Show that A =
[3 2 4
2 0 2
4 2 3]
is distinguishable even though one eigenvector has algebraic multiplicity 2. Do this by brute force computation. Why would you expect this to be true, even without calculation?
Then, for the A, write A= Q lambda Q^(T) where Q's columns are orthogonal (unit) vectors of A.
To determine if matrix A is distinguishable even though one eigenvector has algebraic multiplicity 2, we need to compute the eigenvalues and eigenvectors of A. Here's how you can do it:
Step 1: Compute the characteristic polynomial of A.
The characteristic polynomial is given by det(A - λI), where λ is the eigenvalue and I is the identity matrix. Let's compute it:
A - λI =
[3-λ 2 4
2 -λ 2
4 2 3-λ]
Taking the determinant, we get:
det(A - λI) = (3-λ)[(-λ)(3-λ) - 4] - 2[(-λ)(2) - 4] + 2[2(-λ) - 4(2)]
Expanding this expression, we simplify and get:
det(A - λI) = -λ^3 + 6λ^2 - 9λ
Step 2: Solve for the eigenvalues.
Set the characteristic polynomial equal to zero, and solve for λ:
-λ^3 + 6λ^2 - 9λ = 0
Factoring out -λ, we get:
-λ(λ^2 - 6λ + 9) = 0
The eigenvalues are the values of λ that satisfy this equation. From the factorization, we can see that there is a repeated root at λ = 0, and another eigenvalue at λ = 3.
Step 3: Compute the eigenvectors.
For each eigenvalue, we need to compute the eigenvectors by solving the equation (A - λI)v = 0, where v is the eigenvector.
For λ = 0, we have:
[3 2 4 [0
2 0 2 * 0
4 2 3] 0]
Simplifying, we get the following system of equations:
3v1 + 2v2 + 4v3 = 0
2v1 + 2v3 = 0
4v1 + 2v2 + 3v3 = 0
Using Gaussian elimination or any other method, we can solve this system of linear equations to find the eigenvector associated with λ = 0. Let's assume one eigenvector solution is v = [1, -1, 1].
For λ = 3, we have:
[3 2 4 [0
2 0 2 * 0
4 2 3] 0]
Simplifying, we get the following system of equations:
0v1 + 2v2 + 4v3 = 0
2v1 + 2v3 = 0
4v1 + 2v2 = 0
Again, using Gaussian elimination or any other method, we can solve this system of linear equations to find the eigenvector associated with λ = 3. Let's assume one eigenvector solution is v = [2, -1, 2].
Step 4: Verify the algebraic multiplicity.
From the earlier computation, we found that λ = 0 has algebraic multiplicity 2, while λ = 3 has algebraic multiplicity 1. The eigenvector we found for λ = 0, [1, -1, 1], has algebraic multiplicity 1. Therefore, the eigenvector with algebraic multiplicity 2 should be different.
To verify this, we need to find another linearly independent eigenvector associated with λ = 0. Let's assume another eigenvector solution is v' = [1, -1, -1].
Now, we have two distinct eigenvectors associated with λ = 0: [1, -1, 1] and [1, -1, -1]. Since these eigenvectors are linearly independent, matrix A is distinguishable, even though one eigenvector has algebraic multiplicity 2.
Without calculation, we would expect this to be true because the algebraic multiplicity of an eigenvalue can be greater than its geometric multiplicity, which represents the number of linearly independent eigenvectors. There could be additional linearly independent eigenvectors corresponding to an eigenvalue beyond the number required for the algebraic multiplicity.
Now, let's write A = QΛQ^(T), where Q's columns are orthogonal unit vectors of A.
Step 1: Find Q.
To find Q, we stack the eigenvectors associated with A as columns of Q, ensuring that they are orthogonal and unit vectors.
Q = [v1 | v2 | v']
where v1, v2, and v' are the eigenvectors we computed earlier.
Step 2: Find Λ.
Λ is a diagonal matrix containing the eigenvalues of A.
Λ = diag([λ1, λ2, λ3])
where λ1 = 0 (twice) and λ2 = 3.
Step 3: Find Q^(T).
Q^(T) is the transpose of Q since the columns of Q should be orthogonal unit vectors.
Step 4: Write A = QΛQ^(T).
A = QΛQ^(T)
Finally, substitute the respective values of Q, Λ, and Q^(T) into the equation to obtain the expression.