# Common eigenvector of two linear transformation matrices

I have two linear transformation matrices

\begin{pmatrix} 3 & 2 \\ -2 & 1 \end{pmatrix}

and

\begin{pmatrix} 1-a & -a \\ a & 1 \end{pmatrix}

How to find out what the value of $a$ should be for them to have a common eigenvector?

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find eigenvectors for both matrices and then compare them to obtain $a$? – TZakrevskiy May 14 '13 at 10:34
Note that when $a=-2$ the two matrices coincide. – Andrea Mori May 14 '13 at 10:37
and, after all, if $a=-2$, the two matrices are equal. – TZakrevskiy May 14 '13 at 10:37
If $a=-2$, then we get the first matrix. Then it will have the same eigenvector. Is this acceptable as an answer? – Bhavish Suarez May 14 '13 at 10:43

If two matrices $\begin{pmatrix} 3 & 2 \\ -2 & 1 \end{pmatrix}$ and $\begin{pmatrix} 1-a & -a \\ a & 1 \end{pmatrix}$ have common eigen vector the we will have :(suppose $(x_1,x_2)$ be that eigen vector):

$\begin{pmatrix} 3 & 2 \\ -2 & 1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}=\lambda_1 I \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}=\lambda_1\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$

and

$\begin{pmatrix} 1-a & -a \\ a & 1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}=\lambda_2 \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$

then we will have :

$3x_1+2x_2=\lambda_1 x_1=\frac{\lambda_1}{\lambda_2}\lambda_2 x_1=\frac{\lambda_1}{\lambda_2}((1-a)x_1-ax_2)$ $\to$

$(1-a)=\frac{\lambda_2}{\lambda_1}3$

and

$-a=\frac{\lambda_2}{\lambda_1}2$

so $a=-3\frac{\lambda_2}{\lambda_1}+1$ and $a=-2\frac{\lambda_2}{\lambda_1}$ so $-3\frac{\lambda_2}{\lambda_1}+1=-2\frac{\lambda_2}{\lambda_1}$ $\to$ $\frac{\lambda_2}{\lambda_1}=1$ and so $a=-2$

since a=-2 sarisfy in $-2x_1+x_2=\frac{\lambda_1}{\lambda_2}(ax_1+x_2)$ too they this can accept as unic solution.

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Find the eigenvectors of the first matrix.

Its characteristic polynomial is $(3-x)(1-x)+4=x^2-4x+7=(x-2)^2+3$, so it has no real root, that is, the matrix --viewed over $\Bbb R$ -- doesn't have any eiginvectors.

Well, over $\Bbb C$, it does have two roots, namely $\lambda_{1,2}=2\pm\sqrt3i$, and you can calculate the corresponding eigenvectors (with necessary complex entries), say they are $v_1$ and $v_2\ \in\Bbb C^2$. Then write up the eigen-equation $$\pmatrix{1-a&-a\\a&1}v_i\,\parallel\,v_i$$ for both $i=1,2$. Of course, $a=-2$ will be a solution (see comments), but there might be more.

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