Question

Consider the following linear operator L : R3 ! R3. Find the matrix representation showing your work
(i.e. ‘inspection’ is not enough):

Answer 1

@samjordon

Answer 2

hmmm it R^3 to R^3 wait a sec

Answer 3

@Mr.Math

Answer 4

its pretty simple like cuz its r^3 to R^3 but i just forgot the setup and my notes r upstairs give me a sec

Answer 5

thanks pip

Answer 6

@Jemurray3 can u come help?

Answer 7

Sure. I mean ... inspection would be perfectly valid but I suppose you can do something like this:

Answer 8

\[\left(\begin{matrix}a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{matrix}\right) \left(\begin{matrix}x_1 \\x_2 \\ x_3\end{matrix}\right) = \left(\begin{matrix} a_1x_1 + a_2x_2 + a_3x_3 \\ b_1x_1 + b_2x_2 + b_3x_3 \\ c_1x_1 + c_2x_2 + c_3x_3\end{matrix}\right)\]

Answer 9

Then compare that to the transformation you want. E.g.
\[a_1 = 0 \]
\[a_2 = -2\]
\[a_3 = 1\]

Answer 10

hahaha amistre was the mysterious guest

Answer 11

Linear operators have to satisfy 2 conditions

L{a+b} = L{a} + L{b}

L{ka} = k L{a} for any constant k

yes i was lol

Answer 12

my take on this is:

Answer 13

What is the \(L:\mathbb{R^3}\longrightarrow \mathbb{R^3}\) operation for? Is it for cubing?

Answer 14

its in 3 space 3 dimensions

Answer 15

No, that just means it maps R3 to R3, i.e. it can be represented by a 3x3 matrix.

Answer 16

Ohh, thanks.

Answer 17

set it up for all xs in each row

\[L(x)=\begin{pmatrix}
0x_1-2x_2+1x_3\\
1x_1+0x_2-1x_3\\
1x_1-1x_2+0x_3\end{pmatrix}\]

\[L(x)=\begin{pmatrix}
0x_1\\1x_1\\1x_1\end{pmatrix}+\begin{pmatrix}
-2x_2\\0x_2\\-1x_2\end{pmatrix}+\begin{pmatrix}
1x_3\\
-1x_3\\
0x_3\end{pmatrix}\]

\[L(x)=x_1\begin{pmatrix}
0\\1\\1\end{pmatrix}
+x_2\begin{pmatrix}
-2\\0\\-1\end{pmatrix}+x_3\begin{pmatrix}
1\\
-1\\
0\end{pmatrix}\]

Answer 18

as such:

L(x) = x1v1+x2v2+x3v3 which is definitionly matrix at heart i believe

Answer 19

but this is just what i read from it, right or wrong

Answer 20

You already wrote the matrix, just look

Answer 21

The first line you wrote there is clearly a matrix times a vector, just drop the x's and you have it.

Answer 22

i know that, but this was spose to be more defined than a vague sighted thing

Answer 23

That's not vague. Comparison is valid, they just want to avoid an answer like "By inspection, *here is the correct answer*".

Answer 24

thanks to pippa, Jemurray, and amistre

Answer 25

youre welcome, hope it helped