If $A=\begin{bmatrix} 1 & 2 \\ 5 & 4 \end{bmatrix}$, find a non singular matrix $P$ such that $P^{-1}AP=\begin{bmatrix} 6 & 0 \\ 0 & -1\end{bmatrix}$.
I let $P=\begin{bmatrix}a & b \\ c & d\end{bmatrix}$, so that $P^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ and I put it in the equation, but as expected, it will get tedious because of two degree terms. I couldn't think of any other method to solve it. I then tried to pre-muliply with $P$ in the equation to get $AP=P\begin{bmatrix} 6 & 0 \\ 0 & -1\end{bmatrix}$ and then letting again $P=\begin{bmatrix}a & b \\ c & d\end{bmatrix}$, but here though I am getting simple linear equations, but they are only two instead of four, namely $5a=2c$ and $b=-d$. So, this too fails.
I wanted to express $P$ solely in terms of $A$, $A^{-1}$ by some kind of manipulation in the original equation but couldn't think of any. I am sure there would exist a simple method for this. So, any hints?
Best Answer
Notice that $A$ has eigenvalues $\lambda_1=6$ and $\lambda_2=-1$. This suggests proceeding by diagonalizing $A$. For a $n\times n$ matrix $A$, if one has eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_n$ (they don't have to be distinct) with $n$ linearly independent eigenvectors, then one has that $$\begin{bmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{bmatrix}=P^{-1}AP,$$ where $P=\begin{bmatrix} \mathbf{v}_1 & \mathbf{v}_2 & \cdots & \mathbf{v}_n \end{bmatrix}$ is a matrix consisting of eigenvectors $\mathbf{v}_1,\mathbf{v}_2,\dots,\mathbf{v}_n$ corresponding to $\lambda_1, \lambda_2,\dots,\lambda_n$ respectively.
Hence, for our specific problem, we have that (this is one specific choice of eigenvectors, one could also take any non-zero scalar multiple) $$\mathbf{v}_1=\begin{bmatrix} 2 \\ 5 \end{bmatrix},\quad \mathbf{v}_2=\begin{bmatrix} -1 \\ 1 \end{bmatrix}.$$ Hence a possible matrix $P$ is given by $$P=\begin{bmatrix} 2 & -1 \\ 5 & 1 \end{bmatrix}.$$
Alternatively, proceeding by your method, you obtained the linear system (which was done correctly) $$\begin{cases} 5a=2c, \\ b=-d, \end{cases}$$ which has infinitely many solutions (it is a consistent linear system of nullity $2$), but if you fix for instance $c=5$ and $d=1$, you will get the same $P$ I have obtained. In fact, you could pick $P$ to be any matrix of the form (you can obtain this by fixing $c=\alpha$ and $d=\beta$) $$P=\begin{bmatrix} 2\alpha & -\beta \\ 5\alpha & \beta \end{bmatrix},$$ where $\alpha,\beta\neq 0$.