how can i find the matrix $P$ that diagonalizes the matrix $A$?

Question

I want to find matrix $P$ that diagonalizes the matrix $A$: $$ \begin{bmatrix} 4 & 0 & 1 \\ 2 & 3 & 2 \\ 1 & 0 & 4 \\ \end{bmatrix} $$

Answer 1

You need to calculate the eigenvectors which are (1,2,1), (-1,0-1) and (0,1,0). So we have the passage matrix \begin{equation}P=\left(\begin{array}{ccc} 1 & -1& 0\\ 2 & 0 & 1\\ 1 & 1 & 0 \end{array}\right) \end{equation}

\begin{equation}P^{-1}=\left(\begin{array}{ccc} 1/2 & 0& 1/2\\ -1/2 & 0 & 1/2\\ -1 & 1 & -1 \end{array}\right) \end{equation} The inverse of P is \begin{equation}D=P^{-1}AP=\left(\begin{array}{ccc} 5 & 0& 0\\ 0 & 3 & 0\\ 0 & 0 & 3 \end{array}\right) \end{equation} So we have the diagonalization

Answer 2

Hint: The columns of $P$ are precisely the eigenvectors of $A$.

Answer 3

Suppose that $A$ has eigenvalues $\lambda_1,\lambda_2,\lambda_3$. Let $v_1,v_2,v_3$ be the associated eigenvectors. Let $D$ be the diagonal matrix with diagonal entries $\lambda_1,\lambda_2,\lambda_3$. Let $P$ be the matrix whose columns are $v_1,v_2,v_3$ in that order.

Then, we have $P^{-1}AP = D$, which is to say that $P$ diagonalizes $A$.

Answer 4

The general procedure to diagonalize $A$ is to solve $$A\vec x=\lambda \vec x$$

Then find the eigenvalues $\lambda$ by solving the characteristic equation $$\det(A-\lambda I)=0$$ where $I$ is the identity matrix.

Now for each eigenvalue $\lambda_i$, find the associated eigenvector $x_i$, by solving the homogeneous equation $$(A-\lambda_i I)x_i=0$$ Construct the matrix of eigenvectors $$S=(x_1,x_2,x_3,\cdots , x_n)$$

The last step is $$P=S^{-1}AS$$ where $$P=\mathrm{diag}(\lambda_1,\lambda_2,\lambda_3,\cdots , \lambda_n)$$ is the diagonalized matrix with the eigenvalues along the main diagonal.

In response to the comment below if: $$B= \begin{bmatrix} \lambda-4 & 0 & 7\\ 2 & \lambda & 0 \\ 0 & 2\lambda & 1 \\ \end{bmatrix} $$ then

$$\det(B-\lambda I)=0\implies\det\left( \begin{bmatrix} \lambda-4 & 0 & 7\\ 2 & \lambda & 0 \\ 0 & 2\lambda & 1 \\ \end{bmatrix}-\begin{bmatrix} \lambda & 0 & 0\\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \\ \end{bmatrix}\right)=0 $$

$$\implies \det \begin{pmatrix} -4 & 0 & 7\\ 2 & 0 & 0 \\ 0 & 2\lambda & 1-\lambda \\ \end{pmatrix}=0 $$