Help finding a matrix P that diagonalizes A

Question

$$ A = \begin{bmatrix}2&0&-2\\0&3&0\\0&0&3\end{bmatrix}$$ I know the eigenvalues are $\lambda= 2,3,3$

I solved for: $$\begin{bmatrix} {2-\lambda}&0&-2\\ 0&{3-\lambda}&0\\ 0&0&3-\lambda \end{bmatrix}x = 0$$ for each eigenvalue. I got $x = \begin{bmatrix}3\\0\\0\end{bmatrix}$ for $\lambda=2$ and $\begin{bmatrix}-2\\1\\1\end{bmatrix}$ for $\lambda=3$

I have two bases now for each eigenspace put that means my P matrix has only 2 vector columns. Can someone help me figure out a matrix P with the X columns I found? Since 3 is listed twice, how does that affect my matrix P? I just learned about diagonalization so I appreciate the help.

Answer 1

Let $A$ be a square matrix of order $n$. Assume that $A$ has $m$ distinct eigenvalues. Then A is diagonalizable if the algebraic multiplicity of each eigenvalue is equal to the geometric multiplicity, Moreover, if P is the matrix with the columns $C_1, C_2,\dots, C_n$, the $n$ independent eigenvectors of A, then the matrix $P^{-1}AP$ is a diagonal matrix. In other words, the matrix $A$ is diagonalizable.

So basically just find the eigenvectors corresponding to the eigenvalues then you will get $P$, In this problem you will have two eigenvectors corresponding to the eigenvalue $3$ and $1$ eigenvector corresponding to the eigenvalue $2$

For eigenvalue $3$ we have $(-2,1,1)$ and $(-2,0,1)$ as two independent eigenvectors and for eigenvalue equal to $2$ we have $(3,0,0)$ as an eigenvector

So make these $3$ independent vectors the columns of $P$, then you will get your required matrix.

Answer 2

Let $v_1 = \begin{bmatrix}3\\0\\0\end{bmatrix}$, $v_2 = \begin{bmatrix}-2\\1\\1\end{bmatrix}$, and $v_3 = \begin{bmatrix}-4\\0\\2\end{bmatrix}$. The three vectors are linearly independent.

Then, we have: $$Av_1=2v_1$$ $$Av_2=3v_2$$ $$Av_3=3v_3$$

Let $P=\begin{bmatrix}v_1&v_2&v_3\end{bmatrix}$.

Combine to get: $$\begin{array}{rcl} A\begin{bmatrix}v_1&v_2&v_3\end{bmatrix} &=& \begin{bmatrix}2v_1&3v_2&3v_3\end{bmatrix} \\ A\begin{bmatrix}v_1&v_2&v_3\end{bmatrix} &=& \begin{bmatrix}v_1&v_2&v_3\end{bmatrix} \begin{bmatrix}2&0&0\\0&3&0\\0&0&3\end{bmatrix} \\ A P &=& P \begin{bmatrix}2&0&0\\0&3&0\\0&0&3\end{bmatrix} \\ A &=& P \begin{bmatrix}2&0&0\\0&3&0\\0&0&3\end{bmatrix} P^{-1} \\ \end{array}$$

The invertibility of $P$ follows form the linearly independence of $v_1$, $v_2$, and $v_3$.

Hence, $P$ is your desired matrix.