I"m not sure exactly how to strategy this problem. The examples I"ve come across on the internet present how to discover the adjust of collaborates matrix native a matrix to another matrix, such as B come C (for example).

You are watching: Find the change of coordinates matrix from b to the standard basis

I come up with solution but I"m not certain if it"s correct.

I started out with the matrix v those 3 vectors mentioned above:

3 2 10 2 -26 -4 3Then I uncovered the inverse i beg your pardon is the following:

1/21 5/21 1/72/7 -1/14 -1/72/7 -4/7 -1/7Then i multiplied by the conventional basis that a 3x1 vector:

1/21 5/21 1/7 12/7 -1/14 -1/7 * 02/7 -4/7 -1/7 0And come up through the following answer:

1/212/72/7Is this correct? Something tells me I"m missing something or perhaps I approached the entirety thing incorrectly.


linear-algebra matrices
re-publishing
cite
monitor
edited Feb 10 "19 in ~ 22:08
*

Glorfindel
3,6751010 gold badges2222 silver- badges3434 bronze badges
inquiry Aug 22 "14 in ~ 21:26
*

Peter GriffinPeter Griffin
19333 yellow badges44 silver badges1111 bronze title
$\endgroup$
add a comment |

2 answers 2


active earliest Votes
2
$\begingroup$
First, make sure you recognize what it method to create a vector $v$ in the basis $B$. In the standard basis, $S$, the vector $(1,2,3)_S$ is the straight combination$$1\cdot \left<\beginarrayc1\\0\\0\endarray\right> + 2\cdot \left<\beginarrayc0\\1\\0\endarray\right> + 3\cdot \left<\beginarrayc0\\0\\1\endarray\right> $$ i beg your pardon is the exact same as the matrix multiplication problem: $$ \left< \beginarrayccc 1&0&0\\0&1&0\\0&0&1 \endarray\right> \left<\beginarrayc1\\2\\3\endarray\right>.$$

In the basis $B$, the vector $(1,2,3)_B$ is the straight combination$$1\cdot \left<\beginarrayc3\\0\\6\endarray\right> + 2\cdot \left<\beginarrayc2\\2\\-4\endarray\right> + 3\cdot \left<\beginarrayc1\\-2\\3\endarray\right> = \left< \beginarrayccc 3&0&6\\2&2&-4\\1&-2&3 \endarray\right> \left<\beginarrayc1\\2\\3\endarray\right>.$$

In general, a the matrix $T$ of a basis have the right to be offered to readjust a vector $v_T$ in the basis to the standard basis $S$ via $T\cdot v_T = v_S$.

(This agrees v the fact that $I\cdot v_S = v_S$.)

In this case, a vector stood for in both $S$ and $B$ would certainly satsify $B\cdot v_B = i \cdot v_S$ and also $v_B = B^-1 \cdot v_S$.

See more: The Flash Season 3 Episode 2 Online (Full Episodes), The Flash Season 3

This shows just how $B$ and $B^-1$ room the matrices to go back and forth from $B$ to the traditional basis.