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Hello there, I am currently tackling this Linear Algebra question and I'm struggling to get a grasp on it.

So far I have figured out a) using a theorem from my notes that says that addition is associate and multiplication is distributive and hence I can split:

$$T(x^2 - 2x + 1)= T(x^2)-2T(x) + 1$$

And achieve through substitution:

$$T(x^2 - 2x +1)= 3x^2 - 5x - 1$$

And here is where I start to struggle.

Any pointers/help would be greately appreciated.

  • $\begingroup$ Deleted my answer, thought you were struggling on a)... $\endgroup$
    – user635953
    Nov 6 '20 at 21:06
  • $\begingroup$ Yeah was a bit confused about your answer. I understand a) but I am struggling on the further sections. In my notes for changing basis etc the linear map T has a section where T(p) = a function on p and I can understand changing the basis in response to this factor but I don't understand how to do this without a condition. $\endgroup$
    – xyz
    Nov 6 '20 at 21:08
  1. In part b, you've been given the map in functional notation i.e. $T(v)=\text{something}$, and are asked for a matrix. The matrix "associated" with a linear map is the one such that left-multiplying by that particular matrix is equivalent to applying that map, so you need to work backwards from the map outputs you've been given, along with the rules for matrix multiplication, to find the matrix elements. Notice also that the elements of the set $B$ are the same as the map inputs you've been given. Another hint is that you don't have to compute anything new to answer this - only arrange expressions (or pieces of expressions) you already have in the right order.
  2. This is a question of the conditions that make a set a basis for a given vector space - are there are enough elements for the full dimension of the space (in this case, $3$), and are all of the elements of the set linearly independent (i.e. not scalar multiples of) all of the others?
  3. $[C]_B$ is (if I'm reading right) the matrix that takes the components of some vector expressed in the standard basis $B$ and outputs the components necessary to express that the same "actual" function in terms of elements of $C$, e.g. the function $x^2-x$ as a vector is written $(0, -1, 1)$ in the standard basis, and needs to be transformed into $(1,0,0)$ by the map you calculate. $[B]_C$ is the matrix that does the reverse. This also implies that $[C]_B [B]_C = I$, so they're inverses of each other. These should be relatively easy to calculate if you take advantage of the reasoning from part B to split the problem of computing the matrix's $9$ elements into smaller independent subsets.
  4. $[T]_C$ is (AIUI) a transform that "has the same effect" as the original $T$, but written in terms of coordinates from the basis $C$ rather than the standard $B$. But you already know how to transform one basis into another from the previous question - simply glue that transform in front (to the right of) the original $T$, and the product of the two is the answer. I'm assuming that this has some sort of nice form because the $([T]_C)^N$ part is asking you to write what this matrix looks like multiplied by itself repeatedly.
  5. This is similarly glueing together existing pieces. You can answer the first part by taking the $([T]_C)^N$ from the previous question and transforming it back to the standard basis. (You already have the function you have to apply to do that) You can then take this combined result and use it to transform the function $x^2+1=(1,0,1)$

Hope that helps!

  • $\begingroup$ so I've worked through your guidance and worked out up to half way of 4e) but I'm not sure what the ([T] c) ^N part means. Can you help me out any further please? $\endgroup$
    – xyz
    Nov 7 '20 at 21:24
  • $\begingroup$ @Taylor $[T]_c$ is a symbol for a matrix applying $T$, but expressed in $C$-coordinates, $([T]_c)^N$ is just that matrix multiplied by itself $N$ times. If you have calculated that matrix and found its 0 everywhere except along the diagonal, then $([T]_c)^N$ is just those diagonal entries raised to $N$ $\endgroup$
    – redroid
    Nov 10 '20 at 18:41

Let $B=\{1,x,x^2\}$ then $\forall p \in P_2(R)$ $\exists c_0,c_1,c_2$: $p=c_0+c_1x+c_2x^2$

$T(p)=c_0T(1)+c_1T(x)+c_2T(x^2)$ $\Rightarrow$ $[T(p)]_B=A\begin{pmatrix} c_0 \\ c_1 \\ c_2 \end{pmatrix}$

$A_1=[T(1)]$ ,$A_2=[T(x)]$,$A_3=[T(x^2)]$ where $A_1,A_2,A_3$ are column vectors of $A$

$A=\begin{pmatrix} -1 & 0 & 0\\ -2 & 2 & 1 \\ 2 & 0 & 1 \end{pmatrix}$

We know that dimension of space is 3.All we have to do to prove (c) is show that they are linearly independent

$c_0(x^2-x)+c_1x+c_2(-x^2+x+1)=0$ $\Rightarrow$ $(c_2-c_0+c_1)x+c_2+(c_0-c_2)x^2=0$

$(c_2-c_0+c_1)x+c_2+(c_0-c_2)x^2=0$ $\Rightarrow$ $c_2-c_0+c_1=0$ ,$c_2=0$,$c_0-c_2=0$

which means $c_0=c_1=c_2=0$ so they are linearly independent

Let $C=\{x_2-x,x,-x^2+x+1\}$ and $p=c_0(x^2-x)+c_1x+c_2(-x^2+x+1)=n_0+n_1x+n_2x^2$ then

$c_2-c_0+c_1=n_1$ , $c_2=n_0$,$c_0-c_2=n_2$ so

$\begin{pmatrix} 0 & 0 & 1 \\ 1 & -1 & 1 \\ 1 & 0 &-2\end{pmatrix}$$[p]_B=[p]_C$

  • $\begingroup$ Thank you very much for your help. Could you explain where you got the last matrix from though as I'm not quite following that part? $\endgroup$
    – xyz
    Nov 7 '20 at 18:15
  • 1
    $\begingroup$ I actually did same thing with part (b) last matrix is obtained from the system $c_2=n_0$ $c_'2-c_0+c_1=n_1$ $c_0-c_2=n_'$ we know that the system of lineaar equations can be showed as $AX=B$ where A is $3x3$ matrix $\endgroup$ Nov 7 '20 at 18:51
  • $\begingroup$ Ah okay thank you. $\endgroup$
    – xyz
    Nov 7 '20 at 18:58
  • $\begingroup$ Sorry could you explain this second matrix further. I'm trying to work out what you did from your comment but I'm really struggling to see the connection. $\endgroup$
    – xyz
    Nov 7 '20 at 20:24

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