Diagonalizable :
A square matrix A is said to be diagonalizable if A is similar to a diagonal matrix. That is, if A = PDP-1 where P is invertible and D is a diagonal matrix.
When is A diagonalizable? (The answer lies in examining the eigenvalues and eigenvectors of A.)
Note that
Altogether
Equivalently,
An n x n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors.
In fact, A = PDP-1, with D a diagonal matrix, if and only if the columns of P are n linearly independent eigenvectors of A.
In this case, the diagonal entries of D are eigenvalues of A that correspond, respectively, to the eigenvectors in P.
Example 1 :
Diagonalize the following matrix, if possible.
Solution :
Step 1 :
Find the eigenvalues of A.
= (2 - λ)2(1 - λ) = 0
Eigen values of A : λ = 2 and λ = 1.
Step 2 :
Find three linearly independent eigenvectors of A.
By solving
(A - λI)x = 0,
for each value of λ, we obtain the following :
Step 3 :
Construct P from the vectors in step 2.
Step 4 :
Construct D from the corresponding eigenvalues.
Step 5 :
Check your work by verifying that AP = PD.
Example 2 :
Diagonalize the following matrix, if possible.
Solution :
Since this matrix is triangular, the eigenvalues are λ1 = 2 and λ2 = 4. By solving (A - λI)x = 0 for each eigenvalue, we would find the following :
Every eigenvector of A is a multiple of v1 or v2 which means there are not three linearly independent eigenvectors of A and by Theorem, A is not diagonalizable.
Example 3 :
Solution :
Since A has three eigenvalues :
λ1 = 2, λ2 = 6, λ3 = 1
and since eigenvectors corresponding to distinct eigenvalues are linearly independent, A has three linearly independent eigenvectors and it is therefore diagonalizable.
Example 4 :
Diagonalize the following matrix, if possible.
Solution:
Eigenvalues : -2 and 2 (each with multiplicity 2).
Solving (A -λI)x = 0 yields the following eigenspace basis sets.
{v1, v2, v3, v4} is linearly independent.
---> P = [v1 v2 v3 v4] is invertible
---> A = PDP-1, where
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