Linear algebra Q

[mang73]

Could someone help me here please?

 

Let V and W be two vector spaces and T: V -> W be a linear map. Define

 

T(inv) (0) = { u element of V l T(u) = 0 }

 

where 0 is the zero element in W. Also define,

 

T(V) = { T(u) l u element of V},

 

the image of V under T. Show that T(inv) (0) is a subspace of V and T(V) is a subspace of W.

 

Your help is much appreciated!

[Qfwfq]

These two definitions are often called the kernel of T, ker T, and the image of T, img T.

 

Let's show that ker T is a subspace of V, using linearity of T it isn't hard. 0 belongs to ker T because T(0) = 0, trivial. Let's show that if u and v belong to ker T so does au + bv. The hypothesis means T(u) = 0 and T(v) = 0, so by linearity au + bv = a0 + b0 = 0, so au + bv belongs to ker T.

 

Now for img T. Does 0 belong to it? This is also trivial, using again T(0) = 0. If u and v both belong to img T, does au + bv too? The hypothesis means there will be two elements in V that map to u and v. The linear combination of these belongs to V and its image through T will, by linearity, be the corresponding linear combination of u and v which, therefore, belongs to img T.

 

Helpful? Or is some detail not too clear?

[maddog]

Could someone help me here please?

 

Let V and W be two vector spaces and T: V -> W be a linear map. Define

 

T(inv) (0) = { u element of V l T(u) = 0 }

 

where 0 is the zero element in W. Also define,

 

T(V) = { T(u) l u element of V},

 

the image of V under T. Show that T(inv) (0) is a subspace of V and T(V) is a subspace of W.

 

Your help is much appreciated!

I liked Qfwfq explanation quite well. I was impressed that he got you were looking for ker T.

 

I see you have V, W two vector space where T: V -> W is a linear map. Next line I get confused.

Do you mean

 

T^ -1 (0) = { u element of V | T(u) = 0 } or that T(inv) (0) means T(0) is an involution ? I wasn't sure.

 

Maddog

[mang73]

Hey thanks for the detailed reply. I am just confused as to is this the proof itself or is it guidlines and I need to inspect it further? :xx:

[mang73]

yes inverse= ^ -1

[Qfwfq]

Do you mean

 

T^ -1 (0) = { u element of V | T(u) = 0 } or that T(inv) (0) means T(0) is an involution ? I wasn't sure.

Actually, without being fussy, it seemed obvious that T(inv) (0) meant the anti-image of 0.

 

Strictly, T isn't invertible if this anti-image has more than the null vector but the idea was clear enough, you can consider it a thing similar to the inverse; it's kind of the idea behind Green functions.

[Qfwfq]

I'm glad you found it helpful...

is this the proof itself or is it guidlines and I need to inspect it further?

I didn't want to be too prolix. It can be said slightly more in detail, especially the second part. To pass a test you would probably need to but it can be a matter of working it out. Choose names for the "two elements in V that map to u and v" like:

 

T(p) = u

T(q) = v

 

then you can write T(ap + bq) = aT(p) + bT(q) by linearity. This is the image of an element of V and it is also au + bv. Done along these lines it should be good enough to pass a test. :xx: