How to diagonalize a matrix

Question

So I am trying to diagonalize this matrix $$ A=\begin{pmatrix}2&0&\!\!-2\\1&3&\;2\\0&0&\;3\end{pmatrix} $$ so that those are the rows of the matrix. I know the eigen values are 2 and 3. I don't think that this matrix can be diagonalized, until wolfram alpha said the vectors are {-1,1,0}, {-2,0,1}, and {0,1,0}, respectively. I don't understand how one can get a different eigen vectors from the same eigen value. Please explain how I get that last eigen vector

Answer 1

Recall that an eigenvector $v$ with eigenvalue $\lambda$ solves the equation $(A-\lambda I)v = 0$.

If you have a single eigenvalue, then the associated eigenvector lies in the null space of $A-\lambda I$. Any multiple of this eigenvector will also be in the nullspace, so if you compute $(A-\lambda I) \alpha v$ for any value of $\alpha$, you'll still get the zero vector as an answer.

In the case of a $3\times 3$ matrix, we think of the eigenvector associated with the non-repeated eigenvalue as drawing a line in 3-D space. Picking any point on that line will result in the zero vector!

Now, when we have a twice-repeated eigenvalue, we might expect that we get a two-dimensional structure describing our nullspace. And that's exactly what happens. If you compute $A-\lambda I$ when $\lambda = 3$, you get a $3 \times 3$ rank-1 matrix. The nullspace has dimension two. it may not be immediately apparent, but it should be clear that the two eigenvectors associated with this repeated eigenvalue form a basis for the nullspace! These two vectors form a plane in 3-D space, and if you pick any point on the plane, say $z$, then $(A-\lambda I)z = 0$. Of course, $z$ can be described as the sum of multiples of the eigenvectors, $z = a v_1 + bv_2$.

If we have a repeated eigenvalue, we should examine the rank of the matrix $A-\lambda I$ (and consequently, the dimensionality of its null space). The dimension of the null space will give us the number of associated eigenvectors.

Answer 2

We have

$$A=\begin{pmatrix}2&0&\!\!-2\\1&3&\;2\\0&0&\;3\end{pmatrix}\implies |xI-A|=\begin{vmatrix}x-2&0&\;2\\\!\!-1&x-3&\!\!-2\\0&0&x-3\end{vmatrix}=(x-2)(x-3)^2$$

Now calculate eigenvector(s) for the bigger eigenvalue:

$$\begin{cases}\;\;x+2z=0\\{}\\-x-2z=0\end{cases}\implies\;x=-2z$$

and we get that the solution system above has dimension two and this means the eigenspace corresponding to $\;\lambda=3\;$ has dimension $\,2\,$, so now just choose one eigenvector corr. to $\;\lambda=2\;$ and two linearly independent eigenvector corr. to $\;\lambda =3\;$ and with them form a matrix that will diagonalize $\;A\;$ ...