Diagonalizability

In Linear Algebra, Diagonalizability refers to the ability to diagonalize a matrix $A$ , i.e. the ability to find an invertible matrix $P$ and a diagonal matrix $D$ such that

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

How to diagonalize a matrix

Given a arbitrary $n$ times $n$ matrix $A$ , one calculate the Eigenvalue of $A$ , by finding solutions to the equation

$\text{det} (A - \lambda I_n)=0$

where $I_n$ denotes the $n$ times $n$ Identity matrix and $\text{det}$ stands for the determinant of the matrix.

The polynomial $\text{det} (A - \lambda I_n)$ is known as the matrix's Characteristic polynomial.

Now, order the eigenvalues (they may be complex). There should be $n$ eigenvalues counting multiplicities, $\lambda _1 , \lambda _2 , \dots , \lambda _n$ . Then for each eigenvalue find its corresponding eigenvector. If you cannot find $n$ linear independent eigenvectors, then the next step would fail ( $P$ would not be invertible) and $A$ would be not diagonalizable. On the contrary, suppose one obtains $n$ linear independent eigenvectors, one for each eigenvalue (if there are double eigenvalues the double eigenvalues should generate two linear independent eigenvectors), $v_1, v_2, \dots , v_n$ corresponding to the eigenvalues (This is very very important). Then the matrix

$P= \begin{bmatrix} v_1 & v_2 & \dots & v_n \\ \end{bmatrix}$

and the diagonal matrix

$D= \begin{bmatrix} \lambda _1 & 0 & 0 & \dots & 0 \\ 0 & \lambda _2 & 0 & \dots & 0 \\ 0 & 0 & \lambda _3 & \dots & 0 \\ \dots & \dots & \dots & \dots & \dots \\ 0 & 0 & 0 & \dots & \lambda _n \\ \end{bmatrix}$

satisfy the desired $A=P D P^{-1}$ . (Note that $v_i$ are column vectors)

It is best to illustrate with an example. Consider the matrix

$A= \begin{bmatrix} 7 & 2 \\ -4 & 1 \\ \end{bmatrix}$

(Note that we're using small 2 times 2 matrices since it is easier. In practice, most of the time you will encounter 3 times 3 matrices)

We compute its characteristic equation:

$\text{det} (A - \lambda I_2 )=0$

$\text{det} \begin{bmatrix} 7 - \lambda & 2 \\ -4 & 1 - \lambda \\ \end{bmatrix} = 0$

$\implies (7- \lambda )(1- \lambda ) - 2 (-4)=0$ $\implies \lambda ^2 -8 \lambda +15=0$

$\implies \lambda = 3,5$

We obtain eigenvalues of 5 and 3. Thus, we have $D= \begin{bmatrix} 5 & 0 \\ 0 & 3 \\ \end{bmatrix} = 0$ Now we calculate eigenvectors.

To do so, we first calculate the matrix $A- \lambda _1 I_2$ . We see that would be

$\begin{bmatrix} 2 & 2 \\ -4 & -4 \\ \end{bmatrix}$

for the eigenvalue of $5$ . We calculate possible solutions to the equation

$(A- \lambda I_2) v=0$

(the solution is for the vector $v$ ). Normally, this would involve a row reduction of an augmented matrix, but in this case it is simple, so one can see a solution by looking. We see $\begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}$ is a eigenvector for $5$ . Similarly, we can find rather easily that $\begin{bmatrix} 1 \\ -2 \\ \end{bmatrix}$ is a eigenvector for the eigenvalue $3$ . Adjoining those eigenvector, we see that

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

for matrices

$P= \begin{bmatrix} 1 & 1 \\ -1 & -2 \\ \end{bmatrix}$

and

$D= \begin{bmatrix} 5 & 0 \\ 0 & 3 \\ \end{bmatrix}$

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Diagonalizability

How to diagonalize a matrix