# Kernel & Image of Linear Transformations Watch

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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!

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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#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.

I don't know what you mean by "since ". What is A supposed to be here? You need to be more specific.

Spoiler:

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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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(Original post by

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.

I don't know what you mean by "since ". What is A supposed to be here? You need to be more specific.

**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.

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**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.

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#5

(Original post by

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 ?

**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 ?

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(Original post by

You could, more simply, use the rank nulity formula and use your value for the dimension of the image.

**james22**)You could, more simply, use the rank nulity formula and use your value for the dimension of the image.

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#7

(Original post by

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

**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

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