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pezola
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[SOLVED] Prove A is Diagonalizable (Actual Question)
Suppose that A [tex]\in[/tex] M[tex]^{nxn}[/tex](F) and has two distinct eigenvalues, [tex]\lambda[/tex][tex]_{1}[/tex] and [tex]\lambda[/tex][tex]_{2}[/tex], and that dim(E(subscript [tex]\lambda[/tex][tex]_{1}[/tex] ))= n-1. Prove that A is diagonalizable.
So far, I know that dim(E subscript [tex]\lambda[/tex]2) [tex] \geq1[/tex]
and that
dim(E subscript [tex]\lambda[/tex]1) + dim(E subscript [tex]\lambda[/tex]2) [tex] \leq[/tex] n.
So dim(E subscript [tex]\lambda[/tex]2) = 1.
I am not exactly sure how this helps me to show A is digonalizable. Maybe I am thinking of something else and don't need this to prove A is diagonalizable. Please help.
(Also, sorry about my prevous blank post; I am new)
Homework Statement
Suppose that A [tex]\in[/tex] M[tex]^{nxn}[/tex](F) and has two distinct eigenvalues, [tex]\lambda[/tex][tex]_{1}[/tex] and [tex]\lambda[/tex][tex]_{2}[/tex], and that dim(E(subscript [tex]\lambda[/tex][tex]_{1}[/tex] ))= n-1. Prove that A is diagonalizable.
The Attempt at a Solution
So far, I know that dim(E subscript [tex]\lambda[/tex]2) [tex] \geq1[/tex]
and that
dim(E subscript [tex]\lambda[/tex]1) + dim(E subscript [tex]\lambda[/tex]2) [tex] \leq[/tex] n.
So dim(E subscript [tex]\lambda[/tex]2) = 1.
I am not exactly sure how this helps me to show A is digonalizable. Maybe I am thinking of something else and don't need this to prove A is diagonalizable. Please help.
(Also, sorry about my prevous blank post; I am new)