I have to figure out if it is possible to diagonalize matrix via new basis, also find this basis and appropriate matrix for the basis. Original matrix is:

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

First of all find eigenvalues:

$$\det|A-\lambda I| = \begin{vmatrix} -1-\lambda & 3 & -1 \\ 3 & 5-\lambda & -1\\ -3 &3 & 1-\lambda \end{vmatrix} = \ldots = -\lambda^3+5\lambda^2+10\lambda-32$$


$-\lambda^3+5\lambda^2+10\lambda-32 = 0$ we finally have:

$$\lambda_1= 2 \\ \lambda_2=\frac{3}{2}+\frac{\sqrt{73}}{2}\\ \lambda_3=\frac{3}{2}-\frac{\sqrt{73}}{2}$$

For each eigenvalue we have to solve appropriate linear systems of equations and get $3$ eigenvectors:

$$a_1=(1,1,1) \\ a_2=(1,1,0)\\ a_3=(1,0,-3)$$

$\mathbf{Conclusion 1}:$ $\langle a_1,a_2,a_3\rangle$ - are basis, because they are linear independent

So, to find matrix in given basis I should write down transformation matrix:

$$ T =\left(\begin{array}{ccc|ccc} -1 & 3 & -1 & 1 & 1 & 1 \\ 3 & 5 & -1 & 1 & 1 & 0 \\ -3 & 3 & 1 & 1 & 0 & -3 \end{array}\right)$$

And get Identity matrix at left, so I will get appropriate matrix in the new basis, at right?

And how can I check that matrix can be diagonalized via new basis?

  • $\begingroup$ After you have the transformation matrices, multiply them out and check whether they are diagonal. Assuming $T$ is your transformation matrix, check whether $T^{-1}AT$ has diagonal form. $\endgroup$
    – B.Swan
    Jun 28 '17 at 15:08

It is diagonalised in the new basis, since it is a basis of eigenvectors, by definition of eigenvectors.


Any $n \times n$ matrix with $n$ distinct eigenvalues is diagonlizable, so the answer is yes.

Regarding the how - seems to me you did a fine job yourself. simply put the eigenvectors as column vectors of new matrix $P$, and then check that $P^{-1}AP$ is diagonal, with the eigenvalues on the diagonal.

  • $\begingroup$ What does "distinct" mean in this cintext? $\endgroup$
    – M.Mass
    Jun 28 '17 at 15:15
  • $\begingroup$ It means that the algebraic multiplicty of the root is 1. In our case, this simply means that all the lambdas are different. This is not always the case. Look for example at the eigenvalues of the identity matrix of size 3 by 3. It has 3 eigenvalues, but all of them are $1$. Having $n$ distinct eigenvalues is sufficient to ensure diagonlization, but it is not necessary. $\endgroup$ Jun 28 '17 at 15:21

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