I am taking a Linear Algebra course and I came across this problem.
Find a 4 × 4 invertible real matrix P and a 4 × 4 real diagonal matrix Λ such that:
$P^{−1}AP = Λ$.
A note is given that $P^{-1}$ does not need to be calculated.
I am having difficulty with it, I have come across the form $PAP^{−1} = Λ$ before but never $P^{−1}AP= Λ$. Any advice would be greatly appreciated.
Edit: I missed part of the question that provides context. Full question below.
In following the following problem u,v,w will be the following vectors:
$$u = \begin{pmatrix}1 \\\ 0\\3 \end{pmatrix}, v = \begin{pmatrix}2 \\-1\\1 \end{pmatrix}, q = \begin{pmatrix}3 \\\ 1\\2 \end{pmatrix}$$If a 3×3 matrix M is such that Mu=−5u, Mv=0 and Mw=3w,
Find a 4 × 4 invertible real matrix P and a 4 × 4 real diagonal matrix Λ such that:
P^{−1}AP = Λ.
A note is given that $P^{-1}$ does not need to be calculated.
I attempted to solve this problem, but I do think not that my attempts are correct:
Starting out I used the vectors u, v,w to form an invertible matrix (and checked for linear independence).
That gave the matrix P:
$ \begin{pmatrix}
1&2&3\\\
0&-1&1\\\
3&1&2\\\
\end{pmatrix} $Using the info from Mu = -5u, Mv = 0v, Mw = 3w
=$\lambda = -5, 0, 3$ to make a real diagonal matrix M:
$ \begin{pmatrix}
-5&0&0\\\
0&0&0\\\
0&0&3\\\
\end{pmatrix} $$\Rightarrow P^{-1}MP = \Lambda $
$ \begin{pmatrix}
1&2&3\\\
0&-1&1\\\
3&1&2\\\
\end{pmatrix}^{-1} \cdot \begin{pmatrix}
-5&0&0\\\
0&0&0\\\
0&0&3\\\
\end{pmatrix} \cdot \ \begin{pmatrix}
1&2&3\\\
0&-1&1\\\
3&1&2\\\
\end{pmatrix} = $ $\Lambda $
Best Answer
For $$P^{−1}AP= Λ$$ your matrix $P$ is the matrix whose columns are the eigenvectors of $A$ and Λ is the diagonal matrix of eigenvalues.
Note that we need linearly independent eigenvectors in order for the matrix $P$ to be invertible.