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Math  /  Algebra

QuestionLet T:R2→R2T: R^{2} \rightarrow R^{2} be the linear operator defined by T([x1x2])=[5x1−x25x1+x2]T\left(\left[\begin{array}{c} x_{1} \\ x_{2} \end{array}\right]\right)=\left[\begin{array}{c} 5 x_{1}-x_{2} \\ 5 x_{1}+x_{2} \end{array}\right] and let B={u1,u2}B=\left\{\boldsymbol{u}_{1}, \boldsymbol{u}_{2}\right\} be the basis for which u1=[15],u2=[−10]\boldsymbol{u}_{1}=\left[\begin{array}{l}1 \\ 5\end{array}\right], \boldsymbol{u}_{2}=\left[\begin{array}{c}-1 \\ 0\end{array}\right]. Find [T]B[T]_{B}. [T]B=(□[T]_{B}=(\square

Studdy Solution

STEP 1

What is this asking? We need to find the matrix representation of a linear transformation TT with respect to a non-standard basis BB. Watch out! Don't forget to express the transformed basis vectors in terms of the given basis BB!

STEP 2

1. Transform the basis vectors.
2. Express transformed vectors in terms of the basis.
3. Construct the matrix.

STEP 3

Alright, let's **transform** our first basis vector, u1=[15]\boldsymbol{u}_1 = \begin{bmatrix} 1 \\ 5 \end{bmatrix}, using our transformation TT.
Remember, TT takes a vector and does some cool stuff to it!
Plugging u1\boldsymbol{u}_1 into TT, we get: T(u1)=[5⋅1−55⋅1+5]=[010]. T(\boldsymbol{u}_1) = \begin{bmatrix} 5 \cdot 1 - 5 \\ 5 \cdot 1 + 5 \end{bmatrix} = \begin{bmatrix} 0 \\ 10 \end{bmatrix}. So, TT takes u1\boldsymbol{u}_1 and sends it to the vector [010]\begin{bmatrix} 0 \\ 10 \end{bmatrix}.

STEP 4

Now, let's do the same thing for our second basis vector, u2=[−10]\boldsymbol{u}_2 = \begin{bmatrix} -1 \\ 0 \end{bmatrix}.
Applying the transformation TT, we get: T(u2)=[−5−5].T(\boldsymbol{u}_2) = \begin{bmatrix} -5 \\ -5 \end{bmatrix}. So, TT transforms u2\boldsymbol{u}_2 into [−5−5]\begin{bmatrix} -5 \\ -5 \end{bmatrix}.

STEP 5

We want to express T(u1)=[010]T(\boldsymbol{u}_1) = \begin{bmatrix} 0 \\ 10 \end{bmatrix} as a linear combination of u1\boldsymbol{u}_1 and u2\boldsymbol{u}_2.
Let's find scalars c1c_1 and c2c_2 such that c1u1+c2u2=T(u1)c_1 \boldsymbol{u}_1 + c_2 \boldsymbol{u}_2 = T(\boldsymbol{u}_1): c1[15]+c2[−10]=[010]c_1 \begin{bmatrix} 1 \\ 5 \end{bmatrix} + c_2 \begin{bmatrix} -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 10 \end{bmatrix} This gives us the system of equations: c1−c2=0c_1 - c_2 = 0 and 5c1=105c_1 = 10.
Solving, we find c1=105=2c_1 = \frac{10}{5} = \textbf{2} and c2=c1=2c_2 = c_1 = \textbf{2}.
So, T(u1)=2u1+2u2T(\boldsymbol{u}_1) = \textbf{2}\boldsymbol{u}_1 + \textbf{2}\boldsymbol{u}_2.

STEP 6

Now, let's express T(u2)=[−5−5]T(\boldsymbol{u}_2) = \begin{bmatrix} -5 \\ -5 \end{bmatrix} in terms of u1\boldsymbol{u}_1 and u2\boldsymbol{u}_2.
We want to find scalars d1d_1 and d2d_2 such that d1u1+d2u2=T(u2)d_1 \boldsymbol{u}_1 + d_2 \boldsymbol{u}_2 = T(\boldsymbol{u}_2): d1[15]+d2[−10]=[−5−5]d_1 \begin{bmatrix} 1 \\ 5 \end{bmatrix} + d_2 \begin{bmatrix} -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -5 \\ -5 \end{bmatrix} This gives us d1−d2=−5d_1 - d_2 = -5 and 5d1=−55d_1 = -5.
Solving, we get d1=−55=-1d_1 = \frac{-5}{5} = \textbf{-1} and d2=d1+5=−1+5=4d_2 = d_1 + 5 = -1 + 5 = \textbf{4}.
So, T(u2)=-1u1+4u2T(\boldsymbol{u}_2) = \textbf{-1}\boldsymbol{u}_1 + \textbf{4}\boldsymbol{u}_2.

STEP 7

The matrix [T]B[T]_B is formed by the coefficients we just found.
The coefficients of T(u1)T(\boldsymbol{u}_1) form the first column, and the coefficients of T(u2)T(\boldsymbol{u}_2) form the second column.
Thus, [T]B=[2−124] [T]_B = \begin{bmatrix} 2 & -1 \\ 2 & 4 \end{bmatrix}

STEP 8

[T]B=[2−124][T]_B = \begin{bmatrix} 2 & -1 \\ 2 & 4 \end{bmatrix}

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