Question

Find a basis B for the domain of T such that the matrix T relative to B is

diagonal.
a. T: R3 ⟶ R3; T(x, y, z) = (−2x + 2y − 3z, 2x + y − 6z, −x − 2y)
b. T: P1 ⟶ P1; T(a + bx) = a + (a + 2b)x

Answer 1

To find a basis B for the domain of T such that the matrix T relative to B is diagonal, we need to find the eigenvectors of the transformation matrix of T. This can be done by finding the eigenvalues and eigenvectors of the given matrix.

Step-by-step explanation:

To find a basis B for the domain of T such that the matrix T relative to B is diagonal, we need to find the eigenvectors of the transformation matrix of T. Let's start with part a.

The given transformation T: R3 ⟶ R3 can be represented by the matrix

[[-2, 2, -3], [2, 1, -6], [-1, -2, 0]].

We can find the eigenvalues and eigenvectors of this matrix to construct a diagonal matrix. The eigenvectors will form the basis B we're looking for.

Let's move on to part b.

The given transformation T: P1 ⟶ P1 can be represented by the matrix

[[1, 1], [0, 1]].

This is a 2x2 matrix, so we need to find the eigenvalues and eigenvectors of this matrix to construct the diagonal matrix. The eigenvectors will form the basis B for the domain of T.

Answer 2

For Part (a): Eigenvalues and eigenvectors of the matrix representation of T yield the basis B for which the matrix T relative to B is diagonal.

For Part (b): The standard basis
`\(\{1, x\}\)` for
`\(P_1\)` forms the basis B for which the matrix representation of T is already diagonal.

To find a basis B for the domain of T such that the matrix T relative to B is diagonal:

Part (a):

For the linear transformation
`\(T: \mathbb{R}^3 \rightarrow \mathbb{R}^3\)` given by
`\(T(x, y, z) = (-2x + 2y - 3z, 2x + y - 6z, -x - 2y)\)`:

Approach:

We need to find eigenvectors of T corresponding to distinct eigenvalues. The eigenvectors will form a basis B for which the matrix representation of T will be diagonal.

1. Finding Eigenvalues and Eigenvectors:

- Compute the eigenvalues
`\(\lambda\)` and corresponding eigenvectors v of the matrix representation of T.

- Form a basis B using the eigenvectors corresponding to distinct eigenvalues.

Let's proceed with finding the eigenvalues and eigenvectors of T:

The matrix representation of T is:

`\[ [T] = \begin{bmatrix} -2 & 2 & -3 \\ 2 & 1 & -6 \\ -1 & -2 & 0 \end{bmatrix} \]`

Using this matrix, compute the eigenvalues and eigenvectors.

Part (b):

For the linear transformation
`\(T: P_1 \rightarrow P_1\)` given by
`\(T(a + bx) = a + (a + 2b)x\)`:

Approach:

In this case, the space
`\(P_1\)` represents polynomials of degree at most 1. The transformation is given by
`\(T(a + bx) = a + (a + 2b)x\)`.

Since the transformation is defined explicitly, we can find the matrix representation of T directly. A basis B for the domain can be found by considering the standard basis vectors for
`\(P_1\)`.

For , the standard basis is
`\(\{1, x\}\)`. To check whether the matrix representation of T is diagonalizable or already diagonal, let's represent T in matrix form relative to this basis.

`\[ T(1) = 1 + 0 \cdot x \]`

`\[ T(x) = 0 + 1 \cdot x \]`

The matrix representation of T relative to the basis
`\(\{1, x\}\)` is:

`\[ [T]_B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]`

This matrix is already diagonal, and the basis
`\(B = \{1, x\}\)` forms the required basis for which the matrix representation of T is diagonal.