Normalization and value of eigen vectors for A = [[1 1 0][1 1 0][0 0 1]]

[zak100]

Hi,
I have got the following matrix. I have found the eigen values but in some eq x, y & z terms are vanishing, so how to find the value of eigen vector??? Also why we have to do normalization??

\(\displaystyle \mbox{Given: }\, A\, =\, \left[\begin{array}{ccc}1&1&0\\1&1&0\\0&0&1\end{array} \right]\)


$\displaystyle A\, -\, \lambda\, I\, =\, 0:$

. . . . .$\displaystyle \left[\begin{array}{ccc}1&1&0\\1&1&0\\0&0&1\end{array} \right]\, -\, \left[\begin{array}{ccc}\lambda &0&0\\0& \lambda & 0\\0&0& \lambda \end{array}\right]\, =\, 0$

. . . . .$\displaystyle \left[\begin{array}{ccc}1-\lambda &1&0\\1&1-\lambda & 0\\0&0&1-\lambda \end{array}\right]\, =\, 0$

. . . . .$\displaystyle (1\, -\, \lambda)\left|\begin{array}{cc}1-\lambda & 0\\0&1-\lambda\end{array}\right|\, -\, 1\left|\begin{array}{cc}1&0\\0&1-\lambda \end{array}\right|\, +\, 0\, =\, 0$

. . . . .$\displaystyle (1\, -\, \lambda)\, (1\, -\, \lambda)^2\, -\, (1\, -\, \lambda)\, =\, 0$

Taking the common factor of $\displaystyle (1\, -\, \lambda)$ out front:

. . . . .$\displaystyle (1\, -\, \lambda)\left[(1\, -\, \lambda)^2\, -\, 1\right]\, =\, 0$

First eigen value $\displaystyle \lambda_1\, =\, 1$

Now consider:

. . . . .$\displaystyle \left[(1\, -\, \lambda)^2\, -\, 1\right]\, =\, 0$

. . . . .$\displaystyle 1\, -\, 2\lambda \, +\, {\lambda}^2\, -\, 1\, =\, 0$

. . . . .$\displaystyle -2\lambda\, +\, {\lambda}^2\, =\, 0$

. . . . .$\displaystyle \lambda\, (\lambda\, -\, 2)\, =\, 0$

. . . . .\(\displaystyle \lambda_2\, =\, 0\, \mbox{ and }\, \lambda_3\, =\, 2\)


For $\displaystyle \lambda_1\, =\, 1,\, Ax\, =\, \lambda\, \cdot\, x\, \cdot\, I$

. . . . .$\displaystyle \left[\begin{array}{ccc}1&1&0\\1&1&0\\0&0&1\end{array} \right]\, \left[\begin{array}{c}x\\y\\z\end{array}\right]\, =\, 1\, \cdot\, \left[\begin{array}{c}x\\y\\z\end{array}\right]$

. . . . .\(\displaystyle \begin{array}{lcr}x\, +\, y\, =\, x\, &\,&\mbox{eqn}\, (1)\\x\, +\, y\, =\, y\, &\,&\mbox{eqn}\, (2)\\z\, =\, z\, &\,&\mbox{eqn}\, (3)\end{array}\)


Above x vanishes in eq(1), y vanishes in eq(2) & z vanishes in eq(3).
What is the value of μ1?????


For $\displaystyle \lambda_2\, =\, 0:$

. . . . .$\displaystyle \left[\begin{array}{ccc}1&1&0\\1&1&0\\0&0&1\end{array} \right]\, \cdot\, \left[\begin{array}{c}x\\y\\z\end{array}\right]\, =\, 0\, \cdot\, \left[\begin{array}{c}x\\y\\z\end{array}\right]$

. . . . .\(\displaystyle \begin{array}{lcr}x\, +\, y\, =\, 0&\,&\mbox{eqn}\, (4)\\x\, +\, y\, =\, 0&\,&\mbox{eqn}\, (5)\\z\, =\, 0&\,&\mbox{eqn}\, (6)\end{array}\)

In my view, it should be:

. . . . .$\displaystyle \mu\, =\, \left[\begin{array}{c}-1\\-1\\0\end{array}\right]$

but teacher has got different answer. In addition to this he has done normalization, please guide me what is the need for normalization


For $\displaystyle \lambda\, =\, 2:$

. . . . .$\displaystyle \left[\begin{array}{ccc}1&1&0\\1&1&0\\0&0&1\end{array} \right]\, \left[\begin{array}{c}x\\y\\z\end{array}\right]\, =\, 2\, \cdot\, \left[\begin{array}{c}x\\y\\z\end{array}\right]$

. . . . .\(\displaystyle \begin{array}{lcr}x\, +\, y\, =\, 2x&\,&\mbox{eqn}\, (7)\\x\, =\, -y&\,&\,\\x\, +\, y\, =\, 2y&\,&\mbox{eqn}\, (8)\\y\, =\, x&\,&\,\\z\, =\, 2z&\,&\mbox{eqn}\, (9)\end{array}\)

Therefore:

. . . . .$\displaystyle \mu\, =\, \left[\begin{array}{c}-1\\1\\0\end{array} \right]$

Why we have to normalize the vector?
Somebody please guide me.
Zulfi.

 

[HallsofIvy]

First thing you should understand is that the set of eigenvectors corresponding to a given eigenvalue form a vector space- there are necessarily an infinite number of them. Yes, 1 is an eigenvalue of this matrix and for <x, y, z> an eigenvector corresponding to eigenvalue 1, we must have x+ y= x, x+ y= y, and z= z. The first equation gives x= 0, the second equation gives y= 0, and the third equation is true for any z (z= z does NOT mean "z= 0"). The subspace of eigenvectors corresponding to eigenvalue 1 is the set of all vectors of the form <0, 0, z>. That can be written as z<0, 0, 1>: any multiple of <0, 0, 1> is an eigenvector.

For eigenvalue 0, you have the equations x+ y= 0, x+ y= 0, z= 0. You say "In my view" the eigenvector should be <-1, -1, 0>. Why? x= y= -1 does NOT satisfy x+ y= 0:-1+ -1= -2. From y+ x= 0, we get y= -x. Any eigenvector is of the form <x, -x, 0>. If we take, again arbitrarily, x= 1 then y= -1. < 1, -1, 0> is an eigenvector.

For eigenvalue 2, you have the equation x+ y= 2x. But that does NOT give "y= -x" as you have next. Subtracting x from both sides gives y= x. x+ y= 2y gives also y= x. And, of course, z= 2z gives z= 0. Any eigenvector corresponding to eigenvalue 2 is of the form <x, x, 0> so any multiple or <1, 1, 0>.


As for your question "why would we have to normalize the vector?", in this case the answer is "because your teacher told you to do it!". And if you want a good grade, you had better do what your teacher told you to do. There may be good reasons to normalize vector but there is no reasons why you must do it.

 

[zak100]

First thing you should understand is that the set of eigenvectors corresponding to a given eigenvalue form a vector space- there are necessarily an infinite number of them. Yes, 1 is an eigenvalue of this matrix and for <x, y, z> an eigenvector corresponding to eigenvalue 1, we must have x+ y= x, x+ y= y, and z= z. The first equation gives x= 0, the second equation gives y= 0, and the third equation is true for any z (z= z does NOT mean "z= 0"). The subspace of eigenvectors corresponding to eigenvalue 1 is the set of all vectors of the form <0, 0, z>. That can be written as z<0, 0, 1>: any multiple of <0, 0, 1> is an eigenvector.

For eigenvalue 0, you have the equations x+ y= 0, x+ y= 0, z= 0. You say "In my view" the eigenvector should be <-1, -1, 0>. Why? x= y= -1 does NOT satisfy x+ y= 0:-1+ -1= -2. From y+ x= 0, we get y= -x. Any eigenvector is of the form <x, -x, 0>. If we take, again arbitrarily, x= 1 then y= -1. < 1, -1, 0> is an eigenvector.

For eigenvalue 2, you have the equation x+ y= 2x. But that does NOT give "y= -x" as you have next. Subtracting x from both sides gives y= x. x+ y= 2y gives also y= x. And, of course, z= 2z gives z= 0. Any eigenvector corresponding to eigenvalue 2 is of the form <x, x, 0> so any multiple or <1, 1, 0>.


As for your question "why would we have to normalize the vector?", in this case the answer is "because your teacher told you to do it!". And if you want a good grade, you had better do what your teacher told you to do. There may be good reasons to normalize vector but there is no reasons why you must do it.

Really very good explanation. Thanks a lot. God bless you.

Zulfi.