# 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

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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yagombyeR

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 $h=6$
$\left[\begin{array}{ccccc}0& 1& -3& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0\end{array}\right]$