# Matrix of linear transformation, basis problem

• Jun 28th 2009, 10:28 AM
arbolis
Matrix of linear transformation, basis problem
Let $\displaystyle T$ be a linear transformation in $\displaystyle \mathbb{R}^3$ by $\displaystyle T(x_1,x_2,x_3)=(x_1-x_3, x_1+3x_2,2x_2-x_3)$.
1)Give the matrix of $\displaystyle T$ with respect to the canonical basis $\displaystyle \bold C$.
2)Tell whether $\displaystyle T$ is invertible or no and if it is calculate $\displaystyle T^{-1}$.
3)Give the matrix of $\displaystyle T$ with respect to the basis $\displaystyle \bold {B} =\{ (1,0,1),(-1,2,1),(2,1,1)\}$.
4)Give the matrix of T with respect to the basis $\displaystyle \bold C$, $\displaystyle \bold B$.
5)Give the matrix of $\displaystyle T$ with respect to the basis $\displaystyle \bold B$, $\displaystyle \bold C$.
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My attempt :
1)$\displaystyle [T]_{\bold C} = \begin{bmatrix} 1&0&-1 \\ 1&3&0 \\ 0&2&-1 \end{bmatrix}$.
2)As $\displaystyle [T]_{\bold C}$ is invertible, so is $\displaystyle T$. I calculated $\displaystyle [T]_{\bold C}^{-1}$ which gave me $\displaystyle T^{-1}(x_1,x_2,x_3)=( \frac{3x_1+2x_2-3x_3}{5} , \frac{-x_1+x_2}{5}-\frac{x_3}{10}, \frac{-2x_1+2x_2-3x_3}{5})$.
3) I don't know how to do it! I guess I have to express the vectors of the basis of B as linear combination of the canonical vectors... but I'll have 3 scalars and I don't know what to do with them.
4)I think they ask for $\displaystyle [T]_{\bold B}^{\bold C}$.
I wrote $\displaystyle \left[\begin{array}{ccc|ccc} 1 & -1 & 2 &1&0&-1 \\ 0 & 2 & 1 & 1&3&0 \\ 1 & 1 & 1 & 0&2&-1 \end{array}\right]$ and I reduced the left matrix. At last the right matrix is the one they ask for and I got it to be $\displaystyle \begin{bmatrix} -1& -\frac{1}{2}& -1 \\ 0& \frac{5}{4}&0 \\ 1 & \frac{1}{2}&0 \end{bmatrix}$.
5) If my attempt for 4) is good, I know how to do this one.
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I really need help for part 3), and I'd like to know if I did well what I did. Thanks in advance.
• Jun 28th 2009, 01:14 PM
Random Variable
c)

$\displaystyle T \left(\begin{array}{ccc}1\\0\\1\end{array}\right)$ $\displaystyle = \left(\begin{array}{ccc}0\\1\\\text{-}1\end{array}\right)$

$\displaystyle T \left(\begin{array}{ccc}\text{-}1\\2\\1\end{array}\right)$ $\displaystyle = \left(\begin{array}{ccc}\text{-}2\\5\\3\end{array}\right)$

$\displaystyle T \left(\begin{array}{ccc}2\\1\\1\end{array}\right)$ $\displaystyle = \left(\begin{array}{ccc}1\\5\\1\end{array}\right)$

so the matrix of T with respect to basis B is

$\displaystyle \left(\begin{array}{ccc}0&\text{-}2&1\\1&5&5\\\text{-}1&3&1\end{array}\right)$
• Jun 28th 2009, 01:23 PM
arbolis
Thanks, you saved me (I understand what you did). You don't know how valuable is your help for me. Tomorrow is the final exam which lasts 4 hours and I just can't mess it up because I already messed up the previous one.
• Jun 28th 2009, 01:35 PM
Random Variable
For (d) I would do the following:

$\displaystyle T \left(\begin{array}{ccc}1\\0\\0\end{array}\right) = \Bigg[ \left(\begin{array}{ccc}1\\0\\1\end{array}\right)$ $\displaystyle , \left(\begin{array}{ccc}\text{-}1\\2\\1\end{array}\right) , \left(\begin{array}{ccc}2\\1\\1\end{array}\right) \Bigg] \left(\begin{array}{ccc}\alpha_{1}\\\beta_{1}\\\ga mma_{1}\end{array}\right)$

$\displaystyle T \left(\begin{array}{ccc}0\\1\\0\end{array}\right) = \Bigg[ \left(\begin{array}{ccc}1\\0\\1\end{array}\right)$ $\displaystyle , \left(\begin{array}{ccc}\text{-}1\\2\\1\end{array}\right) , \left(\begin{array}{ccc}2\\1\\1\end{array}\right) \Bigg] \left(\begin{array}{ccc}\alpha_{2}\\\beta_{2}\\\ga mma_{2}\end{array}\right)$

$\displaystyle T \left(\begin{array}{ccc}0\\0\\1\end{array}\right) = \Bigg[ \left(\begin{array}{ccc}1\\0\\1\end{array}\right)$ $\displaystyle , \left(\begin{array}{ccc}\text{-}1\\2\\1\end{array}\right) , \left(\begin{array}{ccc}2\\1\\1\end{array}\right) \Bigg] \left(\begin{array}{ccc}\alpha_{3}\\\beta_{3}\\\ga mma_{3}\end{array}\right)$

then the matrix representation of T with respect to basis C and B is

$\displaystyle \left(\begin{array}{ccc}\alpha_{1}&\alpha_{2}&\alp ha_{3}\\\beta_{1}&\beta_{2}&\beta_{3}\\\gamma_{1}& \gamma_{2}&\gamma_{3}\end{array}\right)$
• Jun 28th 2009, 02:27 PM
arbolis
Quote:

Originally Posted by Random Variable
c)

$\displaystyle T \left(\begin{array}{ccc}1\\0\\1\end{array}\right)$ $\displaystyle = \left(\begin{array}{ccc}0\\1\\\text{-}1\end{array}\right)$

$\displaystyle T \left(\begin{array}{ccc}\text{-}1\\2\\1\end{array}\right)$ $\displaystyle = \left(\begin{array}{ccc}\text{-}2\\5\\3\end{array}\right)$

$\displaystyle T \left(\begin{array}{ccc}2\\1\\1\end{array}\right)$ $\displaystyle = \left(\begin{array}{ccc}1\\5\\1\end{array}\right)$

so the matrix of T with respect to basis B is

$\displaystyle \left(\begin{array}{ccc}0&\text{-}2&1\\1&5&5\\\text{-}1&3&1\end{array}\right)$

I just called a friend and he told me it's not right...
He told me that what you did here is to find $\displaystyle [T]_{\bold C}^{\bold B}$ while the exercise asks for $\displaystyle [T]_{\bold B}^{\bold B}$.
He told me that to find the columns of $\displaystyle [T]_{\bold B}^{\bold B}$, I have to find $\displaystyle a$, $\displaystyle b$ and $\displaystyle c$ such that $\displaystyle (0,1,-1)=a(1,0,1)+b(-1,2,1)+c(2,1,1)$ for the first column, then $\displaystyle (-2,5,3)=d(1,0,1)+e(-1,2,1)+f(2,1,1)$ and $\displaystyle (d,e,f)$ is the second column of the matrix they ask for, and so on.

I'm completely confused. I understand both what you did, but I don't know who's right. I'm at a loss!
• Jun 28th 2009, 03:55 PM
Random Variable
First of all, why did I call part (3) "part (c)" and part (4) "part (d)"? I must be seeing things.

Anyways, how did you do part (1)?

$\displaystyle T \left(\begin{array}{ccc}1\\0\\0\end{array}\right) = \left(\begin{array}{ccc}1\\1\\0\end{array}\right)$

$\displaystyle T \left(\begin{array}{ccc}0\\1\\0\end{array}\right) = \left(\begin{array}{ccc}0\\3\\2\end{array}\right)$

$\displaystyle T \left(\begin{array}{ccc}0\\0\\1\end{array}\right) = \left(\begin{array}{ccc}\text{-}1\\0\\\text{-}1\end{array}\right)$
• Jun 28th 2009, 04:04 PM
Random Variable
To find the matrix representation of T with respect to bases B and C, I would do the reverse of what I did before:

$\displaystyle T \left(\begin{array}{ccc}1\\0\\1\end{array}\right) = \Bigg[ \left(\begin{array}{ccc}1\\0\\0\end{array}\right) , \left(\begin{array}{ccc}0\\1\\0\end{array}\right) , \left(\begin{array}{ccc}0\\0\\1\end{array}\right) \Bigg] \left(\begin{array}{ccc}\alpha_{1}\\\beta_{1}\\\ga mma_{1}\end{array}\right)$

$\displaystyle T \left(\begin{array}{ccc}\text{-}1\\2\\1\end{array}\right) = \Bigg[ \left(\begin{array}{ccc}1\\0\\0\end{array}\right) , \left(\begin{array}{ccc}0\\1\\0\end{array}\right) , \left(\begin{array}{ccc}0\\0\\1\end{array}\right) \Bigg] \left(\begin{array}{ccc}\alpha_{2}\\\beta_{2}\\\ga mma_{2}\end{array}\right)$

$\displaystyle T \left(\begin{array}{ccc}2\\1\\1\end{array}\right) = \Bigg[ \left(\begin{array}{ccc}1\\0\\0\end{array}\right) , \left(\begin{array}{ccc}0\\1\\0\end{array}\right) , \left(\begin{array}{ccc}0\\0\\1\end{array}\right) \Bigg] \left(\begin{array}{ccc}\alpha_{3}\\\beta_{3}\\\ga mma_{3}\end{array}\right)$
• Jun 28th 2009, 04:09 PM
arbolis
Quote:

Originally Posted by Random Variable
First of all, why did I call part (3) "part (c)" and part (4) "part (d)"? I must be seeing things.

Anyways, how did you do part (1)?

$\displaystyle T \left(\begin{array}{ccc}1\\0\\0\end{array}\right) = \left(\begin{array}{ccc}1\\1\\0\end{array}\right)$

$\displaystyle T \left(\begin{array}{ccc}0\\1\\0\end{array}\right) = \left(\begin{array}{ccc}0\\3\\2\end{array}\right)$

$\displaystyle T \left(\begin{array}{ccc}0\\0\\1\end{array}\right) = \left(\begin{array}{ccc}\text{-}1\\0\\\text{-}1\end{array}\right)$

I did the same thing as you did here.
And thanks for T with respect to bases B and C... it helps me.
• Jun 28th 2009, 04:19 PM
arbolis
Wait! But
$\displaystyle \left(\begin{array}{ccc}1\\0\\1\end{array}\right) = \Bigg[ \left(\begin{array}{ccc}1\\0\\0\end{array}\right) , \left(\begin{array}{ccc}0\\1\\0\end{array}\right) , \left(\begin{array}{ccc}0\\0\\1\end{array}\right) \Bigg] \left(\begin{array}{ccc}\alpha_{1}\\\beta_{1}\\\ga mma_{1}\end{array}\right)$ gives
$\displaystyle (\alpha _1,\beta _1, \gamma _1)=(1,0,1)$... so it's like if I didn't do anything. (multiplying a vector with the identity matrix).
• Jun 28th 2009, 04:31 PM
Random Variable
Quote:

Originally Posted by arbolis
Wait! But
$\displaystyle \left(\begin{array}{ccc}1\\0\\1\end{array}\right) = \Bigg[$$\displaystyle \left(\begin{array}{ccc}1\\0\\0\end{array}\right) , \left(\begin{array}{ccc}0\\1\\0\end{array}\right) , \left(\begin{array}{ccc}0\\0\\1\end{array}\right) \Bigg] \left(\begin{array}{ccc}\alpha_{1}\\\beta_{1}\\\ga mma_{1}\end{array}\right)$ gives $\displaystyle (\alpha _1,\beta _1, \gamma _1)=(1,0,1)$... so it's like if I didn't do anything. (multiplying a vector with the identity matrix).

I forgot the T.

$\displaystyle T \left(\begin{array}{ccc}1\\0\\1\end{array}\right) = \Bigg[ \left(\begin{array}{ccc}1\\0\\0\end{array}\right) , \left(\begin{array}{ccc}0\\1\\0\end{array}\right) , \left(\begin{array}{ccc}0\\0\\1\end{array}\right) \Bigg] \left(\begin{array}{ccc}\alpha_{1}\\\beta_{1}\\\ga mma_{1}\end{array}\right)$
• Jun 28th 2009, 05:05 PM
arbolis
Quote:

Originally Posted by Random Variable
I forgot the T.

$\displaystyle T \left(\begin{array}{ccc}1\\0\\1\end{array}\right) = \Bigg[ \left(\begin{array}{ccc}1\\0\\0\end{array}\right) , \left(\begin{array}{ccc}0\\1\\0\end{array}\right) , \left(\begin{array}{ccc}0\\0\\1\end{array}\right) \Bigg] \left(\begin{array}{ccc}\alpha_{1}\\\beta_{1}\\\ga mma_{1}\end{array}\right)$

Oh yes, that's it! I get it now.
• Jun 28th 2009, 05:12 PM
Random Variable
• Jun 28th 2009, 05:18 PM
arbolis
Quote:

Originally Posted by Random Variable

Indeed it helps. There's an error in the first page when they give $\displaystyle [T]_{\beta}^{\gamma}$. The first entry of the matrix is 1/3 while it should be -1/3.(Wink)