1
$\begingroup$

This question already has an answer here:

I'm in trouble at the following exercise:

(Ex. 6, page 143 - Um Curso de Álgebra Linear; Coelho, Flávio Ulhoa)
Show that if $A\in M_2(\mathbb{C})$, then $A$ is similar (or conjugated) to a matrix of one of the following forms $$\left(\begin{array}{cc}a&0\\0&b\end{array}\right) \text{ or } \left(\begin{array}{cc}a&0\\1&a\end{array}\right), \text{ with } a,b\in\mathbb{C}.$$

What I've tried: Be $A\in M_2(\mathbb{C})$. Consider $T:\mathbb{C}^2\to\mathbb{C}^2$ the linear operator whose matrix at the standard basis $\mathscr{C}=\{(1,0),(0,1)\}$ of $\mathbb{C}^2$ is $A$, i.e., $[T]_\mathscr{C}=A$. Since $\mathbb{C}$ is algebraically closed and we know that the characteristic polynomial of $T$ is monic and has degree 2, we have that it is of the form $p_T(x)=(x-a)(x-b)$, for some $a,b\in \mathbb{C}$. Of course $a$ and $b$ are eigenvalues of $T$.

Case 1: $a\neq b$; Considering the dimensions of the eigenspaces of $T$, we conclude that $T$ is diagonalizable and, therefore, exists a basis $\mathscr{B}$ of $\mathbb{C}^2$ such that $$[T]_{\mathscr{B}}=\left(\begin{array}{cc}a&0\\0&b\end{array}\right),$$ and, through basis exchange, $A$ is similar to a matrix of the first kind.

Case 2: $a=b$ and $p_T(x)=(x-a)^2$
In this case, if the geometric multiplicity $m_g(a)=2$ then, again, there is a basis $\mathscr{B}$ of $\mathbb{C}^2$ of eigenvectors of $T$ and $$[T]_{\mathscr{B}}=\left(\begin{array}{cc}a&0\\0&a\end{array}\right).$$ Now, if $m_g(a)=1$, then the eigenspace $Eig_T(a)$ spanned by the eigenvectors associated to the eigenvalue $a$ has dimension one, say, $Eig_T(a)=[v]$. And then, completing a basis $\mathscr{B}=\{u,v\}$ of $\mathbb{C}^2$, I only know that $$[T]_{\mathscr{B}}=\left(\begin{array}{cc}a_1&0\\a_2&a\end{array}\right).$$ From here, I've tried to find some suitable $u$ such that $T(u)=au+v$ in a way that $a_1=a$ and $a_2=1$, but solving this system seems impossible (in general) to me! Is this right?

Suggestions? Ideias? Thank you once more, guys...

$\endgroup$

marked as duplicate by Jendrik Stelzner, Dan, hardmath, user296602, user228113 Jan 22 '16 at 2:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Note that the diagonal elements of a lower triangular matrix give the roots of the characteristic polynomial, so you know $a_1 = a$ right away. So $a_2$ is the only part that takes any real work. $\endgroup$ – Dustan Levenstein Jan 21 '16 at 20:58
0
$\begingroup$

I think I've got. Just like @DustanLevenstein has said in the comment above, $a_1=a$.
Observe now that $(v)_\mathscr{B}=(0,1)$.
And, by definition of $[T]_{\mathscr{B}}$, $T(u)=au+a_2v$ and, therefore $$au-T(u)=-av$$ and writing this on its coordinates relative to $\mathscr{B}$, $$[aI-T]_\mathscr{B}[u]_\mathscr{B}=-[v]_\mathscr{B}$$ $$\iff \left(\left(\begin{array}{cc}a&0\\0&a\end{array}\right)-\left(\begin{array}{cc}a&0\\a_2&a\end{array}\right)\right)\left(\begin{array}{c}1\\0\end{array}\right)=\left(\begin{array}{c}0\\-1\end{array}\right)$$ $$\iff \left(\begin{array}{cc}0&0\\-a_2&0\end{array}\right)\left(\begin{array}{c}1\\0\end{array}\right)=\left(\begin{array}{c}0\\-1\end{array}\right)$$ $$\iff a_2=1$$ and we are done. ^_^

THIS IS ALL WRONG!

CORRECTION:
By definition of $[T]_\mathscr{B}$ we have that $T(u)=au+a_2v$, and $T(v)=av$. See that if $a_2=0$ the matrix is of the first kind. Suppose then $a_2\neq 0$ and consider $\mathscr{B}'=\{u,a_2v\}$. Is not hard to see that $\mathscr{B}'$ is basis of $\mathbb{C}^2$ and, since $T(u)=av+1\cdot(a_2v)$ and $T(a_2v)=a_2T(v)=a_2av=a(a_2v)$, we have $$[T]_{\mathscr{B}'}=\left(\begin{array}{cc}a&0\\1&a\end{array}\right)$$ and, NOW, we are done. Thank you @DustanLevenstein for correcting me!

$\endgroup$
  • $\begingroup$ Why do you have $au-T(u)=-v$? It seems you've assumed $a_2=1$ to get that. Note that it is entirely possible to have $a_2 \neq 1$ - you will need to find a possibly new basis to get that coordinate to be $1$. $\endgroup$ – Dustan Levenstein Jan 21 '16 at 21:38
  • $\begingroup$ Thank you so much @DustanLevenstein! I'll edit it. $\endgroup$ – Derso Jan 21 '16 at 21:46

Not the answer you're looking for? Browse other questions tagged or ask your own question.