Linear Transformation R3 To R2

Problem 339

Allow $\{\mathbf{v}_1, \mathbf{5}_2\}$ be a ground of the vector space $\R^2$, where
\[\mathbf{v}_1=\begin{bmatrix}
1 \\
one
\end{bmatrix} \text{ and } \mathbf{v}_2=\begin{bmatrix}
i \\
-i
\end{bmatrix}.\] The action of a linear transformation $T:\R^2\to \R^3$ on the basis $\{\mathbf{v}_1, \mathbf{v}_2\}$ is given past
\brainstorm{align*}
T(\mathbf{five}_1)=\begin{bmatrix}
2 \\
4 \\
6
\end{bmatrix} \text{ and } T(\mathbf{v}_2)=\begin{bmatrix}
0 \\
eight \\
10
\terminate{bmatrix}.
\terminate{align*}

Notice the formula of $T(\mathbf{10})$, where
\[\mathbf{x}=\begin{bmatrix}
x \\
y
\end{bmatrix}\in \R^2.\]

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Solution.

We give 2 solutions.

Solution one (linear combination)

Since we know the values of $T$ on the footing vectors $\mathbf{v}_1, \mathbf{five}_2$, if we express the vector $\mathbf{x}$ as a linear combination of $\mathbf{v}_1, \mathbf{v}_2$, we can find $F(\mathbf{10})$ by the linearity of the linear transformation $T$.

So permit us find the scalars $c_1, c_2$ such that
\[\mathbf{ten}=c_1\mathbf{v}_1+c_2\mathbf{v}_2.\] We write this as
\begin{align*}
\begin{bmatrix}
ten \\
y
\finish{bmatrix}=c_1\begin{bmatrix}
1 \\
1
\end{bmatrix}+c_2\begin{bmatrix}
1 \\
-1
\end{bmatrix}
=
\brainstorm{bmatrix}
i & one\\
1& -1
\finish{bmatrix}\brainstorm{bmatrix}
c_1 \\
c_2
\end{bmatrix}.
\end{align*}

The matrix $\begin{bmatrix}
ane & 1\\
1& -1
\cease{bmatrix}$ is invertible (as its determinant is $-2$) and its inverse matrix is
\[\begin{bmatrix}
1 & 1\\
1& -one
\end{bmatrix}^{-1}=\frac{1}{2}\begin{bmatrix}
1 & 1\\
i& -i
\stop{bmatrix}.\]

Thus, we have
\begin{align*}
\begin{bmatrix}
c_1 \\
c_2
\end{bmatrix}&=\begin{bmatrix}
one & i\\
1& -ane
\finish{bmatrix}^{-1}\brainstorm{bmatrix}
x \\
y
\end{bmatrix}\\[6pt] &=\frac{1}{two}\begin{bmatrix}
1 & 1\\
1& -1
\terminate{bmatrix}\begin{bmatrix}
ten \\
y
\end{bmatrix}\\[6pt] &=\frac{1}{2}\begin{bmatrix}
x+y \\
10-y
\end{bmatrix}
\cease{marshal*}
Therefore, nosotros obtain the linear combination
\[\mathbf{x}=\frac{1}{2}(x+y)\mathbf{v}_1+\frac{1}{two}(x-y)\mathbf{v}_2.\]

At present we compute $T(\mathbf{x})$ equally follows.
We accept
\begin{align*}
T(\mathbf{x})&=T\left(\frac{ane}{ii}(x+y)\mathbf{v}_1+\frac{1}{2}(x-y)\mathbf{5}_2 \right)\\[6pt] &=\frac{1}{ii}(10+y)T(\mathbf{five}_1)+\frac{1}{ii}(ten-y)T(\mathbf{v}_2) && \text{by linearity of $T$}\\[6pt] &=\frac{1}{2}(x+y)\begin{bmatrix}
2 \\
4 \\
vi
\end{bmatrix}+\frac{1}{2}(x-y)\begin{bmatrix}
0 \\
8 \\
10
\end{bmatrix} \\[6pt] &=\begin{bmatrix}
x+y \\
6x-2y \\
8x-2y
\end{bmatrix}.
\end{marshal*}
Hence the formula is
\[T(\mathbf{x})=\brainstorm{bmatrix}
ten+y \\
6x-2y \\
8x-2y
\stop{bmatrix}.
\]

Solution two (Matrix representation)

In the second solution, we use the matrix representation for the linear transformation $T$.
Let $A$ be the matrix of $T$ with respect to the standard basis $\{\brainstorm{bmatrix}
1 \\
0
\end{bmatrix}, \brainstorm{bmatrix}
0 \\
i
\stop{bmatrix}\}$ of $\R^2$.
Thus, we have $T(\mathbf{x})=A\mathbf{ten}$ by definition.

To detect the matrix $A$, nosotros compute
\begin{align*}
A\begin{bmatrix}
i & one\\
1& -ane
\end{bmatrix}&=A\brainstorm{bmatrix}
\mathbf{v}_1 & \mathbf{five}_2 \\
\end{bmatrix}\\[6pt] &=\brainstorm{bmatrix}
A\mathbf{v}_1 & A\mathbf{five}_2
\terminate{bmatrix}\\[6pt] &=\begin{bmatrix}
2 & 0 \\
four & 8 \\
6 &ten
\end{bmatrix}
\end{align*}
It follows that nosotros have
\begin{align*}
A&=\begin{bmatrix}
2 & 0 \\
4 & 8 \\
six &ten
\cease{bmatrix}\begin{bmatrix}
1 & ane\\
one& -1
\end{bmatrix}^{-1}\\[6pt] &=\begin{bmatrix}
2 & 0 \\
four & 8 \\
6 &10
\end{bmatrix}\frac{i}{ii}\brainstorm{bmatrix}
1 & 1\\
1& -ane
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
1 & ane \\
half dozen & -2 \\
8 &-2
\end{bmatrix}.
\stop{marshal*}

We at present be able to discover $T(\mathbf{x})$ as follows.
Nosotros have
\begin{marshal*}
T(\mathbf{x})&=A\mathbf{x}\\[6pt] &=\begin{bmatrix}
1 & 1 \\
6 & -2 \\
8 &-2
\end{bmatrix}\begin{bmatrix}
10 \\
y
\cease{bmatrix}\\[6pt] &=\begin{bmatrix}
x+y \\
6x-2y \\
8x-2y
\end{bmatrix},
\finish{align*}
which is, of course, the same formula that we obtained in solution one.

A similar problem for a linear transformation from $\R^3$ to $\R^3$ is given in the post "Make up one's mind linear transformation using matrix representation".

Instead of finding the changed matrix in solution 1, we could have used the Gauss-Hashemite kingdom of jordan elimination to find the coefficients.
See the post "Give a formula for a linear transformation if the values on basis vectors are known" for a similar trouble and its solution using this alternative method.

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