Intereting Posts

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Find a matrix transformation mapping $\{(1,1,1),(0,1,0),(1,0,2)\}$ to $\{(1,1,1),(0,1,0),(1,0,1)\}$.

Is the answer

$$

\begin{bmatrix}1& 0& -1\\0& 1& 1\\0& 0& 1\end{bmatrix}?

$$

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I understand the concept of Matrix Transformation, I don’t think I’m doing it right.

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We wish to find a $3\times 3$ matrix $T$ such that $TA=B$ where

\begin{align*}

A &=\begin{bmatrix}1 & 0 & 1\\ 1 & 1 & 0 \\ 1 & 0 & 2 \end{bmatrix}

&

B &= \begin{bmatrix} 1 & 0 & 1\\ 1 & 1 & 0\\ 1 & 0 & 1\end{bmatrix}

\end{align*}

Perhaps the quickest way to find $T$ is to multiply the equation $TA=B$ on the right by $A^{-1}$ to obtain

$$

T=BA^{-1}

$$

Can you compute $A^{-1}$ and carry out the matrix multiplication?

The columns of the matrix tell you where it sends the standard basis vectors. For instance if I am interested in the third column then I need to determine what the action of our linear operator is on the column vector ,

$$\left[ \begin{array}{c}0 \\ 0 \\ 1 \end{array}\right].$$

This vector can be written as a linear combination of the vectors used to define the linear operator,

$$\left[ \begin{array}{c}0 \\ 0 \\ 1 \end{array}\right] =

\left[ \begin{array}{c}1 \\ 0 \\ 2 \end{array}\right]

-\left[ \begin{array}{c}0 \\ 1 \\ 0 \end{array}\right]

-\left[ \begin{array}{c}1 \\ 1 \\ 1 \end{array}\right].$$

Multiplying both sides by our linear operator $M$ we get,

$$M\left[ \begin{array}{c}0 \\ 0 \\ 1 \end{array}\right] =

M\left[ \begin{array}{c}1 \\ 0 \\ 2 \end{array}\right]

-M\left[ \begin{array}{c}0 \\ 1 \\ 0 \end{array}\right]

-M\left[ \begin{array}{c}1 \\ 1 \\ 1 \end{array}\right].$$

Note that we know what $M$ does to the vectors on the right so we can just substitute those values in and add,

$$M\left[ \begin{array}{c}0 \\ 0 \\ 1 \end{array}\right] =

\left[ \begin{array}{c}1 \\ 0 \\ 1 \end{array}\right]

-\left[ \begin{array}{c}0 \\ 1 \\ 0 \end{array}\right]

-\left[ \begin{array}{c}1 \\ 1 \\ 1 \end{array}\right]

=\left[ \begin{array}{c}0 \\ -2 \\ 0 \end{array}\right].$$

The resulting vector is the third column of our matrix,

$$ M = \left[\begin{array}{ccc}

\ & \ & 0 \\

\ & \ & -2 \\

\ & \ & 0

\end{array}\right].$$

A similar process will yield the other collumns.

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