# Math Help - Linear transformations

1. ## Linear transformations

If T:P2 -> M22 is a linear transform such that:

T(1) = |1 0|
.........|0 1|

T(1+x) = |1 1|
.............|0 1|

T(1+x+x^2) = |0 -1|
....................|1 0|

Find T(5-3x+2x^2)

There are a couple ways I know of solving this:
One is to use the fact that the three polynomials are a basis of P2:

Thus 5-3x+2x^2 = c_1*(1) + c_2*(1+x) + c_3*(1+x+x^2)

I need to determine c_1, c_2, and c_3.

So, I think:

T(5-3x+2x^2) = c_1*T(1) + c_2*T(1+x) + c_3*T(1+x+x^2)

However, I am having problems equating c1 c2 and c3.

I came up with c_1 =5, c_2=-3, and c_3=2

but I do not recall I equated coefficients on the two sides of the equation properly.

That is:
The coefficient of x on the LHS is -3, and the RHS = c_2 + c_3 thus:

c_2 + c_3 = -3

Once we know those values (c_1, c_2, and c_3):

Then I "think" we need to sub them into:

T(c_1*v_1 + c_2*v_2) = c_1*T(v_1) + c_2*T(v_2)

Then if we can get the values of c_1 c_2 and c_3...
for the vectors v_1 = 1, v_2 = 1+x, and v_3 = 1+x+x^2.

Therefore:
T(c_1*v_1+c_2*v_2+c_3*v_3)
= c_1*T(v_1)+c_2*T(v_2)+c_3*T(v_3)
= c_1*|1 0| + c_2*|1 1| + c_3*|0 -1|
..........|0 1|..........|0 1|..........|1 0|

Then we should be able to determine a solution for what is equal to T(5-3x+2x^2).

So.
Apoogies from me to the many experts for showin' so much work on this one!

Bottom line:
I have problems in this one computing c_1 c_2 and c_3.

I am a strong believer in showing "your" work. In general I typically make problems more convoluted this way...but we do "not need" an bunch of people out there posting questions and looking for "easy" answers.

Please advise. I hope someone sees it and can share a correct way.

Note: I mean,
There is "One other way" to solve this but I am no expert:

Use the values of T(1), T(1+x), and T(1+x+x^2) to find the values of T on the std basis of P2...the polynomials 1,x,x^2.
We would still have to evaluate the values of c_1 c_2 and c_3 by expansion similar above then equate coefficients of powers on both sides our equations.

Anyway, if anyone can illustrate either method and it works same answer great.

Much appreciated of you guys!

2. Originally Posted by orendacl
If T:P2 -> M22 is a linear transform such that:

T(1) = |1 0|
.........|0 1|

T(1+x) = |1 1|
.............|0 1|

T(1+x+x^2) = |0 -1|
....................|1 0|

Find T(5-3x+2x^2)

There are a couple ways I know of solving this:
One is to use the fact that the three polynomials are a basis of P2:

Thus 5-3x+2x^2 = c_1*(1) + c_2*(1+x) + c_3*(1+x+x^2)

I need to determine c_1, c_2, and c_3.

So, I think:

T(5-3x+2x^2) = c_1*T(1) + c_2*T(1+x) + c_3*T(1+x+x^2)

However, I am having problems equating c1 c2 and c3.

I came up with c_1 =5, c_2=-3, and c_3=2

but I do not recall I equated coefficients on the two sides of the equation properly.

That is:
The coefficient of x on the LHS is -3, and the RHS = c_2 + c_3 thus:

c_2 + c_3 = -3

Once we know those values (c_1, c_2, and c_3):

Then I "think" we need to sub them into:

T(c_1*v_1 + c_2*v_2) = c_1*T(v_1) + c_2*T(v_2)

Then if we can get the values of c_1 c_2 and c_3...
for the vectors v_1 = 1, v_2 = 1+x, and v_3 = 1+x+x^2.

Therefore:
T(c_1*v_1+c_2*v_2+c_3*v_3)
= c_1*T(v_1)+c_2*T(v_2)+c_3*T(v_3)
= c_1*|1 0| + c_2*|1 1| + c_3*|0 -1|
..........|0 1|..........|0 1|..........|1 0|

Then we should be able to determine a solution for what is equal to T(5-3x+2x^2).

So.
Apoogies from me to the many experts for showin' so much work on this one!

Bottom line:
I have problems in this one computing c_1 c_2 and c_3.

I am a strong believer in showing "your" work. In general I typically make problems more convoluted this way...but we do "not need" an bunch of people out there posting questions and looking for "easy" answers.

Please advise. I hope someone sees it and can share a correct way.

Note: I mean,
There is "One other way" to solve this but I am no expert:

Use the values of T(1), T(1+x), and T(1+x+x^2) to find the values of T on the std basis of P2...the polynomials 1,x,x^2.
We would still have to evaluate the values of c_1 c_2 and c_3 by expansion similar above then equate coefficients of powers on both sides our equations.

Anyway, if anyone can illustrate either method and it works same answer great.

Much appreciated of you guys!
One way to find $c_1, c_2$ and $c_3$ is just what you say: set $5-3x+2x^2 = c_1*(1) + c_2*(1+x) + c_3*(1+x+x^2)$. Don't worry about "T" yet, just multiply out the right side:
$5-3x+2x^2 = c_1 + c_2+ c_2x) + c_3+ c_3x+c_3x^2$
$5-3x+2x^2 = (c_1+ c_2+ c_3)+ (c_2+ c_3)x+ c_2x^2$
Since that is to be true for all x, we can equate the same coefficients on both sides: we must have $5= c_1+ c_2+ c_3$, $-3= c_2+ c_3$, and $2= c_3$. Obviously the last equation tells us that $c_3= 2$. Putting that into the second equation, $-3= c_2+ 2$ so $c_2= -5$. Putting those values into the first equation, $5= c_1+ (-5)+ (2)$ so $c_1= 8$.

Another way to find those numbers is to use the fact that $5-3x+2x^2 = c_1*(1) + c_2*(1+x) + c_3*(1+x+x^2)$ is true for all values of x. If we set x= 0 we get $5+0+0= 5= c_1+ c_2+ c_3$ as before. If we set x= 1, we get $5-3+2= 4= c_1+ 2c_2+ 7c_3$. If we set x= -1, we get $5+3+2= 10= c_1+ c_3$

Now you know that $5-3x+2x^2 = 8(1) - 5(1+x) + 2(1+x+x^2)$ so $T(5-3x+2x^2) = 8T(1) - 5T(1+x) + 2T(1+x+x^2)$