# Linear Transformation - Basis and dimension of $f(V)$

I'm learning linear algebra, specifically linear transformations, and need help with the following exercise:

Consider the linear transformation $f:\mathbb{R^{3}} \to \mathbb{R^{2}}$ with $f(x,y,z)=(x+y,z)$. Let $V=\{(x,y,z) \in \mathbb{R^{3}}: x+y+z =0\}$ and $W=\{(x,y) \in \mathbb{R^{2}}: 2x =y\}.$ Find a basis and the dimension of $f(V)$ and $f^{-1}(W)$, respectively.

I honestly don't know what to do here. If we consider the standard basis $\mathcal{B}=\{(1,0,0), (0,1,0), (0,0,1)\}$ of $\mathbb{R^{3}}$, since $f$ is a linear transformation, it can be represented by a matrix $A$ whose columns are the images of the standard basis vectors. We have

$$f(1,0,0) = (1,0), \quad f(0,1,0) = (1,0), \quad f(0,0,1) = (0,1).$$

Hence, the matrix representation of $f$ with respect to the basis $\mathcal{B}$ is given by

$$A = \begin{pmatrix}1 & 1 & 0\\0 & 0 & 1\end{pmatrix}.$$

In general, if $f$ is invertible, then the matrix representing the mapping $f^{-1}$ must be the matrix inverse of $A$. But here the $A$ is clearly not invertible.

I think I have not understand the theory well enough for this problem. Maybe someone can point out to me the right direction/steps to follow to solve this problem.

• $f^{-1}(W)$ is the set of vectors $v\in\mathbb R^3$ such that $f(v)\in W$. The map doesn’t have to be invertible for you to find this pre-image of $W$.
– amd
Commented Feb 7, 2017 at 23:46
• Hint for the first part: Find a basis for $V$. Since $f$ is linear, it maps every linear combination of those basis vectors into a corresponding linear combination of their images. For the second part, you’ll need to solve some linear equations to determine which vectors get mapped to elements of $W$.
– amd
Commented Feb 7, 2017 at 23:49

Since a basis for $V$ is given by $u_1=(-1,0,1)$ and $u_2=(0,-1,1)$ (can you work it out?), the subspace $f(V)$ is spanned by $f(u_1)=(-1,1)$ and $f(u_2)=(-1,1)$.
The set $f^{-1}(W)$ is characterized by $$f(x,y,z)=(x+y,z)\in W$$ that is, $2(x+y)=z$ or $2x+2y-z=0$. Can you find a basis?
• Thank you for your answer. We have that $f(V) = \operatorname{Span}\{ f(u_1)=(-1,1), f(u_2)=(-1,1)\} = \operatorname{Span}\{ f(u_1)=(-1,1)\}$. Therefore, a basis for $f(V)$ is given by $\mathcal{B'}=\{(-1,1)\}$ and $\dim (f(V)) = 1$. Similarly, a basis for $f^{-1}(W)$ is given by $\mathcal{B^{*}}=\{(1,0,2),(0,1,2)\}$ so $\dim (f(W)) = 2$. Is that correct? It is still unclear to me how you found that the set $f^{-1}(W)$ is characterized by $2x+2y-z=0$. Would you mind explaining this to me in more details?
• @Elix A vector $(a,b)$ is in $W$ if and only if $2a=b$; apply this to $f(v)$, where $v=(x,y,z)$. Commented Feb 8, 2017 at 9:01