33.5. Example

Today, we're going to understand how to diagonalize a matrix. Can one of you explain what we mean by the characteristic polynomial?

Isn't it something we get by subtracting lambda times the identity matrix from our matrix A?

Exactly! And after we find that, we can determine the eigenvalues. Let's calculate the characteristic polynomial for our example matrix A.

So we set up the equation like this: det(A - λI)?

Correct! Now, who can tell me how to solve the determinant?

We'd expand it, right? I remember there's a formula for 2x2 matrices.

Great! Now, let's compute it together. The result is $\lambda^2 - 7\lambda + 10 = 0$.

To recap, we learned how to derive the characteristic polynomial and why it's essential for finding eigenvalues.

Now that we have our characteristic polynomial, how do we find the eigenvalues?

I think we need to solve the equation setting it to zero!

That's right! Let's solve $\lambda^2 - 7\lambda + 10 = 0$. Can anyone factor that for us?

It factors to $(\lambda - 5)(\lambda - 2) = 0$.

Well done! So, what are our eigenvalues?

Eigenvalues are $\lambda = 5$ and $\lambda = 2$.

Good job! These eigenvalues will help us find the eigenvectors next. Remember, we need these values for diagonalization.

Next, let’s find the eigenvectors corresponding to our eigenvalues. Can anyone remind us how we do that?

We solve $(A - \lambda I)v = 0$ for each eigenvalue?

Exactly! Let's start with $\lambda = 5$. Who wants to set up that equation?

For $\lambda = 5$, we have $(A - 5I)v = 0$.

Great! Now solve that to find the eigenvector.

We find $v = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ from that equation.

Perfect! Now let’s do the same for $\lambda = 2$. What do we get?

We also get an eigenvector of $v = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$!

Wonderful! We now have both eigenvectors we need — this is essential for forming our matrix P for diagonalization.

Now that we have our eigenvalues and eigenvectors, how do we construct matrices P and D?

We put the eigenvectors in P and the eigenvalues in D, right?

You nailed it! Let's arrange them in our matrices:

So, $P = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ and $D = \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix}$?

Exactly! Now, let's check if our diagonalization works by computing $A = PDP^{-1}$.

I can help with that calculation!

Fantastic! When you do, you'll confirm that matrix A is diagonalizable, reinforcing how critical this process is.

So, in summary, we've learned how to diagonalize a matrix by finding its eigenvalues, eigenvectors, and constructing matrices P and D.