Question

can some help me with part c) in this questions?

Answer 1

Problem A I find to:\[F=\left[\begin{matrix}0 & 1&-1 \\ 1 & -1&0\\0&0&1\end{matrix}\right]\]
B i) The map is invertible ii) It is not isometry

Answer 2

Hello again :)

Answer 3

Hallo.. :)

Answer 4

You are a lifesaver..

Answer 5

Don't depend on it. I'm still thinking :D

Answer 6

Okay. Got it.

Answer 7

At the start, can you see wether A is correct?

Answer 8

Yup. A is correct.

Answer 9

Great.

Answer 10

But you should be able to answer that yourself. Try multiplying that matrix to v1, v2, v3 in turn and you should get the products as mentioned in your question.

Answer 11

Urrh I see

Answer 12

Do you really?
Please, if there's something you don't understand, tell me. For one thing, Linear Algebra isn't my thing... so I can sorta get if you're confused. I know I get confused :)

Answer 13

Just a sec

Answer 14

Let me just set this up.
Now for the new matrix, using v1, v2, and v3 as the basis...
Expect another 3x3 matrix okay?
Now...
\[\Large \left[\begin{matrix}? & ? & ? & ? \\? & \color{red}? & \color{red}? & \color{red}?\\? & \color{red}? & \color{red}? & \color{red}?\\? & \color{red}? & \color{red}? & \color{red}? \end{matrix}\right]\]

Only the red part of this array is your matrix, the rest are just guides.
\[\Large \left[\begin{matrix}\color{green}f & \color{blue}{v_1} & \color{blue}{v_2} & \color{blue}{v_3} \\\color{orange}{v_1} & \color{red}? & \color{red}? & \color{red}?\\\color{orange}{v_2} & \color{red}? & \color{red}? & \color{red}?\\\color{orange}{v_3} & \color{red}? & \color{red}? & \color{red}? \end{matrix}\right]\]

Answer 15

So, can we proceed?

Answer 16

I could ask you the same question :P
Now, shall we proceed?

Answer 17

Now, look at this.... the answer to (c) would be the 3x3 matrix represented by the red ? marks.

Now... what is the image of v1?

Answer 18

\[\Large \left[\begin{matrix}\color{green}f & \color{blue}{v_1} & \color{blue}{v_2} & \color{blue}{v_3} \\\color{orange}{v_1} & \color{red}? & \color{red}? & \color{red}?\\\color{orange}{v_2} & \color{red}? & \color{red}? & \color{red}?\\\color{orange}{v_3} & \color{red}? & \color{red}? & \color{red}? \end{matrix}\right]\]

Answer 19

Just write it as <1, 0 , 0>
I'll understand what you mean :P

Answer 20

<1 0 0>

Answer 21

okay. Now what I want you to do is express this as a linear combination of the elements of your basis. That is to say:
\[\Large <1 \ , 0 \ , \ 0> = <\color{blue}av_1 \ , \ \color{blue}bv_2 \ , \color{blue}cv_3>\]

Can you do that?

Answer 22

Wait a sec, the image of v1 is NOT <1,0,0>
What is it?
Refer to your question again.
What I mean by image is f(v1)
What is f(v1) ?

Answer 23

<0 1 0>

Answer 24

Better. Now can you express <0 , 1 , 0> as a linear combination of v1, v2, and v3?

Answer 25

\[\Large \left[\begin{matrix}0 \\1\\0\end{matrix}\right]=\color{blue}a\left[\begin{matrix}1 \\0\\0\end{matrix}\right]+\color{blue}b\left[\begin{matrix}1 \\1\\0\end{matrix}\right]+\color{blue}c\left[\begin{matrix}1 \\1\\1\end{matrix}\right]\]

Answer 26

a= -1
b= 1
c=0

Answer 27

Very good :) so... we replace the column under the blue v1 with the coefficients a, b, and c... like so...
\[\Large \left[\begin{matrix}\color{green}f & \color{blue}{v_1} & \color{blue}{v_2} & \color{blue}{v_3} \\\color{orange}{v_1} & \color{red}{-1} & \color{red}? & \color{red}?\\\color{orange}{v_2} & \color{red}1 & \color{red}? & \color{red}?\\\color{orange}{v_3} & \color{red}0 & \color{red}? & \color{red}? \end{matrix}\right]\]
Now do the same for v2.
Get f(v2) and express it as a linear combination of v1, v2, and v3.

Answer 28

\[\Large \left[\begin{matrix}\color{green}f & \color{blue}{v_1} & \color{blue}{v_2} & \color{blue}{v_3} \\\color{orange}{v_1} & \color{red}{-1} & \color{red}1 & \color{red}0\\\color{orange}{v_2} & \color{red}1 & \color{red}0 & \color{red}-1\\\color{orange}{v_3} & \color{red}0 & \color{red}0 & \color{red}1 \end{matrix}\right]\]

Answer 29

I like your initiative :P
Nothing more to be done; You have your answer :)

Answer 30

So this is the answer to define the vectors v1, v2 and v3 as an alternative (new) basis in R^3. Right?

Answer 31

Yes. And also define their images as linear combinations of the new basis :)

Answer 32

Try multiplying this matrix to v1 :)

Answer 33

\[\left[\begin{matrix}-1 & 1& 0 \\ 1 & 0&-1\\0&0&1\end{matrix}\right]v_1 = \color{red}?\]

Answer 34

\[\left(\begin{matrix}-1 \\ 1\\0 \end{matrix}\right)\]

Answer 35

Was that the answer you were expecting? :)

Answer 36

\[\left[\begin{matrix}-1 & 1& 0 \\ 1 & 0&-1\\0&0&1\end{matrix}\right]v_2 = \left[\begin{matrix}-1 & 1& 0 \\ 1 & 0&-1\\0&0&1\end{matrix}\right]\left(\begin{matrix}1 \\ 1\\0\end{matrix}\right) = \left(\begin{matrix}0 \\ 1 \\0\end{matrix}\right)\]

Answer 37

no. In this representation,
\[\Large v_2 \ne \left[\begin{matrix}1 \\ 1\\0\end{matrix}\right]\]
Gotcha there :P

Answer 38

Remember, we are NO LONGER working with the standard basis, but rather, our basis is now v1, v2 and v3.
Therefore, in this representation...
\[\Large v_2 =0\color{blue}{v_1}+ 1\color{blue}{v_2}+ 0\color{blue}{v_3}= \left[\begin{matrix}0 \\ 1\\0\end{matrix}\right]\]

NEVER EVER forget which basis you are using :)

Answer 39

I know, right? :)
So in fact...
\[\left[\begin{matrix}-1 & 1& 0 \\ 1 & 0&-1\\0&0&1\end{matrix}\right]v_2 = \left[\begin{matrix}-1 & 1& 0 \\ 1 & 0&-1\\0&0&1\end{matrix}\right]\left[\begin{matrix}0 \\ 1\\0\end{matrix}\right] = \left[\begin{matrix}1 \\ 0 \\0\end{matrix}\right]\]
Reverting to standard basis...
\[\Large\left[\begin{matrix}1 \\ 0\\0\end{matrix}\right]_{\text{basis v1,v2 and v3}}=1\color{blue}{v_1}+ 0\color{blue}{v_2}+ 0\color{blue}{v_3}= \left[\begin{matrix}1 \\ 0\\0\end{matrix}\right]_{\text{standard basis}} \]
Which agrees with your given :)

Answer 40

And for heaven's sake, don't delete your comments almost immediately, it puts my immediate replies out of context -.-

Answer 41

Sorry..

Answer 42

Anyway, there you have it, your answer :)
You got it yourself, (with a little help ;) )
So be happy XD

Answer 43

Really practice Linear Algebra, okay?
It can get more confusing than this :)

Answer 44

Signing off for now :)
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Terence out