Kernel on linear transformation proof

[Cristian1]

hi guys :D

im having trouble with this proof, any hints?

let V be a vector space over a field F and let T1, T2: V--->V be linear transformations

prove that

 

[GJA]

Hi Cristian,

If \(\displaystyle v\in \text{ker}(T_{1})\cap\text{ker}(T_{2}),\)

then what can we say about \(\displaystyle T_{1}(v)\) and \(\displaystyle T_{2}(v)\)? With this question answered, we then look at

\(\displaystyle (T_{1}+T_{2})(v)=T_{1}(v)+T_{2}(v)=\ldots\)

I'm trying to give you a few hints so you can fill in the gaps on your own. Let me know if you're still confused/unclear on something.

 

[Cristian1]

thank you!
I see the point but don't see clearly the difference between the zero of the intersection and the zero of the sum

 

[GJA]

I will try to address what I think may be the confusion; if I have misunderstood you, just let me know.

My understanding of your question is that there is a perceived difference in the zero of the intersection and the zero of the sum. This is not the case: there is a single vector space, \(\displaystyle V,\) and only one zero element of \(\displaystyle V.\)

What we are trying to show is ultimately set-theoretic in nature: we want to show that if an element \(\displaystyle v\in\bigcap_{i=1}^{n}\text{ker}(T_{i}),\) then \(\displaystyle v\in \text{ker}\left(\sum_{i=1}^{n}T_{i} \right).\) Let's break down what all this means:

To say that a vector belongs to the kernel of a linear transformation means that the linear transformation sends that vector to zero. So, by assuming \(\displaystyle v\in\bigcap_{i=1}^{n}\text{ker}(T_{i})\) we are saying that \(\displaystyle v\) makes all of the linear transformations zero; i.e.

\(\displaystyle T_{1}(v)=0,~ T_{2}(v)=0,\ldots, T_{n}(v)=0\)

In each case above the zero is the zero element of \(\displaystyle V,\) they are not different zeros. Now, to show that \(\displaystyle v\in \text{ker}\left(\sum_{i=1}^{n}T_{i} \right),\) we must show that when we plug \(\displaystyle v\) into the function/linear transformation \(\displaystyle \sum_{i=1}^{n}T_{i}\) we still get zero. For this we compute:

\(\displaystyle \left( \sum_{i=1}^{n}T_{i}\right)(v)=\sum_{i=1}^{n}T_{i}(v)=T_{1}(v)+T_{2}(v)+\ldots+T_{n}(v)=0+0+\ldots+0=0\)

Hence, \(\displaystyle v\in\text{ker}\left(\sum_{i=1}^{n}T_{i} \right).\) Since \(\displaystyle v\in\bigcap_{i=1}^{n}\text{ker}(T_{i})\) was arbitrary, the proof is finished.

Let's take an example to help us along. Let our vector space be

\(\displaystyle V=\mathbb{R}^{3}=\left\{\begin{bmatrix}x\\ y\\ z \end{bmatrix}: x, \, y, \, z\in\mathbb{R} \right\}\)

and our linear transformations \(\displaystyle T_{1}\) & \(\displaystyle T_{2}\) be given by

\(\displaystyle T_{1}\left(\begin{bmatrix}x\\y\\z \end{bmatrix} \right)=\begin{bmatrix}0\\y\\0 \end{bmatrix}\)

and

\(\displaystyle T_{2}\left(\begin{bmatrix}x\\y\\z \end{bmatrix} \right)=\begin{bmatrix}x\\0\\0 \end{bmatrix}\)

The zero element in this example is \(\displaystyle \begin{bmatrix}0\\0\\0\end{bmatrix}\)

Notice that \(\displaystyle T_{1}\left(\begin{bmatrix}x\\y\\z \end{bmatrix} \right)=\begin{bmatrix}0\\0\\0\end{bmatrix}\)

if and only if \(\displaystyle y=0.\) Hence, \(\displaystyle \text{ker}(T_{1})=xz\) plane. Similarly, \(\displaystyle \text{ker}(T_{2})=yz\) plane. The intersection of these two kernels is the entire \(\displaystyle z\) axis; i.e.

\(\displaystyle \bigcap_{i=1}^{2}\text{ker}(T_{i})=\left\{\begin{bmatrix}0\\0\\z \end{bmatrix}: z\in\mathbb{R} \right\}\)

(Drawing a picture of the two planes to see that their intersection is the \(\displaystyle z\) axis may be helpful)

Now, if we wish to demonstrate the general proof you're working on in this example, we take an element \(\displaystyle \begin{bmatrix}0\\0\\z\end{bmatrix}\in \bigcap_{i=1}^{2}\text{ker}(T_{i})\) and compute using the definitions of \(\displaystyle T_{1}\) & \(\displaystyle T_{2}\)

\(\displaystyle \left(T_{1}+T_{2}\right)\left(\begin{bmatrix}0\\0\\z\end{bmatrix} \right)=T_{1}\left(\begin{bmatrix}0\\0\\z\end{bmatrix} \right)+T_{2}\left(\begin{bmatrix}0\\0\\z\end{bmatrix} \right)=\begin{bmatrix}0\\0\\0 \end{bmatrix} + \begin{bmatrix}0\\0\\0 \end{bmatrix} = \begin{bmatrix}0\\0\\0 \end{bmatrix}
\)

Hence, \(\displaystyle \begin{bmatrix}0\\0\\z\end{bmatrix}\in\text{ker}\left(\sum_{i=1}^{2}T_{i} \right)\) as well. Thus, for this specific example,

\(\displaystyle \bigcap_{i=1}^{2}\text{ker}(T_{i})\subseteq \text{ker}\left(\sum_{i=1}^{2}T_{i} \right),\)

as it should be from your general exercise.

This is a long post, but I hope I have understood and addressed your concern. Let me know if anything is unclear/not quite right.

 

[Cristian1]

thank you very much! :)