Show map is not a isomorphism

Question

Consider $V= \left \{ v= \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} : x_1-x_2+2x_3=0\right \}\subset K^3$ and the linear map $f:V \rightarrow K^2$ defined by

$f\begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix}=\begin{pmatrix} x_1-x_3\\ x_2-3x_3 \end{pmatrix}$ ($K$ is just a field)

Show that $f:U \rightarrow K^2$ is not a isomorphism

My try

I have only been told what a isomorphism is in a general sense; that it has to be "the same" i.e some structure is still the same. For vectors I guess it could be addition, multiplication and the null vector? I am thinking $f$ must be bijective as well. Other than that I have not been taught much more than "it is the same" – which really does not help me in the slightest.

Answer 1

Hint:

By the rank-nullity theorem, $V$ has dimension $2$, like $K^2$. Hence it is enough to prove that $\ker f$ is non- trivial to prove $f$ is non-injective, and a fortiori non-bijective. This means you have to find a non-zero vector $v\in K^3$ such that \begin{cases} x_1-x_2+2x_3=0,\\ x_1-x_3=0,\\ x_2-3x_3=0. \end{cases} You can check that the kernel of $\varphi:K^3\longrightarrow K^2, \enspace\begin{pmatrix}x_1\cr x_2\cr x_3\end{pmatrix}\longmapsto\begin{pmatrix}x_1-x_3\cr x_2-3x_3\end{pmatrix}$ ($\,f$ is its restriction to $V$) is actually contained in $V$, so you only have to determine a basis of $\ker \varphi$.

Answer 2

Yes, what you have been thinking of, is correct. Here, the map $f$ is between the vector space $V$ and the field $K^2$, hence, we are asked to show that $f$ is an isomorphism of vector spaces. (I leave it to you to check that $V$ is indeed a vector space)

Now, an isomorphism between vector spaces is nothing but an injective and a subjective map, i.e. a bijection between them. Here, it is enough to show that it is not either one of them.

Hint: Show that $f$ is not injective. Can you use the relation between $ x_1,x_2,$ and $x_3$ in the definition of $V$, to write the image of $f$ as something that gives you this?

Answer 3

Note that $(1,1,0),(-2,0,1)\in V$, that $f(1,1,0)=(1,1)$, and that $f(-2,0,1)=(-3,3)$. Therefore, $f(-3,-3,0)=f(-2,0,1)\bigl(=(-3,3)\bigr)$ and so $f$ is not injective. It follows that $f$ is not an isomorphism.

Answer 4

Your notion is actually correct. Observe that for all $x \in V, x_1 - x_2 + 2 x_3 = 0$ which implies $x_1 - x_3 = x_2- 3 x_3 $ . Thus, this map is not surjective as well.

Answer 5

Obviously $V$ is the null space of the rank-1 matrix $[1 \hspace{0.3cm} -1 \hspace{0.3cm} 2]$. So $V$ is 2 dimensional vector space and any linearly independent two vectors from $V$ can be a basis. Select $v=(1,1,0)$ and $w=(0,2,1)$ in $V$. They are lin independent so $\{v,w\}$ can be a basis of $V$.

$\textbf{Fact:}$ $\{fv,fw\}$ can be a basis for $K^2$ iff $f$ is isomorphism.

But $fv=f(1,1,0)=(1,1)$ and $fw=f(0,2,1)=(-1,-1)$.So they are linearly $\textbf{dependent}$ and they cannot be a basis. By the fact above we showed $f$ is not be an isomorphism.