 # It can be shown that the algebraic multiplicity of an eigenvalue lambda is always greater than or equal to the dimension of the eigenspace correspondi fortdefruitI 2021-01-24 Answered
It can be shown that the algebraic multiplicity of an eigenvalue lambda is always greater than or equal to the dimension of the eigenspace corresponding to lambda. Find h in the matrix A below such that the eigenspace for lambda = 5 is two-dimensional: $A=\left[\begin{array}{cccc}5& -2& 6& -1\\ 0& 3& h& 0\\ 0& 0& 5& 4\\ 0& 0& 0& 1\end{array}\right]$
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1) Row reduce the augmented matrix for the equation $\left(A-5I\right)x=0$
2) $\left[\begin{array}{ccccc}0& -2& 6& -1& 0\\ 0& -2& h& 0& 0\\ 0& 0& 0& 4& 0\\ 0& 0& 0& -4& 0\end{array}\right]=\left[\begin{array}{ccccc}0& -2& 6& -1& 0\\ 0& -2& h-6& 1& 0\\ 0& 0& 0& 4& 0\\ 0& 0& 0& 4& 0\end{array}\right]=\left[\begin{array}{ccccc}0& 1& -3& 0& 0\\ 0& 0& h-6& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0\end{array}\right]$
3) For a two-dimensional eigenspace, the system above needs two free variables. his happens if and inly if $h=6$
$\left[\begin{array}{ccccc}0& 1& -3& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0\end{array}\right]$