Definition 4.3.1 Diagonalizable
We say that an \(n\times n\) matrix \(A\) is diagonalizable if it is similar (Definition 4.2.2) to a diagonal matrix \(D\text{.}\) That is if there exists an invertible matrix \(P\) such that
We say that an \(n\times n\) matrix \(A\) is diagonalizable if it is similar (Definition 4.2.2) to a diagonal matrix \(D\text{.}\) That is if there exists an invertible matrix \(P\) such that
Consider the matrix
and let's try to find \(D\) and \(P\) such that
Note, this is the same as
so letting
then we get the system of equations
Since the second two equations are the same as the first two we really just need to solve one pair. If we set up an augmented matrix and solve we get:
So, if \(p_3\neq0\text{,}\) then we get \(\lambda = 2\) or \(3\) and \(p_1=p_3\) or \(2\, p_3\text{.}\) Therefore, we get two vectors and two values for \(\lambda\text{,}\) if \(\lambda=2\) then the vector we want looks like
and if \(\lambda=3\) then the vector we want looks like
Finally, this is what we need because
Thus, with
we finally have \(A=PDP^{-1}\text{.}\)
The vectors we found along the way and the values for \(\lambda\) both play a special roll in linear algebra, they are called eigenvectors and eigenvalues; they are the topic of the next section (Subsection 4.3.2).
Insert example of a matrix that can't be diagonalized.
An eigenvector of a matrix \(A\) is a non-zero vector \(\vec{x}\) such that
for some constant \(\lambda\) which is called the corresponding eigenvalue.
Let's revisit Investigation 4.3.1 since we already know the answers we should get we can focus on how we might get them. We are looking at
which is the same as
If \(\vec{x}\neq \vec{0}\text{,}\) then \((A-\lambda I)\) must be singular (a.k.a. non-invertible or a zero divisor) or zero, either way \(det(A-\lambda I)=0\text{.}\) That is,
Solving for \(\lambda\) we get \(\lambda = 2\) or \(3\text{,}\) as before.
If we now take our values for \(\lambda\) and substitute them back into our original equation we can find our values for \(\vec{x}\text{.}\) Let \(\lambda=2\) so that we are solving
Row reducing the matrix we quickly get one free variable and all the possible solutions look like
Similarly, if we let \(\lambda=3\) we need to solve
which yields solutions of the form
If an \(n\times n\) matrix \(A\) has \(n\) distinct eigenvector, \(\lambda_1,\ldots,\lambda_n\text{,}\) and corresponding eigenvectors, \(\vec{x}_1,\ldots,\vec{x}_n\text{,}\) then \(A\) is diagonalizable and in particular
where the columns of \(P\) are the eigenvectors,
and \(D\) is a diagonal matrix with the eigenvalues on the diagonal,
If \(A\) is a diagonalizable matrix so that
for some diagonal matrix \(D\text{,}\) then: