# diagonalize matrix via changing of basis

I have to figure out if it is possible to diaganolize matrix via changing of basis, and find this basis and appropriate matrix for this basis. Original matrix is:

$$\begin{pmatrix} -1 & 3 & -1 \\ -3 & 5 & -1 \\ -3 & 3 & 1 \end{pmatrix}$$

Which steps should I reproduce to compleate the task?

• Do you know how to find the eigenvalues and corresponding eigenvectors of a matrix? – StackTD May 24 '17 at 10:56
• you must find a set of eigenvectors for your matrix $Ax=\lambda x$ where x is an eigenvectgor and lambda is an eigenvalue – Jose Garcia May 24 '17 at 10:56
• @StackTD, of course. it is $$\mathrm{det}|A-{\lambda}I|$$ – M.Mass May 24 '17 at 11:07
• The answer is the matrix is diagonalizable, the comments above show you the right way. You can do it. – Fakemistake May 24 '17 at 12:14

A 3-by-3 matrix is diagonalizable if you can find a basis of $\mathbb{R^3}$ consisting of eigenvector of the matrix. This means you need 3 linearly independent eigenvectors.
You should find $\lambda = 1$ and $\lambda = 2$, a double eigenvalue.
For $\lambda = 2$, you should find two linearly independent eigenvectors.
The three eigenvectors form a basis of $\mathbb{R^3}$ and with respect to this basis, the matrix will become diagonal (having the eigenvalues as diagonal elements).