Kernel & Image of Linear Transformations Watch

Noble.
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#1
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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.

V = \mathbb{M}_{nxn}(\mathbb{R}), and  T : V \rightarrow \mathbb{R} is given by T(A) = tr(A) = \displaystyle\sum_{i=1}^n a_{ii} for A = [a_{ij}] \in V


dim(V) = n^2

ker(T) = \left\{A \in V : T(A) = 0\right\} = \left\{A \in V : tr(A) = 0 \right\}

Since tr(A) = 0 we have that a_{11} + a_{22} + ... + a_{nn} = 0 and thus we can write one element of the main diagonal as a linear combination of the rest, i.e.

 a_{11} = -(a_{22} + a_{33} + ... + a_{nn})

Thus  \ dim(ker(T)) = n(T) = n^2 - 1

im(T) = \left\{w \in \mathbb{R} : T(A) = w \ \mathrm{for\ some} \ A \in V\right\} 

= \left\{w \in \mathbb{R} : tr(A) = w \ \mathrm{for\ some} \ A \in V\right\}

Clearly  \ dim(im(T)) \leq 1 as \ im(T) \subseteq \mathbb{R}

Since A \in V the trace of A is not necessarily zero and hence

dim(im(T)) = 1


Thanks!
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DFranklin
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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.

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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 A \in V". What is A supposed to be here? You need to be more specific.
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Noble.
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(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 A \in V". 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 A \in V" 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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Noble.
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(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 A \in V". 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 a_{ii}, a_{jj} that both can be written as a combination of the other elements in the main diagonal then a_{ii} = a_{jj} = 0?
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james22
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(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 a_{ii}, a_{jj} that both can be written as a combination of the other elements in the main diagonal then a_{ii} = a_{jj} = 0?
You could, more simply, use the rank nulity formula and use your value for the dimension of the image.
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Noble.
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(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
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Mark13
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(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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