Kernel and Imag

[CannedGib]

Hello, can anyone help me with this:

T:R2-->R2, T(a,b)=(a,a+b)

According to myself this can´t be done, hence, de kernel is 0 and the image 1. And this two other problems I have no idea how to solve them

L: R3-->R, T(a,b,c)=0
P:M(2x2)-->M(2x2), P([a c, b d]) =[c d, a b]

Thanks

 

[stapel]

Hello, can anyone help me with this:

T:R2-->R2, T(a,b)=(a,a+b)

Would it be correct to assume, from the subject line, that the "this" with which you are requesting assistance is finding the kernel and image of this linear transformation?

According to myself this can´t be done, hence, de kernel is 0 and the image 1.

Okay; from the above, it seems that perhaps you're not needing to find the kernel or image; instead, you're needing to determine something else. But I can't guess what it is that you're saying "can't be done", since obviously the transformation itself is linear and well-defined.

So please reply with the full and exact text of the exercise, the complete instructions, and a clear listing of your efforts and results, specifying at what point or in what manner "this" "can't be done". Thank you!

 

[CannedGib]

Find the kernel, image and dimension of the following linear transformations and determine which ones are injective or subjective.

That´s what the asking for. Thanks for the reply.

 

[stapel]

Find the kernel, image and dimension of the following linear transformations and determine which ones are injective or subjective.

Since you posted only one transformation, I'll guess that, by "determine which ones are", you mean "determine if this one is".

Please reply with a clear listing of your thoughts and efforts so far. In particular, please show all of the steps which led to your "can't be done" conclusion. Thank you!

 

[Limax]

Since you posted only one transformation

He actually posted three.

Find the kernel, image and dimension of the following linear transformations and determine which ones are injective or subjective.

These are all related:

• A linear transformation \(\displaystyle T:X\to Y\) is injective if and only if its kernel is \(\displaystyle \{0_X\}\).
• It is surjective if and only if its image is \(\displaystyle Y\).
• Its dimension is the dimension of its image.

The kernel of the transformation is \(\displaystyle \{x\in X\,:\,T(x)=0_Y\}\). Its image is \(\displaystyle \{y\in Y\,:\,T(x)=y\ \text{for some}\ x\in X\}\).

With these in mind, let's look at your first transformation.

T:R2-->R2, T(a,b)=(a,a+b)

According to myself this can´t be done, hence, de kernel is 0 and the image 1.

You're right that the kernel is \(\displaystyle \{0\}\) – or more correctly \(\displaystyle \{(0,0)\}\) – but you must explain why, not just make a lucky guess. In other words, explain why T is injective.

However the image is not "1". Try again. (Hint: Is it surjective?)