Tuesday, 26 March 2013

Eigenvalues and eigen vectors part 2

For a square matrix A of order n, the number $\lambda$ is an eigenvalue if and only if there exists a non-zero vector C such that

\begin{displaymath}A C = \lambda C.\end{displaymath}

Using the matrix multiplication properties, we obtain

\begin{displaymath}\Big(A - \lambda I_n \Big) C = 0.\end{displaymath}

This is a linear system for which the matrix coefficient is $A - \lambda I_n$. We also know that this system has one solution if and only if the matrix coefficient is invertible, i.e. $det\Big(A - \lambda I_n \Big) \neq 0$. Since the zero-vector is a solution and C is not the zero vector, then we must have

\begin{displaymath}det\Big(A - \lambda I_n \Big) = 0.\end{displaymath}

Example. Consider the matrix

\begin{displaymath}A = \left(\begin{array}{rrr}

The equation $det\Big(A - \lambda I_n \Big) = 0 $ translates into

1 -\lambda&-2\\
\end{array}\right\vert = (1 -\lambda) (0-\lambda) - 4 = 0\end{displaymath}

which is equivalent to the quadratic equation

\begin{displaymath}\lambda^2 - \lambda - 4 = 0.\end{displaymath}

Solving this equation leads to

\begin{displaymath}\lambda = \frac{1 + \sqrt{17}}{2},\;\; \mbox{and}\;\; \lambda = \frac{1 -\sqrt{17}}{2}.\end{displaymath}

In other words, the matrix A has only two eigenvalues.

In general, for a square matrix A of order n, the equation

\begin{displaymath}det\Big(A - \lambda I_n \Big) = 0 \end{displaymath}

will give the eigenvalues of A. This equation is called the characteristic equation or characteristic polynomial of A. It is a polynomial function in $\lambda$ of degree n. So we know that this equation will not have more than n roots or solutions. So a square matrix A of order n will not have more than n eigenvalues.

Example. Consider the diagonal matrix

\begin{displaymath}D = \left(\begin{array}{rrrr}

Its characteristic polynomial is

\begin{displaymath}det\Big(D - \lambda I_n \Big) = \left\vert\begin{array}{cccc}...\vert = (a-\lambda )(b-\lambda )(c-\lambda )(d-\lambda ) = 0.\end{displaymath}

So the eigenvalues of D are abc, and d, i.e. the entries on the diagonal.

This result is valid for any diagonal matrix of any size. So depending on the values you have on the diagonal, you may have one eigenvalue, two eigenvalues, or more. Anything is possible.

Remark. It is quite amazing to see that any square matrix A has the same eigenvalues as its transpose AT because

\begin{displaymath}det\Big(A - \lambda I_n \Big) = det\Big(A - \lambda I_n \Big)^T = det\Big(A^T - \lambda I_n \Big).\end{displaymath}

For any square matrix of order 2, A, where

\begin{displaymath}A = \left(\begin{array}{rr}

the characteristic polynomial is given by the equation

a-\lambda &b\\
c&d-\lambda \\ ...
...bda )(d-\lambda ) - bc = \lambda^2 - (a+d) \lambda + ad-bc = 0.\end{displaymath}

The number (a+d) is called the trace of A (denoted tr(A)), and clearly the number (ad-bc) is the determinant of A. So the characteristic polynomial of A can be rewritten as

\begin{displaymath}\lambda^2 - tr(A) \lambda + det(A) = 0.\end{displaymath}

Let us evaluate the matrix

B = A2 - tr(AA + det(AI2.

We have

\begin{displaymath}B = \left(\begin{array}{cc}
a^2 + bc&ab+ bd\\
ac+dc&bc +d^2\...\left(\begin{array}{rr}

We leave the details to the reader to check that

\begin{displaymath}B = \left(\begin{array}{rr}

In other word, we have

\begin{displaymath}A^2 - tr(A) A + det(A) I_2 = {\cal O}.\end{displaymath}

This equation is known as the Cayley-Hamilton theorem. It is true for any square matrix A of any order, i.e.

\begin{displaymath}p(A) = {\cal O}\end{displaymath}

where $p(\lambda) = det\Big(A - \lambda I_n \Big) $ is the characteristic polynomial of A.

We have some properties of the eigenvalues of a matrix.

Theorem. Let A be a square matrix of order n. If $\lambda$ is an eigenvalue of A, then:
$\lambda^m$ is an eigenvalue of Am, for $m=1,2,\cdots$
If A is invertible, then $\displaystyle \frac{1}{\lambda}$ is an eigenvalue of A-1.
A is not invertible if and only if $\lambda = 0$ is an eigenvalue of A.
If $\alpha$ is any number, then $\lambda +\alpha$ is an eigenvalue of $A+\alpha I_n$.
If A and B are similar, then they have the same characteristic polynomial (which implies they also have the same eigenvalues).
Post a Comment