please to respond in your answer, and not as comments? $\endgroup$ – Greek - Area 51 Proposal May 28 '14 at 10:42

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