Determine if the matrix A below is diagonalizable
belongs to book: A First Course in Linear Algebra|Robert A. Beezer|| Chapter number:6| Question number:c21
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belongs to book: A First Course in Linear Algebra|Robert A. Beezer|| Chapter number:6| Question number:c21
total answers (1)
A calculator will provide the eigenvalues λ = 2, 2, 1, 0, so we can reconstruct the characteristic polynomial as
pA (x) = (x − 2)2(x − 1)x
so the algebraic multiplicities of the eigenvalues are
αA (2) = 2 αA (1) = 1 αA (0) = 1
Now compute eigenspaces by hand, obtaining null spaces for each of the three eigenvalues by constructing the correct singular matrix (Theorem EMNS),
From this we can compute the dimensions of the eigenspaces to obtain the geometric multiplicities,
γA (2) = 2 γA (1) = 1 γA (0) = 1
For each eigenvalue, the algebraic and geometric multiplicities are equal and so by Theorem DMFE we now know that A is diagonalizable. The construction in Theorem DC suggests we form a matrix whose colmns are eigenvectors of A
Since det (S) =-1 ≠ 0, we know that S is nonsingular (Theorem SMZD), so the columns of S are a set of 4 linearly independent eigenvectors of A. By the proof of Theorem SMZD we know
is a diagonal matrix with the eigenvalues of A along the diagonal, in the same order as the associated eigenvectors appear as columns of S.
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