Verify my attempt to diagonalize matrix

Question

Decide whether matrix $A$ is diagonalizable. If so, find $P$ such that $P^{-1}AP$ is diagonal.

We are given: $A = \begin{bmatrix}1 & 0 & 0 \\ 1 & 2 & 1 \\ 0 & 0 & 1\end{bmatrix}$

We set up and and solve: $|A - \lambda I| = 0$, which yields:

$$\left|\begin{matrix}1-\lambda & 0 & 0 \\ -1 & 2-\lambda & -1 \\ 0 & 0 & 1-\lambda\end{matrix}\right| = 0$$

This yields a characteristic polynomial and eigenvalues as:

$$-\lambda^3 + 4 \lambda^2 - 5 \lambda + 2 = (1-\lambda)^2 (2-\lambda) = 0 ~~~\rightarrow ~~~ \lambda_1 = 1, \lambda_{2,3} = 2$$

We have multiplicities of $2$ and $1$ for those eigenvalues.

So, for $\lambda_1 = 1$, we have:

$[A- I]v_1 = \begin{bmatrix}0 & 0 & 0 \\ -1 & 1 & -1 \\ 0 & 0 & 0\end{bmatrix}v_1 = 0$

Thus, $v_1 = (1,1,0)$ and $v_2 = (0,1,1)$ .

Repeating this same process for the second eigenvalue, we have as RREF:

$\begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}v_3 = 0$

This yields an eigenvector of $v_3 = (0,1,0)$.

Putting all of this together, we have the eigenvalue/eigenvector pairs:

• $\lambda_1 = 1, v_1 = (1,1,0)$
• $\lambda_2 = 1, v_2 = (0,1,1)$
• $\lambda_3 = 2, v_3 = (0,2,0)$

So P=\begin{bmatrix}1 & 0 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 0\end{bmatrix}

Is this right? Do I need to show more work? Do I need to take any additional steps after I find $P$?

Answer 1

I think you have mistake since $|A-\lambda I|=\left|\begin{matrix}1-\lambda & 0 & 0 \\ 1 & 2-\lambda & 1 \\ 0 & 0 & 1-\lambda\end{matrix}\right| = 0$. the polynom is by minors $p(x)=(1-\lambda)^2(2-\lambda)$ and $\lambda_{1,2}=1,\lambda_3=2$. as a result of the little mistake in the matrix,your eigenvectors are not correct.so for $\lambda=1$ we can write: $$v_{1}=(-1,0,1),v_2=(0,1,-1)$$ and for $\lambda=2,$ we can get $$v_3=(0,1,0)$$so P is $$P=\begin{bmatrix}-1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & -1 & 0\end{bmatrix} $$

Answer 2

I still cannot understand why so many books and prof's still define the characteristic pol. as $\,\det(A-\lambda I)\;$ instead of the usually much simpler (imho) $\,\det(xI-A)\,$ : in your determinant you have the middle row's signs wrong, which causes you to have the eigenvectors wrong...

$$\det(xI-A)=\begin{vmatrix}x-1&0&0\\\!-1&x-2&\!-1\\0&0&x-1\end{vmatrix}=(x-1)^2(x-2)$$

$$\underline{\lambda=1:}\;\;\; -x-y-z=0\implies y=-x-z\implies \begin{pmatrix}1\\\!\!-1\\0\end{pmatrix}\;,\;\;\begin{pmatrix}0\\\!\!-1\\1\end{pmatrix}$$

$$\underline{\lambda=2:}\;\;\; \begin{align*}x&=0\\z&=-x=0\end{align*}\;\;\implies\begin{pmatrix}0\\1\\0\end{pmatrix} \;\ldots\ldots\;\text{and etc.}$$