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The linear transformation $A:\mathbb{R}^2\to \mathbb{R}^2$ is given by the images of basis vectors:
$A((1,1))=(2,1)$ and $A((1,0))=(0,3)$.

  1. Find a matrix of linear transformation $A$ in the basis $(1,1), (1,0)$.

  2. Find $A((3,2))$.

  3. Find vector $x=(x_1,x_2)$ such that the matrix $\begin{pmatrix}-6 &-6\\ 3 &4\end{pmatrix}$ is matrix of the linear transformation $A$ in the basis $x$, $(0,3)$.

Please help me about this.

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    I think that 1. is [2 1; 0 3]. Right?2011-12-17
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    2. (3,2)= a(1,1)+b(1,0) = (a+b,a) . From this I have a=2, b=1. (3,2)=2(1,1)+(1,0) A((3,2)) = A(2(1,1)) + A((1,0)) = 2A((1,1))+A((1,0)) = 2(2,1) + (0,3) = (4,2)+(0,3) = (4,5)2011-12-17
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    is this good? Correct me if not.2011-12-17
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    The image of the basis vectors, are the columns of the transformation matrix, so you can trivially do 1.2011-12-17
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    For (1) you already know how to do it as Paxinum has told you above. For (2), I think when you ask to find $A((3,2))$ it is technically not correct. This is because your matrix $A$ has columns in a basis different from the standard basis (I am assuming that $(3,2) = 3e_1 + 2e_2$, $\{e_1,e_2\}$ the standard basis of $\mathbb{R}^2$.2011-12-17
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    Therefore, you need to ask what is $(3,2)$ in the basis you have specified above? You will need to solve a system of two equations in two unknowns to get this.2011-12-17
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    I did it. I wrote it in comment above. can u check it, please?2011-12-17
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    And I think that Plaxinum is not right.2011-12-17
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    Are you sure the second column in the matrix in part 3. is correct? (Something is wrong in my deleted answer if it is.)2011-12-17
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    yes, it's [-6 -6; 3 4]2011-12-17
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    I got (x1,x2)=(-1,3)2011-12-17
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    @Gorcia You are correct. For part 1., things depend on what you mean by "matrix of linear transformation" and if you are using the given basis in both the domain and range (which is what I assume in my answer).2011-12-17

1 Answers 1

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If $\cal B=\{ {\bf v},{\bf w}\}$ is an ordered basis of $\Bbb R^2$, then the matrix representation, $M$, of $A:\Bbb R^2\rightarrow\Bbb R^2$ with respect to this basis (I assume you want both the domain and range to have basis $\cal B$) is the $2\times 2$ matrix that has as its first column the coordinates of $A{\bf v}$ with respect to $\cal B$ and as its second column the coordinates of $A{\bf w}$ with respect to $\cal B$.

What does this mean? Well, if you write a vector ${\bf x}$ in terms of this basis $${\bf x}= c_1{\bf v}+c_2{\bf w},$$ then, setting, $[{\bf x}]_{\cal B}=[{c_1\atop c_2}]$ $$ \tag {1}[A {\bf x } ]_{\cal B} = M[{\bf x }]_{\cal B}. $$

That is, for $\bf x$ written in the standard basis, the coordinates of $A{\bf x}$ with respect to $\cal B$ are given by the product of the matrix $M$ with the coordinate matrix of $\bf x$ with respect to $\cal B$.


For part 1.:

The matrix representation of $A$ is easily found, since you were told what $A\bigl((1,1)\bigr)$ and $A\bigl((1,0)\bigr)$ were.

We need to write $(2,1)$ and $(0,3)$ in terms of the basis $\cal B=\{(1,1),(1,0)\}$.

$$(2,1)= 1 (1,1)+1(1,0)$$ and $$ (0,3)= 3(1,1)-3(1,0) $$ The matrix $M$ is $$M=\Bigl[\, \underbrace{1\atop 1}_{ [A(1,1)]_{\cal B} } \ \underbrace{3\atop -3}_{ [A(1,0)]_{\cal B} }\,\Bigr].$$


For part 2.:

You need to write $(3,2)$ in terms of $\cal B$: $$ (3,2)=2(1,1)+1(1,0). $$

Using the matrix representation of $A$, $$[A\bigl((3,2 )\bigr)]_{\cal B}=\Bigl [\, {1\atop 1}\ {3\atop -3}\,\Bigr ]\Bigl[ {2\atop 1}\Bigr]=\Bigl[{5\atop -1} \Bigr]. $$

This gives the coordinates of $A((3,2))$ with respect to $\cal B$, so $$ A((3,2))= 5(1,1)+(-1)(1,0)=(4,5). $$


For part 3.: Let $\cal B'=\{(x_1,x_2), (0,3)\}$

You know the matrix $$W= \Bigl[\,{-6\atop 3}\ {-6\atop4}\,\Bigr ] $$ is the matrix representation of $A$ with respect to $\cal B'$.

The second column of $W$ is $[ A\bigl((0,3)\bigr)]_{\cal B'}$.

So, $$\tag{2}A((0,3))=-6(x_1,x_2)+4(0,3)=(-6x_1, -6x_2+12).$$

But you can compute $A\bigl((0,3)\bigr)$ using the matrix representation from part 1.

We find $[A\bigl((0,3)\bigr)]_{\cal B}$ first. Towards this end, we write $(0,3)$ in terms of the basis $\cal B$ first. Solve:
$$ (0,3) = c_1(1,1)+c_2(1,0) $$ to obtain $$ \eqalign{ c_1&=3\cr c_2&=-3. } $$ Then: $$ [A\bigl((0,3)\bigr)]_{\cal B}=\Bigl [\, {1\atop 1}\ {3\atop -3}\,\Bigr ]\Bigl[ {3\atop -3}\Bigr]=\Bigl[{ -6\atop 12}\Bigr]. $$

So, the coordinates of $A((0,3))$ with respect to $\cal B$ are $(-6,12)$. So, $$ \tag{3}A((0,3))=-6(1,1)+12(1,0)= (6, -6) $$ Comparing equations (2) and (3) gives $$ \eqalign{ 6&=-6x_1\cr -6&=-6x_2+12 } $$

This gives $x_1=-1$ and $x_2=3$.

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    Are u sure for the part 1? I consulted with friend and book :D Shouldn't I express images (2,1) and (0,3) over basis vectors (1,1), (1,0)? I got this: (2,1)=(1,1)+(1,0) (0,3)=3(1,1)-3(1,0) so the matrix of A in basis (1,1), (1,0) is [1 3; 1 -3]. I'm sorry, I don't know how to write matrix on this site.2011-12-17