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Question 1182632: Given the representation matrix of a linear transformation over the basis B, how to find a basis for a different representation matrix for the same linear transformation in which a certain condition is required?
For example:
V = R2[x] (all the polynomials of power 2 or less over Real numbers)
Given the following basis:
B= {1, 1+x, x^2 -1}
Defining T: V -> V a linear transformation as following:
T(1) = 4 - 3x^2
T(1+x) = 4 + 4x + 3x^2
T(x^2 -1) = x^2 - 4
The representative matrix of T is:
1 3 -3
0 4 0 = [T]B
-3 3 1
If given that T is diagonalizable, and the eigenvalues of it are { 4, 4, -2}
Find a basis of V for which the representation matrix is a diagonal matrix.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to find a basis for which the representation matrix of T is diagonal, given that T is diagonalizable:
**1. Eigenvalues and Eigenvectors:**
You're given the eigenvalues: λ₁ = 4, λ₂ = 4 (repeated), and λ₃ = -2. Since T is diagonalizable, there must be two linearly independent eigenvectors associated with the eigenvalue λ = 4.
* **For λ = 4:** Solve the equation (T - 4I)v = 0, where v is the eigenvector. In matrix form, using the given representation matrix [T]B:
```
[ 1-4 3 -3 ] [x] [0]
[ 0 4-4 0 ] [y] = [0]
[ -3 3 1-4 ] [z] [0]
```
```
[ -3 3 -3 ] [x] [0]
[ 0 0 0 ] [y] = [0]
[ -3 3 -3 ] [z] [0]
```
This simplifies to -3x + 3y - 3z = 0, or x = y - z. We have two free variables. Let z = t and y = s. Then x = s - t. The eigenvectors associated with λ = 4 can be written as:
```
[ s - t ]
[ s ]
[ t ] = s[1] + t[-1]
[ 1 ] [ 0 ]
[ 0 ] [ 1 ]
```
So, two linearly independent eigenvectors corresponding to λ = 4 (in the B basis) are [1, 1, 0]ᵀ and [-1, 0, 1]ᵀ.
* **For λ = -2:** Solve (T + 2I)v = 0:
```
[ 1+2 3 -3 ] [x] [0]
[ 0 4+2 0 ] [y] = [0]
[ -3 3 1+2 ] [z] [0]
```
```
[ 3 3 -3 ] [x] [0]
[ 0 6 0 ] [y] = [0]
[ -3 3 3 ] [z] [0]
```
From this, we get y = 0 and 3x - 3z = 0, or x = z. Let x = t, then z = t. The eigenvector is:
```
[ t ]
[ 0 ]
[ t ] = t [ 1 ]
[ 0 ]
[ 1 ]
```
So, an eigenvector corresponding to λ = -2 (in the B basis) is [1, 0, 1]ᵀ.
**2. Change of Basis:**
Now we have the eigenvectors in the B basis. We need to convert them to the standard basis (or any other basis you prefer) to get the eigenvectors in terms of polynomials. Recall that B = {1, 1+x, x²-1}.
* **Eigenvector [1, 1, 0]ᵀ (for λ = 4):**
1*(1) + 1*(1+x) + 0*(x²-1) = 1 + 1 + x = 2 + x
* **Eigenvector [-1, 0, 1]ᵀ (for λ = 4):**
-1*(1) + 0*(1+x) + 1*(x²-1) = -1 + x² - 1 = x² - 2
* **Eigenvector [1, 0, 1]ᵀ (for λ = -2):**
1*(1) + 0*(1+x) + 1*(x²-1) = 1 + x² - 1 = x²
**3. Diagonalizing Basis:**
The desired basis for which T is represented by a diagonal matrix is the set of eigenvectors we just found:
{2 + x, x² - 2, x²}
In this basis, the matrix representation of T will be diagonal, with the eigenvalues on the diagonal:
```
[ 4 0 0 ]
[ 0 4 0 ]
[ 0 0 -2 ]
```
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