Question
Let . Find [T]B, [T]B' and P where
.
Answer :
Word Count : 1083
We are asked to compute the matrix representations of a linear transformation (T: \mathbb{C}^3 \to \mathbb{C}^3) in two bases (B) and (B'), and the change of basis matrix (P). Let’s solve this manually, step by step. We are given: [ T\begin{bmatrix}x\y\z\end{bmatrix} = \begin{bmatrix}x + 2y - iz \ 2y + iz \ ix + z - 2z\end{bmatrix} = \begin{bmatrix}x + 2y - iz \ 2y + iz \ ix - z\end{bmatrix}, ] [ B = \left{ \begin{bmatrix}0\i\0\end{bmatrix}, \begin{bmatrix}i\1\-1\end{bmatrix}, \begin{bmatrix}0\0\2\end{bmatrix} \right}, \quad B' = \left{ \begin{bmatrix}1\-i\1\end{bmatrix}, \begin{bmatrix}0\0\1\end{bmatrix}, \begin{bmatrix}1\i\0\end{bmatrix} \right}. ] --- ### Step 1: Compute ([T]_B) ([T]_B) is found by expressing (T(v_j)) in the basis (B) for each (v_j \in B). Let’s denote (B = {b_1, b_2, b_3}). * (b_1 = \begin{bmatrix}0\i\0\end{bmatrix}) [ T(b_1) = T\begin{bmatrix}0\i\0\end{bmatrix} = \begin{bmatrix}0 + 2i - i*0 \ 2i + i*0 \ i*0 - 0\end{bmatrix} = \begin{bmatrix}2i\2i\0\end{bmatrix}. ] Now, write (T(b_1) = c_1 b_1 + c_2 b_2 + c_3 b_3). Let’s solve: [ c_1 \begin{bmatrix}0\i\0\end{bmatrix} + c_2 \begin{bmatrix}i\1\-1\end{bmatrix} + c_3 \begin{bmatrix}0\0\2\end{bmatrix} = \begin{bmatrix}2i\2i\0\end{bmatrix}. ] Compare coordinates: 1. First row: (0*c_1 + i*c_2 + 0*c_3 = 2i \implies i c_2 = 2i \implies c_2 = 2) 2. Second row: (i*c_1 + 1*c_2 + 0*c_3 = 2i \implies i c_1 + 2 = 2i \implies i c_1 = 2i - 2 \implies c_1 = 2 - 2i) 3. Third row: (0*c_1 + (-1)*c_2 + 2*c_3 = 0 \implies -2 + 2c_3 = 0 \implies c_3 = 1) So, first column of ([T]_B) is (\begin{bmatrix}2-2i\2\1\end{bmatrix}). --- * (b_2 = \begin{bmatrix}i\1\-1\end{bmatrix}) [ T(b_2) = \begin{bmatrix}i + 2*1 - i(-1) \ 2*1 + i(-1) \ i*i -1 \end{bmatrix} = \begin{bmatrix}i + ____ ___ ____ __________ ____ ___ ______.
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We are asked to compute the matrix representations of a linear transformation (T: \mathbb{C}^3 \to \mathbb{C}^3) in two bases (B) and (B'), and the change of basis matrix (P). Let’s solve this manually, step by step. We are given: [ T\begin{bmatrix}x\y\z\end{bmatrix} = \begin{bmatrix}x + 2y - iz \ 2y + iz \ ix + z - 2z\end{bmatrix} = \begin{bmatrix}x + 2y - iz \ 2y + iz \ ix - z\end{bmatrix}, ] [ B = \left{ \begin{bmatrix}0\i\0\end{bmatrix}, \begin{bmatrix}i\1\-1\end{bmatrix}, \begin{bmatrix}0\0\2\end{bmatrix} \right}, \quad B' = \left{ \begin{bmatrix}1\-i\1\end{bmatrix}, \begin{bmatrix}0\0\1\end{bmatrix}, \begin{bmatrix}1\i\0\end{bmatrix} \right}. ] --- ### Step 1: Compute ([T]_B) ([T]_B) is found by expressing (T(v_j)) in the basis (B) for each (v_j \in B). Let’s denote (B = {b_1, b_2, b_3}). * (b_1 = \begin{bmatrix}0\i\0\end{bmatrix}) [ T(b_1) = T\begin{bmatrix}0\i\0\end{bmatrix} = \begin{bmatrix}0 + 2i - i*0 \ 2i + i*0 \ i*0 - 0\end{bmatrix} = \begin{bmatrix}2i\2i\0\end{bmatrix}. ] Now, write (T(b_1) = c_1 b_1 + c_2 b_2 + c_3 b_3). Let’s solve: [ c_1 \begin{bmatrix}0\i\0\end{bmatrix} + c_2 \begin{bmatrix}i\1\-1\end{bmatrix} + c_3 \begin{bmatrix}0\0\2\end{bmatrix} = \begin{bmatrix}2i\2i\0\end{bmatrix}. ] Compare coordinates: 1. First row: (0*c_1 + i*c_2 + 0*c_3 = 2i \implies i c_2 = 2i \implies c_2 = 2) 2. Second row: (i*c_1 + 1*c_2 + 0*c_3 = 2i \implies i c_1 + 2 = 2i \implies i c_1 = 2i - 2 \implies c_1 = 2 - 2i) 3. Third row: (0*c_1 + (-1)*c_2 + 2*c_3 = 0 \implies -2 + 2c_3 = 0 \implies c_3 = 1) So, first column of ([T]_B) is (\begin{bmatrix}2-2i\2\1\end{bmatrix}). --- * (b_2 = \begin{bmatrix}i\1\-1\end{bmatrix}) [ T(b_2) = \begin{bmatrix}i + 2*1 - i(-1) \ 2*1 + i(-1) \ i*i -1 \end{bmatrix} = \begin{bmatrix}i + ____ ___ ____ __________ ____ ___ ______.
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Related Questions
Which of the following statements are true and which are false? Give reasons for your answer.
i) If V is a finite dimensional vector space and is a diagonalisable linear operator, then there is a basis, unique up to order of the elements, with respect to which the matrix of T is diagonal.
ii) Up to similarity, there is a unique matrix with minimal polynomial (x - 1)2(x - 2).
iii) If is the eigenvalue of a matrix A with characteristic polynomial f(x),
and
, then the geometric multiplicity of
is at most k.
iv) If , then
as
.
v) If N is nilpotent, eN is also nilpotent.
vi) The sum of two normal matrices of the order n is normal.
vii) If P and Q are positive definite operators, P + Q is a positive definite operator.
viii) Generalised inverse of an matrix need not be unique.
ix) All the entries of a positive definite matrix are non-negative.
x) The SVD of any matrix is unique.
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.
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and D-1 exists, then show that C is similar to -C. Hence show that the eigenvalues of C must come in plus-minus pairs.
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, where
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