Objectives: Topics
-
Similar matrices
-
Diagonal, similar, and diagonalizable matrices
-
Diagonalizing matrices
Print preview
Section 5.3 Diagonalization
Handout 5.3 Diagonalization
Motivation: It can be useful to take large powers of matrices, for example \(A^k\) for large \(k\text{.}\)
Question: Is there a more efficient way to compute \(A^k\text{?}\)
Objectives: Goals
-
Determine whether a matrix can be diagonalized, and if possible diagonalize a square matrix.
-
Apply diagonalization to compute matrix powers.
Definition 5.27. Similarity.
Two \(n\times n\) matrices \(A\) and \(B\) are similar if there exists an invertible matrix \(P\) such that \(A=PBP^{-1}.\)
Matrix similarity is an equivalence relation meaning:
-
Reflexive Property: Every matrix \(A\) is similar to itself.
-
Transitive Property: If \(A\) is similar to \(B\) and \(B\) is similar to \(C\text{,}\) then \(A\) is similar to \(C\text{.}\)
Theorem 5.28. Characteristic Polynomial of Similar Matrices.
Notes:
-
Similar matrices have exactly the same eigenvalues with exactly the same algebraic multiplicity for each eigenvalue and exactly the same geometric multiplicity for each eigenvalue.
-
The converse of Theorem 5.28 is not true; that is, just because two matrices have the same characterisitc polynomial, that does not mean that they are similar. For example,\begin{equation*} \begin{bmatrix} 0 \amp 1\\ 0 \amp 0 \end{bmatrix} \quad\text{and}\quad \begin{bmatrix} 0 \amp 0\\ 0 \amp 0 \end{bmatrix} \end{equation*}have the same characteristic polynomial, but are not similar because the eigenvalue \(\lambda = 0\) does not have the same geometric multiplicty for each matrix.
Example 5.29.
True or false: If \(A \) is similar to the identity matrix, then \(A\) is equal to the identity matrix.
Definition 5.30. Diagonal Matrix.
A matrix is diagonal if the only nonzero entries (if any) lie on the main diagonal. In this course, we work only with square diagonal matrices.
Examples of diagonal matrices:
\begin{equation*}
\begin{bmatrix}2 \amp 0\\ 0 \amp 2\end{bmatrix},\quad
\begin{bmatrix}2\end{bmatrix},\quad
I_n,\quad
\begin{bmatrix}0 \amp 0\\ 0 \amp 0\end{bmatrix}.
\end{equation*}
If \(A\) is diagonal, then \(A^k\) is easy to compute by raising diagonal entries to the \(k\)-th power.
Example 5.31. Example: Powers of a Diagonal Matrix.
(a)
Let
\begin{equation*}
A=\begin{bmatrix} 3 \amp 0\\ 0 \amp 0.5 \end{bmatrix}\text{.}
\end{equation*}
Compute \(A^2\) and \(A^k\text{.}\)
(b)
Is this as easy to do if \(A\) is not diagonal?
(c)
What about if \(A\) is similar to a diagonal matrix?
Definition 5.32. Diagonalizable Matrix.
A matrix \(A\in\mathbb{R}^{n\times n}\) is diagonalizable if it is similar to a diagonal matrix \(D\text{;}\) that is, there exists an invertible matrix \(P\) such that
\begin{equation*}
A=PDP^{-1}.
\end{equation*}
Theorem 5.33. Diagonalization Criterion.
A matrix \(A\) is diagonalizable if and only if \(A\) has \(n\) linearly independent eigenvectors. Equivalently,
\begin{equation*}
A = \bigl[\vec v_1\ \vec v_2\ \cdots\ \vec v_n\bigr]\,
\begin{bmatrix}
\lambda_1 \amp \amp \amp \\
\amp \lambda_2 \amp \amp \\
\amp \amp \ddots \amp \\
\amp \amp \amp \lambda_n
\end{bmatrix}
\bigl[\vec v_1\ \vec v_2\ \cdots\ \vec v_n\bigr]^{-1},
\end{equation*}
where \(\vec v_1,\dots,\vec v_n\) are linearly independent eigenvectors and \(\lambda_1,\dots,\lambda_n\) are the corresponding eigenvalues (in order).
Theorem 5.34. Algebraic vs. Geometric Multiplicity.
Suppose:
-
\(A\) is \(n\times n\text{,}\)
-
\(A\) has distinct eigenvalues \(\lambda_1,\ldots,\lambda_k\) with \(k\le n\text{,}\)
-
\(a_i\) is the algebraic multiplicity of \(\lambda_i\text{,}\)
-
\(d_i\) is the geometric multiplicity (dimension of the \(\lambda_i\)-eigenspace).
Then:
-
\(d_i \le a_i\) for all \(i\text{.}\)
-
\(A\) is diagonalizable \(\ \iff \ \sum_i d_i = n \ \iff \ d_i=a_i\) for all \(i\text{.}\)
-
\(A\) is diagonalizable \(\iff\) the eigenvectors for all eigenvalues together form a basis of \(\mathbb{R}^n\text{.}\)
Theorem 5.35. Diagonalization with Distinct Eigenvalues.
Example 5.36.
Diagonalize if possible:
\begin{equation*}
A = \begin{bmatrix}
3 \amp 1\\
0 \amp 3
\end{bmatrix}.
\end{equation*}
Example 5.37.
Diagonalize if possible:
\begin{equation*}
A = \begin{bmatrix}
2 \amp 6\\
0 \amp -1
\end{bmatrix}.
\end{equation*}
Example 5.38.
The eigenvalues of \(A\) are \(\lambda=3,1\text{.}\) If possible, construct \(P\) and \(D\) such that \(A=PDP^{-1}\text{.}\) Verify by checking that \(AP = PD\text{.}\)
\begin{equation*}
A=\begin{bmatrix}
7 \amp 4 \amp 16\\
2 \amp 5 \amp 8\\
-2 \amp -2 \amp -5
\end{bmatrix}.
\end{equation*}
Example 5.39.
Diagonalize the following matrix, if possible.
\begin{equation*}
A=\begin{bmatrix}
2 \amp 4 \amp 3\\
-4 \amp -6 \amp -3\\
3 \amp 3 \amp 1
\end{bmatrix}.
\end{equation*}
Example 5.40. Diagonalization and a Famous Sequence.
Note that
\begin{equation*}
\vec x_k=
\begin{bmatrix}
0 \amp 1\\
1 \amp 1
\end{bmatrix}\vec x_{k-1},\quad
\vec x_0=
\begin{bmatrix}
0\\
1
\end{bmatrix},\quad
k=1,2,3,\ldots
\end{equation*}
generates a well-known sequence of numbers. Use diagonalization to find a matrix expression that gives the \(n\)-th number in this sequence.
