An n×n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors.
In fact, A=PDP−1 where D is some diagonal matrix if and only if the columns of of P are n linearly independent eigenvectors of A. In this case the diagonal entries of D are eigenvalues of A that correspond to the eigenvectors in P.
Ex. Is A=⎝⎜⎛500−80017−2⎠⎟⎞ diagonalizable? If yes, express A as PDP−1 with D diagonal.
A is a triangular matrix, so its eigenvalues are λ1=5, λ2=0, λ3=−2. Since A is a 3×3 matrix and it has 3 distinct eigenvalues, we know it is diagonalizable.
Now all we have to do is calculate the eigenvectors of A to construct P.
For λ=5:
[A−5I∣0]=⎝⎜⎛000−8−5017−7000⎠⎟⎞x3=0, x2=0, x1 freeNul(A−5I)=span{⎝⎜⎛100⎠⎟⎞}
For λ=0:
[A∣0]=⎝⎜⎛500−80017−2000⎠⎟⎞x3=0, x2 free, x1=58x2Nul(A)=span{⎝⎜⎛5810⎠⎟⎞}
For λ=−2:
[A+2I∣0]=⎝⎜⎛700−820170000⎠⎟⎞x3 free, x2=2−7x3, x1=7−29x3Nul(A+2I)=span{⎝⎜⎛7−292−71⎠⎟⎞}
PDA=⎝⎜⎛10058107−292−71⎠⎟⎞=⎝⎜⎛50000000−2⎠⎟⎞=PDP−1.
What happens if a matrix doesn’t have n distinct eigenvalues?
Theorem.
Let A be an n×n matrix whose distinct eigenvalues are λ1,...,λp.
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For 1≤k≤p, the dimension of the eigenspace for λk is less than or equal to the multiplicity of λk.
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A is diagonalizable iff if the sum of the dimensions of the eigenspaces is n. This occurs iff either of the following occur:
- The characteristic polynomial factors completely into distinct linear factors.
(This is the case we dealt with before; this is true because each value’s eigenspace will have a multiplicity & dimension of 1)
- The dimension of the eigenspace for λk equals the multiplicity of λk in the characteristic polynomial.
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If A is diagonalizable and Bk is a bsais for the eigenspace corresponding to λk for each k, then the total collection of vectors in the sets B1,...,Bp forms a basis for Rn.
Ex. Determine if A=(1101) is diagonalizable.
Adet(A−λI)λA−1INul(A−1I)=(1101)=∣∣∣∣∣1−λ101−λ∣∣∣∣∣=(1−λ)2=0=1=(0100)=span{(01)} A is not diagonalizable, as λ has multiplicity 2 in the characteristic equation, but the dimension of its eigenspace is only 1.