# How to prove that A is diagonalizable?

So I was given this question in my exam, and it is by no mean a homework.

Let A = \begin{pmatrix} 5 & 1 & 0 \\ 1 & 5 & 0 \\ 0 & 0 & 6 \end{pmatrix}

(a) Find the eigenvalues and a basis for each of the eigenspaces of A answer: $\lambda_1 = \lambda_3 = 6$ and $\lambda_2 = 4$

The corresponding basis are: For $\lambda_1 = \lambda_3 = 6$:

Span = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} and \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}

For $\lambda_2= 4$: Span = \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}

now for the part (b): Show that $A$ is diagonalizable. Find an invertible matrix $P$ and a diagonal matrix $D$ such that $P^{-1} A P = D$ (do not verify).

So I assumed that since there is 2 distinct eigenvalues and 3 linearly independent eigenvectors, $A$ is diagonalizable. How do I continue? Should I assume that the entries of the diagonal matrix $D$ is the eigenvalues and proceed from there?

Thank you.

• You have done the work. Form $P$ from the 3 LI column eigenvectors you derived. Then $P^{-1} A P= D,$ where $D=\text{diag}(6,6,4).$ – matt biesecker May 11 '15 at 21:16

The eigenvectors are the columns of the matrix $P$.
You have: $$A=PDP^{-1}$$ with: $$P= \pmatrix{-1&0&1\\1&0&1\\0&1&0}$$ $$D= \pmatrix{4&0&0\\0&6&0\\0&0&6}$$
$$P^{-1}= \pmatrix{-1/2&1/2&1\\0&0&1\\1/2&1/2&0}$$
You have done the hard part. You need only organize it correctly. To do so, simply note that $$A=P\Lambda P^{-1}$$ where \begin{align*} P &= \begin{pmatrix}-1&0&1\\1&0&1\\0&1&0\end{pmatrix} & \Lambda &=\begin{pmatrix}4&0&0\\0&6&0\\0&0&6\end{pmatrix} \end{align*} The columns of $P$ are the eigenvectors of $A$ and $\Lambda$ is the diagonal matrix whose diagonal consists of the eigenvalues of $A$ ordered to coincide with the eigenvectors listed in $P$.