# Kernel & Image of Linear TransformationsWatch

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#1
I just wanted to check if this was correct.

Describe the kernel and image of each of the following linear transformations, and in each case find the rank and nullity.

, and is given by for

Since we have that and thus we can write one element of the main diagonal as a linear combination of the rest, i.e.

Thus

Clearly as

Since the trace of is not necessarily zero and hence

Thanks!
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6 years ago
#2
Your proof that dim(ker T)) = n^2 - 1 doesn't entirely convince me. (Because, for example, if you could write two elements of the main diagonal as a combination of the rest, the dimension would only be n^2 - 2, and you haven't proved you can't do that). So you've only really shown the dimension is <= n^2 - 1. So for completeness you should also argue the other way to show the dimension is at least n^2 - 1.

Spoiler:
Show
I would do something like: There are n^2-1 elements other than the bottom right corner. If we choose these arbitrarily, we can choose a_nn s.t. Tr(A) = 0. So the kernel has dimension at least n^2 - 1

I don't know what you mean by "since ". What is A supposed to be here? You need to be more specific.
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#3
(Original post by DFranklin)
Your proof that dim(ker T)) = n^2 - 1 doesn't entirely convince me. (Because, for example, if you could write two elements of the main diagonal as a combination of the rest, the dimension would only be n^2 - 2, and you haven't proved you can't do that). So you've only really shown the dimension is <= n^2 - 1. So for completeness you should also argue the other way to show the dimension is at least n^2 - 1.

Spoiler:
Show
I would do something like: There are n^2-1 elements other than the bottom right corner. If we choose these arbitrarily, we can choose a_nn s.t. Tr(A) = 0. So the kernel has dimension at least n^2 - 1

I don't know what you mean by "since ". What is A supposed to be here? You need to be more specific.
Ah right yeah, I neglected to consider the possibility the dimension could be <= n^2 -1. By "since " I just meant because A is an nxn matrix, the trace of it doesn't necessarily equal zero, to try and rule out the dimension of the image being 0, since the dimension is either 0 or 1.
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#4
(Original post by DFranklin)
Your proof that dim(ker T)) = n^2 - 1 doesn't entirely convince me. (Because, for example, if you could write two elements of the main diagonal as a combination of the rest, the dimension would only be n^2 - 2, and you haven't proved you can't do that). So you've only really shown the dimension is <= n^2 - 1. So for completeness you should also argue the other way to show the dimension is at least n^2 - 1.

Spoiler:
Show
I would do something like: There are n^2-1 elements other than the bottom right corner. If we choose these arbitrarily, we can choose a_nn s.t. Tr(A) = 0. So the kernel has dimension at least n^2 - 1

I don't know what you mean by "since ". What is A supposed to be here? You need to be more specific.
Also, in regards to showing the dimension is at least n^2 - 1, is it sufficient to show that if we have that both can be written as a combination of the other elements in the main diagonal then ?
0
6 years ago
#5
(Original post by Noble.)
Also, in regards to showing the dimension is at least n^2 - 1, is it sufficient to show that if we have that both can be written as a combination of the other elements in the main diagonal then ?
You could, more simply, use the rank nulity formula and use your value for the dimension of the image.
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#6
(Original post by james22)
You could, more simply, use the rank nulity formula and use your value for the dimension of the image.
Yeah, which is why I stated the dimension of V at the start, but I feel it's cheating a bit when they say to describe the ker(T) (and the nullity) and I just give the definition of the kernel and say n(T) = dim(V) - r(T) = n^2 - 1
0
6 years ago
#7
(Original post by Noble.)
Yeah, which is why I stated the dimension of V at the start, but I feel it's cheating a bit when they say to describe the ker(T) (and the nullity) and I just give the definition of the kernel and say n(T) = dim(V) - r(T) = n^2 - 1
The easiest way to show directly that the kernel has dimension at least n^2-1 is what DFranklin suggested above - you can freely specify all elements of the matrix apart from a_1,1, and there is a choice of a_1,1 for which the matrix will lie in the kernel of the map. This shows the kernel has a subspace isomorphic to R^{n^2-1}, and so the kernel itself has dimension at least n^2-1.
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