Find the eigen values and normalized eigen vectors of the following matrix.

3      1      4

0      2      6

0      0      5

                        [ 3 2 6

0 2 6

0 0 5 ]

richa singh, 12 years ago

Grade:

1 Answers

Aman Bansal

592 Points

12 years ago

Dear Richa,

The eigenvalue problem is a problem of considerable theoretical interest and wide-ranging application. For example, this problem is crucial in solving systems of differential equations, analyzing population growth models, and calculating powers of matrices (in order to define the exponential matrix). Other areas such as physics, sociology, biology, economics and statistics have focused considerable attention on "eigenvalues" and "eigenvectors"-their applications and their computations. Before we give the formal definition, let us introduce these concepts on an example.

Example. Consider the matrix

$\begin{displaymath}A = \left(\begin{array}{rrr} 1&2&1\\ 6&-1&0\\ -1&-2&-1\\ \end{array}\right).\end{displaymath}$

Consider the three column matrices

$\begin{displaymath}C_1 = \left(\begin{array}{rrr} 1\\ 6\\ -13\\ \end{array}\r... ... = \left(\begin{array}{rrr} 2\\ 3\\ -2\\ \end{array}\right).\end{displaymath}$

We have

$\begin{displaymath}AC_1 = \left(\begin{array}{rrr} 0\\ 0\\ 0\\ \end{array}\ri... ... = \left(\begin{array}{rrr} 6\\ 9\\ -6\\ \end{array}\right).\end{displaymath}$

In other words, we have

$\begin{displaymath}AC_1 = 0 C_1,\; AC_2 = -4 C_2,\; \;\mbox{and}\;\; AC_3 = 3 C_3.\end{displaymath}$

Next consider the matrix P for which the columns are C₁, C₂, and C₃, i.e.,

$\begin{displaymath}P = \left(\begin{array}{rrr} 1&-1&2\\ 6&2&3\\ -13&1&-2\\ \end{array}\right).\end{displaymath}$

We have det(P) = 84. So this matrix is invertible. Easy calculations give

$\begin{displaymath}P^{-1} = \frac{1}{84} \left(\begin{array}{rrr} -7&0&-7\\ -27&24&9\\ 32&12&8\\ \end{array}\right).\end{displaymath}$

Next we evaluate the matrix P^-1AP. We leave the details to the reader to check that we have

$\begin{displaymath}\frac{1}{84} \left(\begin{array}{rrr} -7&0&-7\\ -27&24&9\\ ... ...gin{array}{rrr} 0&0&0\\ 0&-4&0\\ 0&0&3\\ \end{array}\right).\end{displaymath}$

In other words, we have

$\begin{displaymath}P^{-1}AP = \left(\begin{array}{rrr} 0&0&0\\ 0&-4&0\\ 0&0&3\\ \end{array}\right).\end{displaymath}$

Using the matrix multiplication, we obtain

$\begin{displaymath}A = P \left(\begin{array}{rrr} 0&0&0\\ 0&-4&0\\ 0&0&3\\ \end{array}\right)P^{-1}\end{displaymath}$

which implies that A is similar to a diagonal matrix. In particular, we have

$\begin{displaymath}A^{n} = P \left(\begin{array}{rrr} 0&0&0\\ 0&(-4)^n&0\\ 0&0&3^n\\ \end{array}\right)P^{-1}\end{displaymath}$

for $n=1,2,\cdots$ . Note that it is almost impossible to find A⁷⁵ directly from the original form of A.

This example is so rich of conclusions that many questions impose themselves in a natural way. For example, given a square matrix A, how do we find column matrices which have similar behaviors as the above ones? In other words, how do we find these column matrices which will help find the invertible matrix P such that P^-1AP is a diagonal matrix?

From now on, we will call column matrices vectors. So the above column matrices C₁, C₂, and C₃ are now vectors. We have the following definition.

Definition. Let A be a square matrix. A non-zero vector C is called an eigenvector of A if and only if there exists a number (real or complex) $\lambda$ such that

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

If such a number $\lambda$ exists, it is called an eigenvalue of A. The vector C is called eigenvector associated to the eigenvalue $\lambda$ .

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Thanks

Aman Bansal

Askiitian Expert

$\begin{displaymath}A {\cal O} = {\cal O} = \lambda {\cal O}\end{displaymath}$