Some linear transformation

[Bertrandkis]

Question 1
Let T: P2 -> M22 be a linear transformation such that

$$T(1+t)=\left[\begin{array}{cc}1&0\\0&0\end{array} \right];$$
$$T(t+t^{2})=\left[\begin{array}{cc}0&1\\1&0\end{array} \right];$$
$$T(1+t^{2})=\left[\begin{array}{cc}0&1\\0&1\end{array} \right];$$
Then find$$T(1),T(t),T(t^{2})$$

My attempt
All I know is that $$1,t,t^{2}$$ are basis of P2, what do I do next?
How do I find them from given matrices?


Question 2
let dim(v)=n and dim(W)=m and P:V->W be a linear transformation, i.e P(v)=0 for all v in V. Show that the matrix of P with respect to any bases for V and W is the mxn zero matrix.
My attempt
Let S be a basis of V S={v1,v2,...vn}
Let v a vector in v
$$v=c1v1+c2v2+ ...cnvn$$

$$P(v)=c1w1+c2w2+ ...+cnwm=0$$
Because vectors of S are linearly independant c1,c2 ... cn are all 0
So the resultant matrix of P is a zero matrix



Question 3
Let L:V->W be a linear transformation. show that L is one to one if and only if dim(range L)=dim(V)
My attempt:
We know that dim(V)=dim(range L)+dim(ker L) (1)
if dim(V)>dim(range L) then dim(ker L) is not 0 and L is not One to one
if dim(V)=dim(range L) then dim(V)-dim(range L) = dim(ker L)
and dim(ker L)=0 hence L is one to one.

 

[siddharth]

Lets take one at a time. For the first one, notice that T is a linear transformation. What does that imply?

 

[HallsofIvy]

Question 1
Let T: P2 -> M22 be a linear transformation such that

$$T(1+t)=\left[\begin{array}{cc}1&0\\0&0\end{array} \right];$$
$$T(t+t^{2})=\left[\begin{array}{cc}0&1\\1&0\end{array} \right];$$
$$T(1+t^{2})=\left[\begin{array}{cc}0&1\\0&1\end{array} \right];$$
Then find$$T(1),T(t),T(t^{2})$$

You just stated [itex]T(1), T(2), T(t^2)[/tex]!? Don't you mean "find T(p) where p is any member of P²"? As Siddharth said, T is linear. Any member of P² can be written at²+ bt+ c. What is T(at²+ at+ b)?


[/quote]My attempt
All I know is that $$1,t,t^{2}$$ are basis of P2, what do I do next?
How do I find them from given matrices?


Question 2
let dim(v)=n and dim(W)=m and P:V->W be a linear transformation, i.e P(v)=0 for all v in V. Show that the matrix of P with respect to any bases for V and W is the mxn zero matrix.
My attempt
Let S be a basis of V S={v1,v2,...vn}
Let v a vector in v
$$v=c1v1+c2v2+ ...cnvn$$

$$P(v)=c1w1+c2w2+ ...+cnwm=0$$
Because vectors of S are linearly independant c1,c2 ... cn are all 0
So the resultant matrix of P is a zero matrix[/quote]
Your final equaiton, P(v)= c1w1+ c2w2+ ...+ cnwn= 0, is in W- it says NOTHING about "the vectors of S". If it were true that "c1, c2, ..., cn are all 0", then v would be the 0 vector- and that is not, in general true. Remember that you can write a linear transformation, L:V->W, in given bases for V and W by applying L to each basis vector in V in turn, then writing the result in the basis in W. The coefficients then form a column for the matrix. If {v2, v2, ..., vn} is a basis for V, what is P(v1)? What is P(v2)?

Question 3
Let L:V->W be a linear transformation. show that L is one to one if and only if dim(range L)=dim(V)
My attempt:
We know that dim(V)=dim(range L)+dim(ker L) (1)
if dim(V)>dim(range L) then dim(ker L) is not 0 and L is not One to one
if dim(V)=dim(range L) then dim(V)-dim(range L) = dim(ker L)
and dim(ker L)=0 hence L is one to one.

 

[Bertrandkis]

Question 1 is formulated correctly. They want $$T(1);T(t);T(t^{2})$$.
Some one has suggested that :
$$T(1)=1/2( T(1+t) - T(t+t^{2}) + T(1+t^{2}) ))$$
becaused T being a linear transformation when the RHS expression is developed it yields $$T(1)$$. The problem is solved by replacing T(...) in the RHS expression by their given matrices.
In the same way we can find $$T(t)$$ and $$T(t^{2})$$

 

[HallsofIvy]

Question 1 is formulated correctly. They want $$T(1);T(t);T(t^{2})$$.

My mistake. I misread. You are NOT given T(1), T(t), and T(t²) as I thought. You are given T(1+ t), T(1+ t²) and T(t+ t²).

Some one has suggested that :
$$T(1)=1/2( T(1+t) - T(t+t^{2}) + T(1+t^{2}) ))$$
becaused T being a linear transformation when the RHS expression is developed it yields $$T(1)$$. The problem is solved by replacing T(...) in the RHS expression by their given matrices.
In the same way we can find $$T(t)$$ and $$T(t^{2})$$

Yes, that would work, although I would be inclined to wonder HOW you noticed that
$$T(1)=1/2( T(1+t) - T(t+t^{2}) + T(1+t^{2}) ))$$!

Siddharth's original suggestion was to use linearity to say that
$$1: T(1+t)= T(1)+ T(t)= \left[\begin{array}{cc}1&0\\0&0\end{array}\right];$$
$$2: T(t+t^{2})= T(t)+ T(t^2)= \left[\begin{array}{cc}0&1\\1&0\end{array}\right];$$
$$3:T(1+t^{2})=T(1)+ T(t^2)= \left[\begin{array}{cc}0&1\\0&1\end{array}\right];$$
Now treat those as three equations in the three unknown matrices, T(1), T(t), T(t²⁾. For example, adding (1) and (3) gives the equation 2T(1)+ T(t)+ T(t²)= a matrix. Subtracting (2) from that gives 2T(1)= a matrix, giving the equation you have. You can similarly solve for T(t) and T(t²).