Find a Formula for a Linear Transformation

Purdue University Linear Algebra Exam Problems and Solutions

Problem 36

If $L:\R^2 \to \R^3$ is a linear transformation such that
\begin{align*}
L\left( \begin{bmatrix}
1 \\
0
\end{bmatrix}\right)
=\begin{bmatrix}
1 \\
1 \\
2
\end{bmatrix}, \,\,\,\,
L\left( \begin{bmatrix}
1 \\
1
\end{bmatrix}\right)
=\begin{bmatrix}
2 \\
3 \\
2
\end{bmatrix}.
\end{align*}
then

(a) find $L\left( \begin{bmatrix}
1 \\
2
\end{bmatrix}\right)$, and

(b) find the formula for $L\left( \begin{bmatrix}
x \\
y
\end{bmatrix}\right)$.

 

If you think you can solve (b), then skip (a) and solve (b) first and use the result of (b) to answer (a).

(Part (a) is an exam problem of Purdue University.)

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Hint.

  1. Express $\begin{bmatrix}
    1 \\
    2
    \end{bmatrix}$ as a linear combination of $\begin{bmatrix}
    1 \\
    0
    \end{bmatrix}$ and $\begin{bmatrix}
    1 \\
    1
    \end{bmatrix}$.
  2. Use the linearity of linear transformation $L$.
  3. Same for part (b). Replace $\begin{bmatrix}
    1 \\
    2
    \end{bmatrix}$ with the general vector $\begin{bmatrix}
    x \\
    y
    \end{bmatrix}$.

Solution.

(a) Find $L\left( \begin{bmatrix}
1 \\
2
\end{bmatrix}\right)$

Note that the vectors $\begin{bmatrix}
1 \\
0
\end{bmatrix}$ and $\begin{bmatrix}
1 \\
1
\end{bmatrix}$ are a basis of $\R^2$.
We first express $\begin{bmatrix}
1 \\
2
\end{bmatrix}$ as a linear combination of the vectors $\begin{bmatrix}
1 \\
0
\end{bmatrix}$ and $\begin{bmatrix}
1 \\
1
\end{bmatrix}$.
Let $\begin{bmatrix}
1 \\
2
\end{bmatrix}
=c_1 \begin{bmatrix}
1 \\
0
\end{bmatrix} +
c_2 \begin{bmatrix}
1 \\
1
\end{bmatrix}
=
\begin{bmatrix}
c_1+c_2 \\
c_2
\end{bmatrix}$.
Solving this, we have $c_1=-1$ and $c_2=2$.

Then we calculate
\begin{align*}
L\left( \begin{bmatrix}
1 \\
2
\end{bmatrix}\right)
& =L\left( – \begin{bmatrix}
1 \\
0
\end{bmatrix} +
2 \begin{bmatrix}
1 \\
1
\end{bmatrix}
\right)
= -L\left( \begin{bmatrix}
1 \\
0
\end{bmatrix} \right) +
2 L \left(\begin{bmatrix}
1 \\
1
\end{bmatrix}
\right) \\
&= -\begin{bmatrix}
1 \\
1 \\
2
\end{bmatrix}
+2 \begin{bmatrix}
2 \\
3 \\
2
\end{bmatrix}
=\begin{bmatrix}
3 \\
5 \\
2
\end{bmatrix},
\end{align*}
where the second equality follows from the linearity of $L$.

Thus we have
\[L\left( \begin{bmatrix}
1 \\
2
\end{bmatrix}\right)=\begin{bmatrix}
3 \\
5 \\
2
\end{bmatrix}.\]

(b) Find the formula for $L\left( \begin{bmatrix}
x \\
y
\end{bmatrix}\right)$.

 We generalize the proof of (a).
Let $\begin{bmatrix}
x \\
y
\end{bmatrix}
=c_1 \begin{bmatrix}
1 \\
0
\end{bmatrix} +
c_2 \begin{bmatrix}
1 \\
1
\end{bmatrix}
=
\begin{bmatrix}
c_1+c_2 \\
c_2
\end{bmatrix}$ be a linear combination. Solving this, we have $c_1=x-y, c_2=y$.

Hence the linear combination is
\[ \begin{bmatrix}
x \\
y
\end{bmatrix}
=(x-y) \begin{bmatrix}
1 \\
0
\end{bmatrix} +
y \begin{bmatrix}
1 \\
1
\end{bmatrix}.
\] Using the linearity of $L$, we compute
\begin{align*}
L\left( \begin{bmatrix}
x \\
y
\end{bmatrix}\right)
&=
L\left( (x-y) \begin{bmatrix}
1 \\
0
\end{bmatrix} +
y \begin{bmatrix}
1 \\
1
\end{bmatrix}\right)
=(x-y) L\left( \begin{bmatrix}
1 \\
0
\end{bmatrix} \right) +
y L\left( \begin{bmatrix}
1 \\
1
\end{bmatrix}\right) \\
&=(x-y) \begin{bmatrix}
1 \\
1 \\
2
\end{bmatrix}+y \begin{bmatrix}
2 \\
3 \\
2
\end{bmatrix}
=
\begin{bmatrix}
x+y \\
x+2y \\
2x
\end{bmatrix}.
\end{align*}
Therefore the formula is
\[ L\left( \begin{bmatrix}
x \\
y
\end{bmatrix}\right)
= \begin{bmatrix}
x+y \\
x+2y \\
2x
\end{bmatrix}.\]


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