What Are the Values of T(1), T(t), and T(t^2) in These Linear Transformations?

[Bertrandkis]

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

<br /> T(1+t)=\left[\begin{array}{cc}1&amp;0\\0&amp;0\end{array}<br /> \right];
<br /> T(t+t^{2})=\left[\begin{array}{cc}0&amp;1\\1&amp;0\end{array}<br /> \right];
<br /> T(1+t^{2})=\left[\begin{array}{cc}0&amp;1\\0&amp;1\end{array}<br /> \right];
Then findT(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 independent 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

<br /> T(1+t)=\left[\begin{array}{cc}1&amp;0\\0&amp;0\end{array}<br /> \right];
<br /> T(t+t^{2})=\left[\begin{array}{cc}0&amp;1\\1&amp;0\end{array}<br /> \right];
<br /> T(1+t^{2})=\left[\begin{array}{cc}0&amp;1\\0&amp;1\end{array}<br /> \right];
Then findT(1),T(t),T(t^{2})

You just stated T(1), T(2), T(t^2)[/tex]!? Don&#039;t you mean &quot;find T(p) where p is any member of P²&quot;? As Siddharth said, T is linear. Any member of P² can be written at²+ bt+ c. What is T(at²+ at+ b)?<br /> <br /> <br /> [/quote]<b>My attempt</b><br /> All I know is that 1,t,t^{2} are basis of P2, what do I do next? <br /> How do I find them from given matrices?<br /> <br /> <b><br /> Question 2</b><br /> let dim(v)=n and dim(W)=m and P:V-&gt;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.<br /> <b>My attempt</b><br /> Let S be a basis of V S={v1,v2,...vn}<br /> Let v a vector in v<br /> v=c1v1+c2v2+ ...cnvn<br /> <br /> P(v)=c1w1+c2w2+ ...+cnwm=0<br /> Because vectors of S are linearly independent c1,c2 ... cn are all 0<br /> So the resultant matrix of P is a zero matrix[/quote]<br /> Your final equaiton, P(v)= c1w1+ c2w2+ ...+ cnwn= 0, is in W- it says NOTHING about &quot;the vectors of S&quot;. If it were true that &quot;c1, c2, ..., cn are all 0&quot;, then <b>v</b> would be the 0 vector- and that is not, in general true. Remember that you can write a linear transformation, L:V-&gt;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)?<br /> <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> <b><br /> Question 3</b><br /> Let L:V-&gt;W be a linear transformation. show that L is one to one if and only if dim(range L)=dim(V)<br /> <b>My attempt:</b><br /> We know that dim(V)=dim(range L)+dim(ker L) (1)<br /> if dim(V)&gt;dim(range L) then dim(ker L) is not 0 and L is not One to one<br /> if dim(V)=dim(range L) then dim(V)-dim(range L) = dim(ker L) <br /> and dim(ker L)=0 hence L is one to one. </div> </div> </blockquote>

 

[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&amp;0\\0&amp;0\end{array}\right];
2: T(t+t^{2})= T(t)+ T(t^2)= \left[\begin{array}{cc}0&amp;1\\1&amp;0\end{array}\right];
3:T(1+t^{2})=T(1)+ T(t^2)= \left[\begin{array}{cc}0&amp;1\\0&amp;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).