Question

Math

Posted 5 months ago

```
In the problems below, $S=\{1,2,3,4\}$.
Calculate the matrix $\mathrm{M}(T)$ representing the given linear transformations (relative to standard bases)
$T: \mathbb{R}^{4} \rightarrow \operatorname{Fun}(S)$ defined by
$T(a, b, c, d)=a\left(3 \chi_{1}+\chi_{2}-2 \chi_{3}\right)+b\left(\chi_{1}-2 \chi_{3}\right)+c\left(\chi_{1}-2 \chi_{3}-\chi_{4}\right)$
```

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Answer from Sia

Posted 5 months ago

Solution by Steps

step 1

To find the matrix $M(T)$ representing the linear transformation $T$, we need to apply $T$ to the standard basis vectors of $\mathbb{R}^4$

step 2

The standard basis vectors in $\mathbb{R}^4$ are $e_1 = (1, 0, 0, 0)$, $e_2 = (0, 1, 0, 0)$, $e_3 = (0, 0, 1, 0)$, and $e_4 = (0, 0, 0, 1)$

step 3

Apply $T$ to $e_1$: $T(1, 0, 0, 0) = 3\chi_1 + \chi_2 - 2\chi_3$. This gives us the first column of $M(T)$: $(3, 1, 1, 0)$

step 4

Apply $T$ to $e_2$: $T(0, 1, 0, 0) = \chi_1 - 2\chi_3$. This gives us the second column of $M(T)$: $(0, 0, 0, 0)$

step 5

Apply $T$ to $e_3$: $T(0, 0, 1, 0) = \chi_1 - 2\chi_3 - \chi_4$. This gives us the third column of $M(T)$: $(1, -2, -2, 0)$

step 6

Apply $T$ to $e_4$: $T(0, 0, 0, 1) = -\chi_4$. This gives us the fourth column of $M(T)$: $(-2, 0, -1, 0)$

step 7

Combine the results from steps 3 to 6 to form the matrix $M(T)$:

$$
M(T) = \begin{pmatrix}
3 & 0 & 1 & -2 \\
1 & 0 & -2 & 0 \\
1 & 0 & -2 & -1 \\
0 & 0 & 0 & 0
\end{pmatrix}
$$

Answer

The matrix $M(T)$ representing the linear transformation $T$ is:
$M(T) = \begin{pmatrix}
3 & 0 & 1 & -2 \\
1 & 0 & -2 & 0 \\
1 & 0 & -2 & -1 \\
0 & 0 & 0 & 0
\end{pmatrix}$

Key Concept

Matrix representation of a linear transformation

Explanation

The matrix representing a linear transformation with respect to the standard bases is constructed by applying the transformation to each standard basis vector and using the resulting vectors as the columns of the matrix.

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