Kernel & Image of Linear Transformations

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\} [br]= \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!

Reply 1

DFranklin

18

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

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.

Reply 2

Noble.

OP

17

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

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.

Reply 3

Noble.

OP

17

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

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

?

Reply 4

james22

16

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.

Reply 5

Noble.

OP

17

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

Reply 6

Mark13

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.

Kernel & Image of Linear Transformations

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