Is This Matrix Diagonalizable?

Question

Consider the matrix A below: $$A = \begin{pmatrix} 3 & 2 & 2 \\ -1 & 0 & -2 \\ 1 & 2 & 4\end{pmatrix}$$

Is the matrix A diagonalizable? If so, then find a diagonal matrix D and an invertible matrix P such that $P^{-1}AP=D$.

I know it is supposed to be diagonalizable but I have tried to solve it and haven't succeeded. The format I'm used to is $A=PDP^{-1}$.

Thanks.

Answer 1

Doing some regular computations for eigenvalues and eigenvectors you should find

$$\matrix{ {{\lambda _1} = 2} \hfill & \to \hfill & {{X_1} = \left[ {\matrix{ { - 2} \cr 0 \cr 1 \cr } } \right]} \hfill & {{X_2} = \left[ {\matrix{ { - 2} \cr 1 \cr 0 \cr } } \right]} \hfill \cr {{\lambda _2} = 3} \hfill & \to \hfill & {{X_3} = \left[ {\matrix{ 1 \cr { - 1} \cr 1 \cr } } \right]} \hfill & {} \hfill \cr } $$

and hence your matrix $P$ will be

$$P = \left[ {\matrix{ {{X_1}} & {{X_2}} & {{X_3}} \cr } } \right] = \left[ {\matrix{ \hfill { - 2} & \hfill { - 2} & \hfill 1 \cr \hfill 0 & \hfill 1 & \hfill { - 1} \cr \hfill 1 & \hfill 0 & \hfill 1 \cr } } \right]$$

and finally

$$D = {P^{ - 1}}AP = \left[ {\matrix{ 2 & 0 & 0 \cr 0 & 2 & 0 \cr 0 & 0 & 3 \cr } } \right]$$

will be the diagonal matrix where its diagonal elements are the eigenvalues. However, as $\lambda_1=2$ had multiplicity $2$, it will appear twice on the diagonal here! :)

Answer 2

The characteristic polynomial $p(X)=\det(A-XI_3)$ is $$ p(X)=12-16X+7X^2-X^3=(3-X)(2-X)^2 $$ Now, can you check the geometric multiplicity of the eigenvalue $2$?

The rank of $A-2I_3$ is $1$, so the geometric multiplicity is $3-1=2$. So yes, the matrix is diagonalizable. Find a basis for the eigenspace relative to $3$ and a basis for the eigenspace relative to $2$ and you've found the diagonalizing matrix.

Answer 3

Assuming $A$ is an $n\times n$ matrix, $A$ diagonalizable if $P_A(\lambda)$ has $n$ distinct real roots or if, for each eigenvalue, the algebraic multiplicity equals the geometric multiplicity. So let's find out:

$$P_A(\lambda) = det(A - \lambda I) = det \begin{pmatrix} 3-\lambda & 2 & 2 \\ -1 & -\lambda & -2 \\ 1 & 2 & 4-\lambda\end{pmatrix}$$ $$ = (3-\lambda)((-\lambda)(4-\lambda) + 4) - 2(\lambda - 4 + 2) + 2(-2+\lambda) = 0$$ $$\Rightarrow -\lambda^3+7 \lambda^2-16 \lambda+12 = 0$$ $$\Rightarrow -(\lambda-3) (\lambda-2)^2 = 0$$ $$\lambda = 3 \text{ (multiplicity $1$)}$$ $$\lambda = 2 \text{ (multiplicity $2$)}$$

We see that $\lambda=2$ has algebraic multiplicity 2, so we need to check for its geometric multiplicity.

Geometric multiplicity is equal to $n - Rank(A - \lambda I)$ $$Rank(A - 2I) = dim\left(col\begin{pmatrix} 1 & 2 & 2 \\ -1 & -2 & -2 \\ 1 & 2 & 2\end{pmatrix}\right) $$ $$ = dim\left(col\begin{pmatrix} 1 & 2 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}\right) = 1$$

So $$\text{geometric multiplicity} = 3 - Rank(A - 2I) = 3 - 1 = 2$$ $$ = \text{algebraic multiplicity when $\lambda=2$}$$

So the matrix is diaginalizable.

Now we can diaginalize it:

Let's find our eigenvectors.
For $\lambda=2$: $$\vec{V_1} + \vec{V_2} = null(A - \lambda I) = null(A - 2I) = null\begin{pmatrix} 1 & 2 & 2 \\ -1 & -2 & -2 \\ 1 & 2 & 2\end{pmatrix}$$ $$ = null\begin{pmatrix} 1 & 2 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}$$ $$ = \begin{pmatrix} -2 \\ 1 \\ 0\end{pmatrix}x_1 + \begin{pmatrix} -2 \\ 0 \\ 1\end{pmatrix}x_2$$ So our eigenpairs so far are $\left(2, \begin{pmatrix} -2 \\ 1 \\ 0\end{pmatrix}\right)$ and $\left(2, \begin{pmatrix} -2 \\ 0 \\ 1\end{pmatrix}\right)$

Now, to find it for $\lambda=3$: $$\vec{V} = null(A-\lambda I) = null(A - 3I) = null\begin{pmatrix} 0 & 2 & 2 \\ -1 & -3 & -2 \\ 1 & 2 & 1\end{pmatrix}$$ $$ = null\begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 0\end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}x_3$$ So our eigenpair is $\left(3, \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}\right)$

We now have everything we need to make our diaginalized matrix:

$$PDP^{-1} = \begin{pmatrix}-2 & -2 & 1 \\ 1 & 0 & -1 \\ 0 & 1 & 1 \end{pmatrix}\begin{pmatrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3\end{pmatrix}\begin{pmatrix}-2 & -2 & 1 \\ 1 & 0 & -1 \\ 0 & 1 & 1 \end{pmatrix}^{-1}$$ $$ = \begin{pmatrix}-2 & -2 & 1 \\ 1 & 0 & -1 \\ 0 & 1 & 1\end{pmatrix}\begin{pmatrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3\end{pmatrix}\begin{pmatrix}1 & 3 & 2 \\ -1 & -2 & -1 \\ 1 & 2 & 2\end{pmatrix}$$