Consider matrices,
$ P = \begin{pmatrix} 3 & 1 \\ 5 & 2 \\\end{pmatrix}$
$ D = \begin{pmatrix} 1 & 0 \\ 0 & 2 \\\end{pmatrix}$
defined as $ A = PDP^{-1} $. The solutions state that the eigenvectors of $A$ can be found without calculation as,
$ v_{1} = \begin{pmatrix} 3 \\ 5 \\\end{pmatrix}$
$ v_{2} = \begin{pmatrix} 1 \\ 2 \\\end{pmatrix}$
I can find the above vectors using the standard method of finding eigenvalues and then equating to a zero vector, but what I would like to understand is how the vectors can be found WITHOUT any calculations?
Thank you.
Best Answer
If you have $A=PDP^{-1}$, then we have
$$AP=PD$$
Which is
$$A\begin{bmatrix}v_1 & v_2 \end{bmatrix}=\begin{bmatrix}d_1 v_1 & d_2v_2 \end{bmatrix}$$
$$\begin{bmatrix}Av_1 & Av_2 \end{bmatrix}=\begin{bmatrix}d_1 v_1 & d_2v_2 \end{bmatrix}$$
Hence columns of $P$ consists of the eigenvectors.