Problem: Let
\[
V = \left\{ (x,y,z)\in \mathbb{R} ^3: x+y+2z = 0 \right\}
\]
and
\[
W = \left\{ (x,y,z)\in \mathbb{R} ^3: x + y = 0 \right\}.
\]
Then
-
Find out an invertible linear transformation $T:\mathbb{R} ^3\to \mathbb{R} ^3$ which maps $V$ onto $W$.
-
Find out a non-invertible linear transformation $T:\mathbb{R} ^3 \to \mathbb{R} ^3$ such that $T(V) \cap W$ is not $\{0\}$ and $1$ is an eigenvalue of $T$.
Solution: Let us find a basis for the spaces $V$ and $W$. Since $V$ is determined by the equation $x+y + 2z = 0$, so
\begin{align*}
V & = \left\{ (x,y,z) \in \mathbb{R} ^3 : x + y + 2z = 0 \right\} \\
& = \left\{ (x,y,z) \in \mathbb{R} ^3 : x + y = -2z \right\} \\
& = \left\{ \left( x, y, \frac{-(x+y)}{2} \right) : x,y\in \mathbb{R} \right\} \\
& = \text{span} \left\{ \left( 2,0,-1 \right), \left( 0,2,-1 \right) \right\} .
\end{align*}
Similarly,
\begin{align*}
W & = \left\{ (x,y,z) \in \mathbb{R} ^3: x+ y = 0 \right\} \\
& = \left\{ (x,-x,z): x,z\in \mathbb{R} \right\} \\
& = \text{span}\left\{ (1,-1,0), (0,0,1) \right\}.
\end{align*}
In order to find the linear transformation we recall the following theorem,
Let $V$ and $W$ be vector spaces with dimensions $n$ and $m$ respectively. Let $\left\{ v_1,v_2,\dots,v_n \right\} $ be a basis of $V$ and $\left\{ w_1,w_2,\dots,w_n \right\} $ be any $m$ elements of $W$. Then there exists a unique linear transformation $T: V \to W$ such that $T(v_i) = w_i$ for $i=1,2,\dots,n$.
-
We want to find an invertible linear transformation $T:\mathbb{R} ^3 \to \mathbb{R} ^3$ which maps $V$ onto $W$. We take a basis of $\mathbb{R} ^3$
\[
\mathcal{B}_{\mathbb{R} ^3} = \left\{ v_1 = (2,0,-1)^T, v_2 = (0,2,-1)^T, v_3 = (1,0,0)^T \right\}
\]
and consider vectors
\[
\mathcal{B} _{\mathbb{R} ^3} = \left\{ w_1 = (1,-1,0)^T, w_2 = (0,0,1)^T, w_3 = (0,1,0)^T \right\}.
\]
Then using the above theorem, there exists a unique linear transformation $T:\mathbb{R} ^3 \to \mathbb{R} ^3$ such that
\[
T(v_1) = w_1, ~T(v_2) = w_2,~T(v_3) = w_3.
\]
Therefore, the above linear transformation has the property that $T(V)= W$. Also, note that $T$ is invertible because we took basis on both sides.
-
We will again apply the same theorem by considering the vectors:
\[
\mathcal{B}_{\mathbb{R} ^3} = \left\{ v_1 = (2,0,-1)^T, v_2 = (0,2,-1)^T, v_3 = (1,0,0)^T \right\}
\]
and
\[
\left\{ w_1 = (1,-1,0)^T, w_2 = (0,2,-1)^T, w_3 = (0,0,0)^T \right\}.
\]
Then the linear transformation with the property
\[
T(v_1) = w_1, T(v_2) = w_2 = v_2, \text{ and } T(v_3) = w_3
\]
has the desired property.