1. ## Similar matrices

Show that $\displaystyle A$ and $\displaystyle B$ are similar by finding $\displaystyle M$ so that $\displaystyle B=M^{-1}AM$.

$\displaystyle A=\left[ \begin{array}{cc} 1 & 0 \\ 1 & 0 \end{array} \right]$

$\displaystyle B=\left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right]$

2. Well, define $\displaystyle \displaystyle M = \left[\begin{matrix}a&b\\c&d\end{matrix}\right]$ so that $\displaystyle \displaystyle M^{-1} = \frac{1}{ad - bc}\left[\begin{matrix}\phantom{-}d&-b\\-c&\phantom{-}a\end{matrix}\right]$.

What is $\displaystyle \displaystyle M^{-1}AM$? Set it equal to $\displaystyle \displaystyle B$ and see if you can solve for $\displaystyle \displaystyle a,b,c,d$.

3. Equivalently, $\displaystyle B= M^{-1}AM$ is the same as $\displaystyle MB= AM$ so set
$\displaystyle \begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}= \begin{bmatrix}1 & 0 \\ 1 & 0\end{bmatrix}\begin{bmatrix}a & b \\ c & d\end{bmatrix}$
and solve for a, b, c, and d.

4. Originally Posted by HallsofIvy
Equivalently, $\displaystyle B= M^{-1}AM$ is the same as $\displaystyle MB= AM$ so set
$\displaystyle \begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}= \begin{bmatrix}1 & 0 \\ 1 & 0\end{bmatrix}\begin{bmatrix}a & b \\ c & d\end{bmatrix}$
and solve for a, b, c, and d.
$\displaystyle M=\begin{bmatrix}a & 0 \\ a & d\end{bmatrix}$