Finding $Im(T)$ and $Ker(T)$ of the following linear transformation

Question

Let $$T:\mathbb{R}^4\to\mathbb{R}^3$$

$$T(x,y,z,w)=(x-y+z-w,x+y,z+w)$$

I need to find $\operatorname{Ker}(T),\operatorname{Im}(T)$ and the basis of them and to show if $T$ is is one-to-one and if it onto $\mathbb{R}^3$

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I'm having hard time finding $\operatorname{Ker}(T),\operatorname{Im}(T)$ and I don't know if I did is correct.

So this is what I did :

for $\operatorname{Ker}(T)$ I need to find null space of $A=\begin{pmatrix}1&-1&1&-1\\ 1&1&0&0\\ 0&0&1&1\end{pmatrix}...\to\begin{pmatrix}1&-1&1&-1\\ 0&2&-1&1\\ 0&0&1&1\end{pmatrix}$

$...$ so we get $\operatorname{Ker}(T)=\operatorname{span}\{(1,-1,-1,1)\}$ and this is also the basis of $\operatorname{Ker}(T)$,$\dim(\operatorname{Ker}(T))=1$ and because $\operatorname{Ker}(T)\ne \operatorname{span}\{(0,0,0,0)\}$ then $T$ is not one-to-one.

To find $\operatorname{Im}(T)$ what I did I found column space of $A'=\begin{pmatrix}1&1&0\\ -1&1&0\\ 1&0&1\\ 1&0&1\end{pmatrix}...\to \begin{pmatrix}2&0&0\\ 0&2&0\\ 0&0&2\\ 0&0&0\end{pmatrix}$

... so $\operatorname{Im}(T)=\operatorname{span}\{(0,0,0)\}$ and this is the basis so $\dim(\operatorname{Im}(T))=1$ and because $\operatorname{Im}(T)\ne \mathbb{R}^3,$ $T$ is not onto.

Is what I did correct , if not what is wrong?

Thanks a lot

Answer 1

You have found that $$A=\begin{pmatrix}1&-1&1&-1\\ 1&1&0&0\\ 0&0&1&1\end{pmatrix}...\to\begin{pmatrix}1&-1&1&-1\\ 0&2&-1&1\\ 0&0&1&1\end{pmatrix}$$Note that the Row Space gives you the rank of $A$ and in your last matrix you have three linearly independent vectors.

That implies the $ Rank(A)=3$ which is the dimension of the $Im(A)$.

Thus the image of $ A$ is a three dimensional subspace of $R^3$ which is $R^3$.

In order to find the $Ker(A)$ , you solve, $AX=0$ which implies $X= t(1,-1,-1,1)^T$.

Therefore the $Ker(A)$ is the one dimensional subspace of $R^4$ spanned by $ t(1,-1,-1,1)^T$.

Answer 2

Some extra comments:

If you have that $\operatorname{Im}(T)=\operatorname{span}\{(0,0,0)\}$ that does not mean that $\dim(Im) = 1$. That would be zero dimension.

As pointed in one of the answers, the rank of $A$ tells you the dimension of $Im(A)$. In this case, since we have dimension $3$ and $Im(A) \subseteq \mathbb{R}^3$ then any basis for $\mathbb{R}^3$ is going to work.

Finally, since $T$ is defined between $\mathbb{R}^4$ and $\mathbb{R}^3$ (different dimension) we cannot have a one-to-one correspondence. That is because when you reduce an $m\times n$ matrix (with $m \ne n$) you can't get the identity matrix. That might be helpful so you can check if your answers make sense :)