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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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  • $\begingroup$ I think that 1. is [2 1; 0 3]. Right? $\endgroup$
    – gov
    Commented Dec 17, 2011 at 10:22
  • $\begingroup$ 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) $\endgroup$
    – gov
    Commented Dec 17, 2011 at 10:32
  • $\begingroup$ is this good? Correct me if not. $\endgroup$
    – gov
    Commented Dec 17, 2011 at 10:39
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    $\begingroup$ The image of the basis vectors, are the columns of the transformation matrix, so you can trivially do 1. $\endgroup$ Commented Dec 17, 2011 at 10:46
  • $\begingroup$ 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 $\mathbb{R}^2$. $\endgroup$
    – user38268
    Commented Dec 17, 2011 at 11:57

1 Answer 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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  • $\begingroup$ 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. $\endgroup$
    – gov
    Commented Dec 17, 2011 at 11:08

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