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I am told that $A\in M_{2}(\mathbb{R})$ s.t $1-i$ is an eigenvalue of $A$ with corresponding eigenvector $\begin{pmatrix}3+2i\\ 1+3i \end{pmatrix}$.

I wish to find a matrix $P$ s.t $P^{-1}AP=\begin{pmatrix}1-i & 0\\ 0 & 1+i \end{pmatrix}$. From what I understand since $\lambda_{1}=1-i$ is an eigenvalue then also $\lambda_{2}=\bar{\lambda}_{2}=1+i$ is an eigenvalue and since $\lambda_{1}\neq\lambda_{2}$ and $A$ is a $2\times2$ matrix I conclude that such $P$ does exist.

My efforts for finding $P$ are: I said that if $v_{1}=\begin{pmatrix}3+2i\\ 1+3i \end{pmatrix}$ is an eigenvector corresponding to $\lambda_{1}$ then $v_{2}=\overline{v_{1}}$ is eigenvector corresponding to $\lambda_{2}$ hence $B=\{v_{1,}v_{2}\}$ is a basis of eigenvectors.

From what I remember $P$ is the matrix that transfers from the standard base to $B$ (or the other way around ?), so I tried writing $\begin{pmatrix}1\\ 0 \end{pmatrix}=\alpha v_{1}+\beta v_{2}$ and failed (I got that such $\alpha,\beta$ does not exist).

Can somene please help me understand what is my mistake and how to find the matrix $P$ ?

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Why do you think that $v_2=\overline v_1$? –  PAD Jun 26 '12 at 22:16
    
So I gather that you want to work on $\,\mathbb C\,$ in spite of the matrix being real...? –  DonAntonio Jun 26 '12 at 23:05
    
$P$ is the matrix whose columns are the eigenvectors of $A$. –  Gerry Myerson Jun 27 '12 at 0:24
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1 Answer

up vote 3 down vote accepted

As the matrix is real, its characteristic polynomial is real as well, so if $\,1-i\,$ is a root of it also $\,1+i\,$ is a root, and thus the char. polynomial is $$p_A(x)=(x-(1-i))(x-(1+i))=x^2-2x+2$$Now you're right: $$Av_1=\lambda v_1\Longrightarrow A\overline{v_1}=\overline{Av_1}=\overline{\lambda v_1}=\overline{\lambda}\overline{v_1}$$as the matrix is real , so $$\overline{v_1}=\overline{\begin{pmatrix}3+2i\\1+3i\end{pmatrix}}=\begin{pmatrix}3-2i\\1-3i\end{pmatrix}$$ is the eigenvector corresponding to the other eigenvalue, and thus $$P:=\begin{pmatrix}3+2i&3-2i\\1+3i&1-3i\end{pmatrix}$$

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