We did a problem in class I didn't understand.
"Find a vector w so that:"
[1 3]w = 2w
[1 -1]
I'm confident I could do it if it equaled only 2, but the 2w is throwing me off...
Let w be the vector
$\displaystyle w = \left [ \begin{matrix} a \\ b \end{matrix} \right ] $
Then we need to find a, b such that the following is true:
$\displaystyle \left [ \begin{matrix} 1 & 3 \\ 1 & -1 \end{matrix} \right ] ~ \left [ \begin{matrix} a \\ b \end{matrix} \right ] = 2 \left [ \begin{matrix} a \\ b \end{matrix} \right ] $
Multiplying this out we get the two equations
$\displaystyle \begin{matrix} a + 3b = 2a \\ a - b = 2b \end{matrix} $
What can a and b be?
-Dan
?????
\[\left[ {\begin{array}{*{20}{c}}
1&3 \\
1&{ - 1}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{x + 3y} \\
{x - y}
\end{array}} \right]
\\\\\left[ {\begin{array}{*{20}{c}}
1&3 \\
1&{ - 1}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
0 \\
0
\end{array}} \right] = 2\left[ {\begin{array}{*{20}{c}}
{0} \\
{0}
\end{array}} \right]\]
It appears that the solution is any vector [w1, w2] where w1/w2=3.
If I am not mistaken, this kind of solution often occurs in vibrations where multiple masses oscillate with any amplitude, however these amplitudes have fixed ratios (eigenvectors). Been a while though, I may be wrong.
If it only equaled two it would be impossible! The left side is a vector while "2" is not.
Write [tex]w= \begin{bmatrix} x \\ y \end{bmatrix}[tex]. The equation becomes
$\displaystyle \begin{bmatrix}1 & 3 \\ 1 & -1 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}x+ 3y \\ x- y \end{bmatrix}= 2\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}2x \\ 2y \end{bmatrix}$.
That is the same as the two equations x+ 3y= 2x and x- y= 2y. It should be easy to see that the only solution is x= y= 0.
A little more "sophisticated" would be to write the equation as $\displaystyle \begin{bmatrix}1 & 3 \\ 1 & -1 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}2 & 0 \\ 0 & 2 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}$.
Then $\displaystyle \begin{bmatrix}1 - 2 & 3- 0 \\ 1- 0 & -1- 2\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}-1 & 3 \\ 1 & -3 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$.
Now, as long as that matrix is invertible, we could multiply both sides by its inverse and get, since the product of any matrix with the 0 vector is the 0 vector, $\displaystyle \begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$.