# Thread: Canonical Basis?

1. ## 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?

Thanks in advance.
Simon DK.

2. 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?

Thanks in advance.
Simon DK.
are these two separate transformations that we are talking about?

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

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

?

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 $\displaystyle \mathbb {R}$, $\displaystyle e_1 = {1 \choose 0}$ and $\displaystyle e_2 = {0 \choose 1}$

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

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

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

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

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

the second one is done similarly

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

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

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

?

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 $\displaystyle \mathbb {R}$, $\displaystyle e_1 = {1 \choose 0}$ and $\displaystyle e_2 = {0 \choose 1}$

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

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

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

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

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

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.

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

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

JakeD

4. 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.

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

$\displaystyle \left(\begin{array}{cc}1/4&3/4\\3/4&1/4\end{array}\right) {-1 \choose 1} = {1/2 \choose -1/2}$
nice. i don't remember what was going through my head at the time i assumed it was a diagonal matrix.... what you did seems so obvious now

5. ## 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

6. 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 $\displaystyle \left(\begin{array}{cc}a&b\\c&d\end{array}\right)$ such that:

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

and

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

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

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

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

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

So we got the system:

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

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

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

7. ## Hi Jhevon and JakeD, thanks.

Thanks for your replies!

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.

8. Originally Posted by sh01by
Thanks for your replies!

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.
Ok, good luck