22
$\begingroup$

Can someone show me step-by-step how to diagonalize this matrix? I'm trying to teach myself differential equations + linear algebra, but I'm stumped on how to do this. I'd really appreciate if someone would take the time to do this with me!

$\begin{bmatrix} 1 & 2 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & -3 \\ \end{bmatrix}$

$\endgroup$
59
$\begingroup$

First step: Find the eigenvalues of your matrix.

Eigenvectors are vectors $\mathbf x$ such that upon being multiplied by a matrix $A$, they are only scaled by a number. That is $A\mathbf x = \lambda \mathbf x$, where $\lambda$ is just a number, called the eigenvalue associated with the eigenvector $\mathbf x$.

The way to do this is to subtract the $\lambda \mathbf x$ from both sides to get $A\mathbf x -\lambda \mathbf x=\mathbf 0$. Now factor out the $\mathbf x$ to get $(A-\lambda I)\mathbf x = \mathbf 0$, where $I$ is the identity matrix -- note: we need the identity matrix because adding a matrix and a scalar is undefined.

This equation, $(A-\lambda I)\mathbf x = \mathbf 0$ has a nontrivial solution (a solution where $\mathbf x \ne 0$) if and only if $\det(A-\lambda I)=0$ (can you prove this?).

So let's look at that determinant: $$\det(A-\lambda I) = \det\left(\begin{bmatrix} 1 & 2 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & -3 \\ \end{bmatrix} - \begin{bmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix}\right)$$ $$= \det\left(\begin{bmatrix} 1 - \lambda & 2 & 0 \\ 2 & 1 - \lambda & 0 \\ 0 & 0 & -3-\lambda \\ \end{bmatrix}\right) = 0$$

Using our formula for the determinant we see that this is equal to $(1-\lambda)((1-\lambda)(-3-\lambda)-0)-2(2(-3-\lambda)-0)=0$. This has solutions (if I didn't make a mistake) of $\lambda =-3$ or $\lambda = -1$ or $\lambda = 3$.

These are the eigenvalues of this matrix.

Second step: Put these numbers into the diagonal of a matrix -- with zeroes everywhere else -- in whichever order you like and you're done! Yippee!

$$\begin{bmatrix} -3 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 3\end{bmatrix}$$ is one of your possible diagonal matrices.


Let's verify our solution, though.

Third step: Find the eigenvectors associated with the eigenvalues of this matrix.

Let's start with the first eigenvalue $\lambda = -3$ and plug it into our matrix $A-\lambda I$ and then try to find all of the solutions to $(A-\lambda I)\mathbf x = \mathbf 0$.

$$(A-\lambda I)\mathbf x = \begin{bmatrix} 4 & 2 & 0 \\ 2 & 4 & 0 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

Now we note that if $A\mathbf x= \mathbf 0$, then $R\mathbf x = \mathbf 0$ where $R$ is the RREF form of $A$. I assume you know how to row reduce a matrix, so I'll just show you the $R\mathbf x=\mathbf 0$ form:

$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}$$

Here we can clearly see that all solutions $\mathbf x$ to this equation are of the form $t\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$, for some scalar $t$. Thus $\begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}$ is the eigenvector associated with the eigenvalue $\lambda =-3$ of the matrix $A$.

Following the exact same procedure: you see that the other two eigenvectors are $\mathbf x_{\lambda_2} = \begin{bmatrix} -1 \\ 1 \\ 0\end{bmatrix}$ and $\mathbf x_{\lambda_{3}} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$.

Fourth step: Verify your diagonal matrix by constructing a block matrix $P$ of your eigenvectors and using the equation $P^{-1}AP=\Lambda$.

Stick your eigenvectors into the columns of a matrix: $$P = \begin{bmatrix} 0 & -1 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0\end{bmatrix}$$

Find the inverse $P^{-1}$ of that matrix: $$P^{-1} = \begin{bmatrix}0 & 0 & 1 \\ -\frac 12 & \frac 12 & 0 \\ \frac 12 & \frac 12 & 0\end{bmatrix}$$

Now multiply out the expression $P^{-1}AP$ and see if you get back your diagonal matrix.

$$P^{-1}AP = \begin{bmatrix}0 & 0 & 1 \\ -\frac 12 & \frac 12 & 0 \\ \frac 12 & \frac 12 & 0\end{bmatrix}\begin{bmatrix} 1 & 2 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & -3 \\ \end{bmatrix}\begin{bmatrix} 0 & -1 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0\end{bmatrix} = \begin{bmatrix} -3 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 3\end{bmatrix}$$

Success!

$\endgroup$
  • $\begingroup$ You're awesome! I'm going to upvote + vote as best once I can! $\endgroup$ – Josh Dec 12 '14 at 1:35
  • 1
    $\begingroup$ @Josh - you can already choose this as the best answer, hover over the green tick on the left under the number. $\endgroup$ – Suzu Hirose Dec 12 '14 at 1:37
  • $\begingroup$ @bye_world Detailed! +1 $\endgroup$ – Swapnil Tripathi Dec 14 '14 at 18:03
  • $\begingroup$ How can you prove that This equation, (A−λI)x=0 has a nontrivial solution (a solution where x≠0) if and only if det(A−λI)=0 $\endgroup$ – octavian May 19 '17 at 13:20
  • $\begingroup$ @octavian The proof of that would require more space than is available in the comments. You should ask it as a separate question. $\endgroup$ – user137731 May 19 '17 at 14:04
2
$\begingroup$

Since Bye_World has done everything, I just want to try proving his statement:

This equation, $(A−\lambda I)x=0$ has a nontrivial solution (a solution where $x\neq 0$) if and only if $\det(A−\lambda I)=0$"

We first assume that $B = A−\lambda I$ is invertible (or $\det(B)≠0$). If this is true, the equation can be rewritten as:

$$B^{-1}Bx = B^{-1}0=0$$

Since $B^{-1}B=I$ (the identify matrix), then $x=0$, which is false. Then B is not invertible or $\det(B)=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.