 # Find a basis for the eigenspace corresponding to each listed eigenvalue. dictetzqh 2021-11-21 Answered

Find a basis for the eigenspace corresponding to each listed eigenvalue.
$A=\left[\begin{array}{cc}5& 0\\ 2& 1\end{array}\right],\lambda =1,5$

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To get the basis for the eigenspace, we first solve the system $\left(A-\lambda l\right)x=0$
For $\lambda =1,A-l=$
$\left[\begin{array}{cc}5-1& 0\\ 2& 1-1\end{array}\right]$
$\left[\begin{array}{cc}4& 0\\ 2& 0\end{array}\right]$
The augment matrix of $\left(A-l\right)x=0$ is
$\left[\begin{array}{ccc}4& 0& 0\\ 2& 0& 0\end{array}\right]$
With ${x}_{1}=0$ and ${x}_{2}$ is free, the solution can be written in the form:
$x={x}_{2}$
$\left[\begin{array}{c}0\\ 1\end{array}\right]$
So $\left[\begin{array}{c}0\\ 1\end{array}\right]$
Is a basis for the eigenspace.
For $\lambda =5,A-5l=$
$\left[\begin{array}{cc}5-5& 0\\ 2& 1-5\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 2& -4\end{array}\right]$
The augmented matrix of (A-5l)x=0 is
$\left[\begin{array}{ccc}0& 0& 0\\ 2& -4& 0\end{array}\right]$
That leads to $2{x}_{1}-4{x}_{2}=0\to {x}_{1}=2{x}_{2}$. The solution can be writtem in the form:
$x={x}_{2}$
$\left[\begin{array}{c}2\\ 1\end{array}\right]$
So
$\left[\begin{array}{c}2\\ 1\end{array}\right]$
is a basis for the eigenspace.