Linear Transformation

[tall brook]

could someone help me with the second Question it ask me to find the Basis of the Kernel and the Image.

[umbral nebula]

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[celest kestrel]

easiest to take the standard basis 1,x,x^2 and express the matrix of f with respect to this basis

@tall brook

basis of the kernel is obtain by solving the homogenous system

and the image is the span of the column vectors of the matrix of f

$$ f(1) = f(1\cdot 1 + 0\cdot x + 0\cdot x^2) = 4 $$

for instance

[lofty kernel]

aL

[celest kestrel]

so the first column of the matrix of f is (4, 0, 0)

[tall brook]

hmm i understand. but i don't understand how did you become (4, 0, 0) 🙏

[celest kestrel]

what are the coordinates of f(1) with respect to basis 1,x,x^2?

[tall brook]

(1, 0 , 0) and also (1, 1, 1) ?

[celest kestrel]

coordinates are unique for a given basis

neither is correct

[tall brook]

okay i'll try now to understand and i'll let you know 🙏

[celest kestrel]

$$ f(x) = f(0 + 1x + 0x^2) = (2\cdot 1 + 0 )x^2 + (4\cdot 0+2\cdot 0) = 2x^2 = 0 + 0x + 2x^2 $$

[lofty kernel]

aL

[celest kestrel]

@tall brookso the coordinates of f(x) are (0,0,2), see if you can figure out what the coordinates of f(x^2) are

[tall brook]

but this one is not related to the gevin function. i think

it's just ()x NOT ()x^2

in the funczion

[celest kestrel]

this is how f is defined

[tall brook]

yes

just a moment i think i get it

yes you'r correct

i have to of this Exercices that's why

[tall brook]

i think here you started with methode (0. 1. 0)

[celest kestrel]

that is correct

[tall brook]

so can also change it and start wit (1, 0, 0)

[celest kestrel]

or (0,0,1)

yes, that's what we want to do, we want to know what the basis elements map to

[tall brook]

so you gived a_0 -> 0 / a_1 -> 1 / a_2-> 0

[tall brook]

i understand

i'll do it now for f(x^2)

f(x²) = f((0x² + 0x + 1)) =(2·0 + 1)x² + (4·0 + 2·1) = x² + 2 = 2 + 0x + 1x²

-> (2, 0, 1)

right??

[celest kestrel]

correct

so now you know the matrix of f and you can find its kernel and image

[tall brook]

okay i can see now

i understand thanks a lot

i'll do it right now

and i'll send it

[celest kestrel]

sure thing

[tall brook]

It's correct until now right?

[celest kestrel]

yes

Call the matrix at the bottom A, solve the homogeneous system Ax =0

the solution space is the kernel of f

[tall brook]

Okay

It's the row 1,2

[celest kestrel]

this is correct so the kernel is generated by (-1/2, -1/2, 1)

now convert it back to polynomial form

[tall brook]

Like that ?

[celest kestrel]

keep the order

(-1/2, -1/2, 1) -> -1/2 - (1/2)x + 1x^2

[tall brook]

oh okay

[celest kestrel]

and the kernel is not just one element

it's a set

right?

[tall brook]

Yes

[tall brook]

Yes

And that's the image

Cause both are linear Independence

[celest kestrel]

span of column vectors

the first vector is definitely correct

i don't know how you got the second one

column 1,3 or column 2,3 for example

or column 1,2

it really doesn't matter

[tall brook]

It's the row 1,2

The are linear Independence

[celest kestrel]

row or column?

[tall brook]

Row

[celest kestrel]

has to be column

image is the linear span of column vectors

[tall brook]

that's what i know i found this in the internet also

[celest kestrel]

this is correct because it's done for A^T

we didn't transpose

so you have to do it for columns

[tall brook]

ooh okay

so it's 2,0,0 - 0,2,0 - 1,1,0 or what ??

[celest kestrel]

what are you talking about?

[tall brook]

the image

[celest kestrel]

what is the rank of this matrix?

[tall brook]

it's 2

[celest kestrel]

hence there are two linearly independent column vectors

[tall brook]

rang is equal to image

[tall brook]

yes

[celest kestrel]

pick any two independent column vectors, that's a basis for the image

[tall brook]

ooh okay

so it's column 1,2

[celest kestrel]

for example

[tall brook]

i didn't understand actually

[celest kestrel]

you picked columns 1 and 2

[tall brook]

yes

[celest kestrel]

convert back to polynomial form, that's a basis for the image of f

you can write

$$\mathrm{Im}f = {4a + b\cdot 2x^2 \mid a,b\in\mathbb R} = {a + bx^2 \mid a,b\in\mathbb R}$$

the cofficients dont matter really

[lofty kernel]

aL

[tall brook]

ooh i understand it

now

i really appreciate it

[celest kestrel]

no worries

[tall brook]

i saved my life

[celest kestrel]

😂

[tall brook]

hahahaha

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[tall brook]

+close

[umbral nebula]

✅ Post closed

[twin lantern]

Hmm

[twin lantern]

Test

[rich goblet]

@twin lantern

[twin lantern]

Make it so that the member can message here

[rich goblet]

they can

i made it so

fr

[twin lantern]

Thanks

[tall brook]

can i also use this to find the image, i think they are the same. but i will become a deferent answer

[celest kestrel]

what image do you get when using that?

[tall brook]

2a + 2bx

not 2a + 2bx^2

[celest kestrel]

also please do not dm helpers about your exercises, its ok if you ping me in the channel

[celest kestrel]

indeed, this is not correct

its a different matrix now thats why

[tall brook]

okay bro cause i couldn't send a message here and i took time until the others open it

that's why

sorry

[tall brook]

yes

[celest kestrel]

for starters you swapped two rows

[tall brook]

it's the same but i simplify it

[tall brook]

yes so i should work with the main matrice

[celest kestrel]

it is not the same, give it some thought

[tall brook]

yes they are not the same

so should i use the main one to find the Image

okay thanks

a lot

[celest kestrel]

when you solve the homogeneous case, you can do elementary row operations to simplify

that has no effect on the solution space

but the initial matrix is THE matrix of f with respect to the given basis

you cant do elementary operations to find image

[tall brook]

okay it's clear now

[celest kestrel]

at least not with rows

[tall brook]

cause i was wondering if i can also use the other one

but it's not correct

[celest kestrel]

you can do column operations, that has no effect on image

[tall brook]

you'r right

appreciate it 🙏

[tall brook]

+close