Matrix - Diagonalization

Diagonalization is the process of finding a diagonal matrix $D$ that is similar to a given square matrix $A$. This is done by finding an invertible matrix $P$ such that:

$$A=PD{P}^{-1}$$

Where:

• $D$ is a diagonal matrix containing the eigenvalues of $A$ on its main diagonal.
• $P$ is a matrix whose columns are the corresponding eigenvectors of $A$.
• $P^{-1}$ is the inverse of matrix $P$.

Finding $P$

The columns of $P$ are the linearly independent eigenvectors of $A$. These columns can be put in any order.

As well, the diagonal elements of $D$ are the corresponding eigenvalues of $A$, arranged in the same order as their respective eigenvectors in $P$.

This process is called Eigenvalue Decomposition.

Conditions for Diagonalization

A matrix $A$ is diagonalizable if and only if it has enough linearly independent eigenvectors to form the matrix $P$. Specifically, for an $n\times n$ matrix, there must be $n$ linearly independent eigenvectors.

Examples

Example 1

Problem Statement

Consider the matrix $A$

$$A=\left[\begin{array}{ccc}1& 3& 3\\ -3& -5& -3\\ 3& 3& 1\end{array}\right]$$

Diagonalize $A$ if possible.

Solution

Starting with the Matrix Characteristic Equation

$$\left|\begin{array}{ccc}1-\lambda & 3& 3\\ -3& -5-\lambda & -3\\ 3& 3& 1-\lambda \end{array}\right|=0$$

Expanding the determinant

$$(1-\lambda )(-5-\lambda )-27-27+9(5+\lambda )+9(1-\lambda )+9(1-\lambda )=0$$

Simplifying yields

$$(1-\lambda )(\lambda +2{)}^2=0$$

Note: $\lambda =-2$ has multicity of 2, meaning its eigenspace has a maximum dimension of 2

Begin with $\lambda =1$:

1. construct the homogenous equation $(A-\lambda I)\vec{x}=\vec{0}$
2. construct the augmented matrix
3. Apply row reduction to bring the matrix to RREF

$$\left[\begin{array}{cccc}0& 3& 3& 0\\ -3& -6& -3& 0\\ 3& 3& 0& 0\end{array}\right]\to \left[\begin{array}{cccc}1& 0& -1& 0\\ 0& 1& 1& 0\\ 0& 0& 0& 0\end{array}\right]$$

Free variable: $x_3=s$

$$\{\begin{array}{l}{x}_1=s\\ x_2=-s\end{array}$$ $$\vec{x}=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}1\\ -1\\ 1\end{array}\right]s$$

The Eigenspace is now the span of these vectors. That is,

$$\mathrm{span}\{\left[\begin{array}{c}1\\ -1\\ 1\end{array}\right]\}$$

Next $\lambda =-2$

performing the same calculation yields the following eigenspace:

$$\mathrm{span}\{\left[\begin{array}{c}-1\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}-1\\ 0\\ 1\end{array}\right]\}$$

Therefore, the matrix $A$ has 3 linearly independent eigenvectors

Conclusion

Thus, we conclude that $A$ is diagonalizable and

$$P=\left[\begin{array}{ccc}1& -1& -1\\ -1& 1& 0\\ 1& 0& 1\end{array}\right],D=\left[\begin{array}{ccc}1& 0& 0\\ 0& -2& 0\\ 0& 0& -2\end{array}\right]P^{-1}=\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 1\\ -1& -1& 0\end{array}\right]$$

Example 2

Problem Statement

Diagonalize the following matrix $A$

$$A=\left[\begin{array}{cc}4& -3\\ 2& -1\end{array}\right],\mathrm{calculate}{A}^8$$

SOlution

$$\left|\begin{array}{cc}4-\lambda & -3\\ 2& -1-\lambda \end{array}\right|=0\Rightarrow (4-\lambda )(-1-\lambda )+6=0$$ $$(\lambda -1)(\lambda -2)=0$$

Therefore, $\lambda =1,2$ both of multiplicity $1$. Then, applying the characteristic equation and setting $\lambda =1$

$$\left[\begin{array}{ccc}3& -3& 0\\ 2& -2& 0\end{array}\right]\to \left[\begin{array}{ccc}1& -1& 0\\ 0& 0& 0\end{array}\right]\Rightarrow \vec{x}=\left[\begin{array}{c}1\\ 1\end{array}\right]s,x_2=s$$ $$\overrightarrow{v_1}=\left[\begin{array}{c}1\\ 1\end{array}\right]$$

And then $\lambda =2$

$$\left[\begin{array}{ccc}2& -3& 0\\ 2& -3& 0\end{array}\right]\Rightarrow \vec{x}=\left[\begin{array}{c}\frac{3}{2}\\ 1\end{array}\right]s,x_2=s$$

For simplicity, we can choose $s=2$, s can be any real number

$$\overrightarrow{x_2}=\left[\begin{array}{c}3\\ 2\end{array}\right]$$

As there are $2$ eigenvectors, $A$ is diagonalizable and:

$$P=\left[\begin{array}{cc}1& 3\\ 1& 2\end{array}\right]\Rightarrow P^{-1}=\left[\begin{array}{cc}-2& 3\\ 1& -1\end{array}\right]$$

Diagonalizing $D=\left[\begin{array}{cc}1& 0\\ 0& 2\end{array}\right];A=PD{P}^{-1}$

Now raising both sides to the power of $8$

$$A^8=(PD{P}^{-1}{)}^8=PD{P}^{-1}PD{P}^{-1}PD{P}^{-1}\dots$$

Any instance of $P\cdot P^{-1}=I$, so the right side can be simplified to:

$$A^8=P{D}^8{P}^{-1}$$

Raising a diagonal matrix to a power is very easy, as in general:

$${\left[\begin{array}{ccc}a& 0& 0\\ 0& b& 0\\ 0& 0& c\end{array}\right]}^k=\left[\begin{array}{ccc}{a}^k& 0& 0\\ 0& b^k& 0\\ 0& 0& c^k\end{array}\right]$$

Therefore:

$$A^8=\left[\begin{array}{cc}1& 3\\ 1& 2\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 256\end{array}\right]\left[\begin{array}{cc}-2& 3\\ 1& -1\end{array}\right]$$