Canonical Basis?

• Jun 4th 2007, 10:20 AM
sh01by
Canonical Basis?
Hi there,

I need to create a matrix given some information, without the matrix it is impossible to continue the assignment - unfortunately, I don't find much relevant information on canonical basis online, well, I have found a few examples, but they are similar to what I'm looking at...

Find the 2x2-matrix, A, belonging to the linear "map"/"projection", T, with regard to the canonical basis of R^2 (real numbers).

A small note e(1) means e sub one.

T(e(1) + e(2)) = e(1) + e(2), T(-e(1)+e(2)) = (1/2)e(1) - (1/2)e(2)

Does anyone have an idea on this matter?

Simon DK.
• Jun 4th 2007, 10:55 AM
Jhevon
Quote:

Originally Posted by sh01by
Hi there,

I need to create a matrix given some information, without the matrix it is impossible to continue the assignment - unfortunately, I don't find much relevant information on canonical basis online, well, I have found a few examples, but they are similar to what I'm looking at...

Find the 2x2-matrix, A, belonging to the linear "map"/"projection", T, with regard to the canonical basis of R^2 (real numbers).

A small note e(1) means e sub one.

T(e(1) + e(2)) = e(1) + e(2), T(-e(1)+e(2)) = (1/2)e(1) - (1/2)e(2)

Does anyone have an idea on this matter?

Simon DK.

are these two separate transformations that we are talking about?

the first: $T(e_1 + e_2) = e_1 + e_2$

and the second: $T(-e_1 + e_2) = \frac {1}{2}e_1 - \frac {1}{2}e_2$

?

I find it difficult to see how one transformtion would do both operations, but then again, i am a bit rusty with this

Here's what i am thinking:

Recall: in $\mathbb {R}$, $e_1 = {1 \choose 0}$ and $e_2 = {0 \choose 1}$

we want $T_1$ such that $T_1 (e_1 + e_2) = e_1 + e_2$. Now $T_1$ must be of the form $\left(\begin{array}{cc}a&0\\0&b\end{array}\right)$

So we have: $\left(\begin{array}{cc}a&0\\0&b\end{array}\right) \left( {1 \choose 0}+{0 \choose 1} \right) = {1 \choose 0}+{0 \choose 1}$

$\Rightarrow \left(\begin{array}{cc}a&0\\0&b\end{array}\right) {1 \choose 1} = {1 \choose 1}$

$\Rightarrow a \cdot 1 = 1 \mbox { and } b \cdot 1 = 1 \Rightarrow a = b = 1$

So, $T_1 = \left(\begin{array}{cc}1&0\\0&1\end{array}\right)$

the second one is done similarly
• Jun 4th 2007, 12:23 PM
JakeD
Quote:

Originally Posted by Jhevon
are these two separate transformations that we are talking about?

the first: $T(e_1 + e_2) = e_1 + e_2$

and the second: $T(-e_1 + e_2) = \frac {1}{2}e_1 - \frac {1}{2}e_2$

?

I find it difficult to see how one transformtion would do both operations, but then again, i am a bit rusty with this

Here's what i am thinking:

Recall: in $\mathbb {R}$, $e_1 = {1 \choose 0}$ and $e_2 = {0 \choose 1}$

we want $T_1$ such that $T_1 (e_1 + e_2) = e_1 + e_2$. Now $T_1$ must be of the form $\left(\begin{array}{cc}a&0\\0&b\end{array}\right)$

So we have: $\left(\begin{array}{cc}a&0\\0&b\end{array}\right) \left( {1 \choose 0}+{0 \choose 1} \right) = {1 \choose 0}+{0 \choose 1}$

$\Rightarrow \left(\begin{array}{cc}a&0\\0&b\end{array}\right) {1 \choose 1} = {1 \choose 1}$

$\Rightarrow a \cdot 1 = 1 \mbox { and } b \cdot 1 = 1 \Rightarrow a = b = 1$

So, $T_1 = \left(\begin{array}{cc}1&0\\0&1\end{array}\right)$

the second one is done similarly

Jhevon, one transformation will do both operations. Why did you assume a diagonal matrix? Letting all 4 elements of the matrix be non-zero leads to 4 equations in 4 unknowns. Here are the results.

$\left(\begin{array}{cc}1/4&3/4\\3/4&1/4\end{array}\right) {1 \choose 1} = {1 \choose 1}$

$\left(\begin{array}{cc}1/4&3/4\\3/4&1/4\end{array}\right) {-1 \choose 1} = {1/2 \choose -1/2}$

JakeD
• Jun 4th 2007, 12:25 PM
Jhevon
Quote:

Originally Posted by JakeD
Jhevon, one transformation will do both operations. Why did you assume a diagonal matrix? Letting all 4 elements be non-zero leads to 4 equations in 4 unknowns. Here are the results.

$\left(\begin{array}{cc}1/4&3/4\\3/4&1/4\end{array}\right) {1 \choose 1} = {1 \choose 1}$

$\left(\begin{array}{cc}1/4&3/4\\3/4&1/4\end{array}\right) {-1 \choose 1} = {1/2 \choose -1/2}$

nice. i don't remember what was going through my head at the time i assumed it was a diagonal matrix...:confused:. what you did seems so obvious now
• Jun 4th 2007, 01:02 PM
sh01by
Thanks, please elaborate on the transformation...
Hi Jhevon and JakeD,

it is always perfect with quick replies...

Yes, Jhevon, they are separate - I tried to make a few spaces, but I guess they were removed, when I posted. : o )

JakeD, I see your solution, and it is right, but...
How is it you find the matrix? I see how it works, how you multiply the matrix with the information in T(e(1)+e(2)) as seen in your post - the first math-input...

Suppose, you're looking at the first case... What I don't get is? Why don't you just choose a matrix of pure 1/2? That would be satisfying...

I guess what I am asking is - how do set up and equation, so you get the condition from the other information too - and end up with your matrix..?

Did that make any sense?

Simon DK
• Jun 4th 2007, 01:21 PM
Jhevon
Quote:

Originally Posted by sh01by
Hi Jhevon and JakeD,

it is always perfect with quick replies...

Yes, Jhevon, they are separate - I tried to make a few spaces, but I guess they were removed, when I posted. : o )

JakeD, I see your solution, and it is right, but...
How is it you find the matrix? I see how it works, how you multiply the matrix with the information in T(e(1)+e(2)) as seen in your post - the first math-input...

Suppose, you're looking at the first case... What I don't get is? Why don't you just choose a matrix of pure 1/2? That would be satisfying...

I guess what I am asking is - how do set up and equation, so you get the condition from the other information too - and end up with your matrix..?

Did that make any sense?

Simon DK

Do you remember how to do matrix multiplication?

we wanted to find a matrix $\left(\begin{array}{cc}a&b\\c&d\end{array}\right)$ such that:

$\left(\begin{array}{cc}a&b\\c&d\end{array}\right) {1 \choose 1} = {1 \choose 1}$

and

$\left(\begin{array}{cc}a&b\\c&d\end{array}\right) {-1 \choose 1} = {1/2 \choose -1/2}$

Then combining the matrices on the left of the first equation, we realize that this means:

$-a + b = \frac {1}{2}$ and $-c + d = - \frac {1}{2}$

Combining the matrices on the left of the second equation, we realize that:

$a + b = 1$ and $c + d = 1$

So we got the system:

$-a + b = \frac {1}{2}$............................(1)
$-c + d = - \frac {1}{2}$..........................(2)
$a + b = 1$ ..............................(3)
$c + d = 1$ ...............................(4)

solving that system gives us the components of the matrix JakeD came up with

Quote:

Originally Posted by sh01by
What I don't get is? Why don't you just choose a matrix of pure 1/2? That would be satisfying...

he wanted a single matrix that does both operations. had he choosen a matrix with just 1/2's everywhere, we would satisfy the first operation, but not the second
• Jun 6th 2007, 11:44 PM
sh01by
Hi Jhevon and JakeD, thanks.

It was a big help. I think, I'd be able to solve a similar assignment, if it is part of the exam, which I'm taking tomorrow.

It makes a little more sense now too, because I've been working on other topics in linear algebra for the last days, so it is all coming together - just not fast enough to ace the test tomorrow, but - maybe next time, for now - I'm just looking to pass.

Have a nice day,
Simon DK.
• Jun 6th 2007, 11:46 PM
Jhevon
Quote:

Originally Posted by sh01by