1
$\begingroup$

Let's assume that a linear operator in basis $$e_1=(1,0,0)$$ $$e_2=(0,1,0)$$ $$e_3=(0,0,1)$$ hast matrix $$A = \begin{pmatrix} -1&2&1\\3&1&0\\1&1&1&\end{pmatrix}$$ I need to find matrix of this operator in new basis $$a_1=(0,1,2)$$ $$a_2=(3,1,0)$$ $$a_3=(0,1,1)$$


Okay, what I do is:


1) I say, that $$T(1,0,0)=(-1,2,1)$$ $$T(0,1,0)=(3,1,0)$$ $$T(0,0,1)=(1,1,1)$$ First question, is it correct (that I write rows of the matrix $A$, or I should write its columns)?


2) I say, that $$T(0,1,2)=T((0,1,0)+2*(0,0,1))=T(0,1,0)+2*T(0,0,1)=(3,1,0)+2(1,1,1)=(5,3,2)$$ $$T(3,1,0)=T((3*(1,0,0)+(0,1,0))=3*T(1,0,0)+T(0,1,0)=3(-1,2,1)+(3,1,0)=(0,7,3)$$ $$T(0,1,1)=T((0,1,0)+(0,0,1))=T(0,1,0)+T(0,0,1)=(3,1,0)+(1,1,1)=(4,2,1)$$


3)Finally, I say that new matrix $$A = \begin{pmatrix} 5&3&2\\0&7&3\\4&2&1&\end{pmatrix}$$


Is my answer correct or I've made a mistake somewhere?

$\endgroup$
0
$\begingroup$

For your first question, you should write columns of $A$ not rows. So we have

$Te_1= (-1,3,1), Te_2=(2,1,1), Te_3= (1,0,1)$. So, we have

$$T(0,1,2) = (10/3)(0,1,2)+(4/3)(3,1,0)+(-11/3)(0,1,1)$$ $$T(3,1,0) = (-19/3)(0,1,2)+(-1/3)(3,1,0)+(50/3)(0,1,1)$$ $$T(0,1,1) = (2)(0,1,2)+(1)(3,1,0)+(-2)(0,1,1)$$ Therefore, the matrix with respect to this basis is

$\begin{bmatrix} 10/3 & -19/3 & 2 \\ 4/3 & -1/3 & 1\\ -11/3 & 50/3 & -2\end{bmatrix}$

$\endgroup$
  • $\begingroup$ How do we find 10/3, 4/3, -11/3 and so on? $\endgroup$ – Николай Журба Dec 4 '17 at 19:29
  • $\begingroup$ $T(0,1,2) = (4,1,3)$. Now write $(4,1,3)$ as linear combination of vectors $\{(0,1,2), (3,1,0), (0,1,1)\}$ which turns out to be $(4,1,3) = (10/3)(0,1,2) + (4/3)(3,1,0) + (-11/3)(0,1,1)$. $\endgroup$ – Mr. X Dec 4 '17 at 19:36
0
$\begingroup$

For (1), you should write it's columns. E.g.

$$ Te_1 = \begin{bmatrix} -1 & 2 & 1 \\ 3 & 1 & 0\\ 1 & 1 & 1\end{bmatrix}\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} -1 \\ 3 \\ 1 \end{bmatrix}. $$

Also, when you're changing basis from the standard basis $\{e_1,e_2,e_3\}$ to the basis $\beta := \{v_1, v_2, v_3\}$, you can always compute the matrix representation of $A$ in $\beta$ by forming a matrix

$$ V = \begin{bmatrix} v_1 & v_2 & v_3 \end{bmatrix}, $$

by concatenating the column vectors in $\beta$. Then

$$ [A]_\beta = V^{-1} A_{\{e_1,e_2,e_3\}} V. $$

See if you can see why this works. For more information, see for example this.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.