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$A = \left(\begin{array}{cc}2 & -1 \\0 & 2\end{array}\right)$

$B = \left(\begin{array}{cc}\lambda & 0 \\1 & \lambda\end{array}\right)$.

I know that the $\lambda = 2$. And $r(1,0)^t$, where $r \in $ Reals are the eigenvectors.

How do you show, via calculation, that $A$ is similar to $B$?

Note: If two matrices are similar, then there is a third matrix $C$, such that $B = C^{-1}AC$.

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That means find an invertible matrix $P$ with \begin{align} B = P^{-1} A P \iff \\ \left( \begin{array}{cc} \lambda & 0 \\ 1 & \lambda \end{array} \right) &= \frac{1}{ad-bc} \left( \begin{array}{cc} d & -b \\ -c & a \end{array} \right) \left( \begin{array}{cc} 2 & -1 \\ 0 & 2 \end{array} \right) \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) \\ &= \frac{1}{ad-bc} \left( \begin{array}{cc} d & -b \\ -c & a \end{array} \right) \left( \begin{array}{cc} 2a-c & 2b-d \\ 2c & 2d \end{array} \right) \\ &= \frac{1}{ad-bc} \left( \begin{array}{cc} 2(ad-bc) - cd & -d^2 \\ c^2 & 2(ad-bc) + cd \end{array} \right) \end{align} Comparing components, we take $d=0$ and continue with \begin{align} \left( \begin{array}{cc} \lambda & 0 \\ 1 & \lambda \end{array} \right) &= \frac{1}{-bc} \left( \begin{array}{cc} -2bc & 0 \\ c^2 & -2bc \end{array} \right) \\ &= \left( \begin{array}{cc} 2 & 0 \\ -c/b & 2 \end{array} \right) \end{align} So we can pick any $a$, any $b \ne 0$ and then choose $c = -b$ and $d = 0$. $$ P = \left( \begin{array}{cc} a & b \\ -b & 0 \end{array} \right) \quad P^{-1} = \left( \begin{array}{cc} 0 & -1/b \\ 1/b & a/b^2 \end{array} \right) $$ For example $$ P = \left( \begin{array}{cc} 1 & 1 \\ -1 & 0 \end{array} \right) \quad P^{-1} = \left( \begin{array}{cc} 0 & -1 \\ 1 & 1 \end{array} \right) $$ should do the job of making $B$ similar to $A$. \begin{align} P^{-1} A P &= \left( \begin{array}{cc} 0 & -1 \\ 1 & 1 \end{array} \right) \left( \begin{array}{cc} 2 & -1 \\ 0 & 2 \end{array} \right) \left( \begin{array}{cc} 1 & 1 \\ -1 & 0 \end{array} \right) \\ &= \left( \begin{array}{cc} 0 & -1 \\ 1 & 1 \end{array} \right) \left( \begin{array}{cc} 3 & 2 \\ -2 & 0 \end{array} \right) \\ &= \left( \begin{array}{cc} 2 & 0 \\ 1 & 2 \end{array} \right) \\ &= B \end{align}

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Finding the transpose of an upper triangular matrix will render it similar to a lower triangular matrix, as introduced by British mathematician Arthur Cayley.

Let $\mathbf U$ be an upper triangular matrix:

\begin{bmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1,n-1} & a_{1n} \\ 0 & a_{22} & a_{23} & \cdots & a_{2,n-1} & a_{2n} \\ 0 & 0 & a_{33} & \cdots & a_{3,n-1} & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & a_{n-1,n-1} & a_{n-1,n} \\ 0 & 0 & 0 & \cdots & 0 & a_{nn} \\ \end{bmatrix}

By definition:

$$\forall a_{ij} \in \mathbf U: i > j \implies a_{ij} = 0$$

We can let $\mathbf{U}^T$ be the transpose of $\mathbf{U}$, so $\mathbf U^\intercal = \left[{b}\right]_n: \forall i \in \left[{1 \,.\,.\, n}\right], j \in \left[{1 \,.\,.\, n}\right]: b_{i j} = a_{j i}$

From this, we get:

$$\forall b_{ji} \in \mathbf U^\intercal: i > j \implies b_{ji} = 0$$

Then, exchange $i$ and $j$ in the notation above to get:

$$\forall b_{ij} \in \mathbf U^\intercal: i < j \implies b_{ij} = 0$$

Use this theorem to show how matrix $A$ is similar to matrix $B$.

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