help real analysis

[knotty radish]

Does anyone know how to do cii)
Help

[midnight patio]

1. Ask your question and show the work you've done so far. If you've posted a screenshot of a question, specify which part you need help with.
2. Wait patiently for a helper to come along.
3. Once someone helps you, say thank you and close the thread with:

   +close
   

4. Feel free to nominate the person for helper of the week in #helper-nominations
5. Do not ping the mods, unless someone is breaking the rules.
6. If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[pliant zenith]

What dimension is the image ?

Looking at it it seems the rank of the matrice is 2?

If so using the rank theorem you have the dimension of the kernel

[knotty radish]

Sorry typo ciii

Very sorry

[pliant zenith]

Oh okay well f here is clearly a projector

[pliant zenith]

Np

I think a certain type of projector works here

Think of one

[knotty radish]

Orthogonal?

[pliant zenith]

Yes

Orthogonal projection onto ker(g) seems to work

You have $im(f)=ker(g)$ and $ker(f)=ker(g)^{\perp}$

[exotic wasp]

Rotor 😑

[knotty radish]

So what does this mean?

[pliant zenith]

Well if f is an orthogonal projection onto ker(g) you have f^2=f and im(f)=ker(g)

And it maps all elements of im(g) to ker(g) ( it can’t be an isomorphic map considering the dimension on V is uneven)

In theory any projector onto ker(g), orthogonal or not works but considering we are working in a Euclidean space why not take the orthogonal projector

You also know what the projector is in the orthogonal case

If $v$ is a vector spanning $ker(g)$ then $x \to \frac{\langle x|v \rangle}{||v||^{2}} v$

[exotic wasp]

Rotor 😑

[pliant zenith]

is the orthogonal projection considering the dimension of ker(g) is 1

(Im assuming so I haven’t checked)

[knotty radish]

Firstly Is my part I and ii correct tho?

It’s not???

[pliant zenith]

Wait lemma think

[pliant zenith]

Sorry I was afk

But I think it’s false sorry to say $im(g) \neq ker(g)^{\perp}$

[exotic wasp]

Rotor 😑

[pliant zenith]

Take $v=(18,-15,9)^{T}$

[exotic wasp]

Rotor 😑

[pliant zenith]

Well $v \in ker(g)$ but $\langle v | (12,9,0)^{T} \rangle \neq 0$

[exotic wasp]

Rotor 😑

[pliant zenith]

But using the rank theorem you know that $ker(g)$ is if dimension 1 thus $ker(g)=Span(v)$

[exotic wasp]

Rotor 😑

[pliant zenith]

You have an equation of the plane $ker(g)^{\perp}$ from $v$ though

[exotic wasp]

Rotor 😑

[pliant zenith]

Which you can use to find a basis of $ker(g)^{\perp}$ then orthonormalise it using gram shmidt

[exotic wasp]

Rotor 😑

[knotty radish]

What is this?

[pliant zenith]

It’s ker(g) mb

I mistyped

Oh wait I’m sorry I misunderstood

Basically if u=(x,y,z) is a vector of $ker(g)^{\perp}$ you have that $\langle v|u \rangle=0$ from this you have a Cartesian equation of the plane

[exotic wasp]

Rotor 😑

[pliant zenith]

from which you can deduce a basis of $ker(g)^{\perp}$

[exotic wasp]

Rotor 😑

[pliant zenith]

Seems correct but in the matrice it’s a -5/3 on the last column

If you represent an orthogonal projection in the canonical basis in a matrice then it’s a symmetric matrice

Or any orthonormal basis for that matter, an orthogonal projection is self adjoint

Also if you want to quickly get a basis of the orthogonal of the kernel I’d recommend considering $I_3-F$

[exotic wasp]

Rotor 😑

[pliant zenith]

Which is the matrice of the orthogonal projector onto the orthogonal of the kernel