Respuesta :
Answer:
∴The eigen values are -1,1.
The eigen vector for 1 is [tex]\left[\begin{array}{c}1\\-1\end{array}\right][/tex] .
The eigen vector for [tex]\lambda[/tex]= - 1 is [tex]\left[\begin{array}{c}1\\1\end{array}\right][/tex] .
Step-by-step explanation:
Given matrix is
[tex]A=\left[\begin{array}{cc}0&1\\1&0\end{array}\right][/tex]
To find the eigen values of the given matrix, we set
[tex]det(A-\lambda I)=0[/tex]
[tex]\left|\begin{array}{cc}0&1\\1&0\end{array}\right|- \lambda \left|\begin{array}{cc}1&0\\0&1\end{array}\right|=0[/tex]
[tex]\left|\begin{array}{cc}0&1\\1&0\end{array}\right|- \left|\begin{array}{cc} \lambda&0\\0& \lambda\end{array}\right|=0[/tex]
[tex]\left|\begin{array}{cc} -\lambda&1\\1& -\lambda\end{array}\right|=0[/tex]
[tex]\Rightarrow \lambda^2-1=0[/tex]
[tex]\Rightarrow \lambda^2=1[/tex]
[tex]\Rightarrow \lambda=\pm1[/tex]
∴The eigen values are -1,1.
For [tex]\lambda=1[/tex]
Let the eigen vector for [tex]\lambda=1[/tex] is
[tex]v_1=\left[\begin{array}{c}x_1\\x_2\end{array}\right][/tex]
∴[tex](A-\lambda I)v_1=O[/tex]
[tex]\left[\begin{array}{cc} 1&1\\1&1\end{array}\right] \left[\begin{array}{c}x_1\\x_2\end{array}\right]=O[/tex]
[tex]\Rightarrow \left[\begin{array}{c}x_1+x_2\\x_1+x_2\end{array}\right]=O[/tex]
[tex]\therefore x_1+x_2=0[/tex]
[tex]\Rightarrow x_2=-x_1[/tex]
let [tex]x_1=1[/tex]
[tex]\therefore x_2=-1[/tex]
The eigen vector for 1 is [tex]\left[\begin{array}{c}1\\-1\end{array}\right][/tex] .
For [tex]\lambda=-1[/tex]
Let the eigen vector for [tex]\lambda=1[/tex] is
[tex]v_2=\left[\begin{array}{c}x_3\\x_4\end{array}\right][/tex]
∴[tex](A-\lambda I)v_2=O[/tex]
[tex]\left[\begin{array}{cc} -1&1\\1&-1\end{array}\right] \left[\begin{array}{c}x_3\\x_4\end{array}\right]=O[/tex]
[tex]\Rightarrow \left[\begin{array}{c}-x_3+x_4\\x_3-x_4\end{array}\right]=O[/tex]
[tex]\therefore- x_3+x_4=0[/tex]
and
[tex]\therefore x_3-x_4=0[/tex]
From the above equations, we get
[tex]x_3=x_4[/tex]
Let [tex]x_3=1[/tex]
Then, [tex]x_4=1[/tex]
The eigen vector for [tex]\lambda[/tex]= - 1 is [tex]\left[\begin{array}{c}1\\1\end{array}\right][/tex] .
We want to find the eigenvalues and eigenvectors of the given matrix.
The eigenvalues are 1 and -1 and the correspondent eigenvectors are:
[tex]\frac{1}{\sqrt{2} } \left[\begin{array}{ccc}1\\-1\end{array}\right][/tex] and [tex]\frac{1}{\sqrt{2} } \left[\begin{array}{ccc}1\\1\end{array}\right][/tex]
We have the matrix:
[tex]A = \left[\begin{array}{ccc}0&1\\1&0\end{array}\right][/tex]
To find the eigenvalues λ, we need to solve:
[tex]det (A - I*\lambda) = det(\left[\begin{array}{ccc}- \lambda &1\\1&-\lambda\end{array}\right]) = 0[/tex]
This gives:
[tex]\lambda^2 - 1 = 0\\\\\lambda = \pm 1[/tex]
Then the eigenvalues are 1 and -1
Now let's get the eigenvectors.
For λ = 1 we must solve:
[tex]\left[\begin{array}{ccc}-1&1\\1&-1\end{array}\right]*\left[\begin{array}{ccc}a\\b\end{array}\right] = 0\\\\\left[\begin{array}{ccc}-1&1\\1&-1\end{array}\right]*\left[\begin{array}{ccc}a\\b\end{array}\right] = \left[\begin{array}{ccc}-a+b\\a-b\end{array}\right] = 0[/tex]
Then a = -b
This eigenvector is:
[tex]\frac{1}{\sqrt{2} } \left[\begin{array}{ccc}1\\-1\end{array}\right][/tex]
(the factor is for normalization)
We do the same thing for the other eigenvalue:
[tex]\left[\begin{array}{ccc}1&1\\1&1\end{array}\right]*\left[\begin{array}{ccc}a\\b\end{array}\right] = 0\\\\\left[\begin{array}{ccc}1&1\\1&1\end{array}\right]*\left[\begin{array}{ccc}a\\b\end{array}\right] = \left[\begin{array}{ccc}a+b\\a+b\end{array}\right] = 0[/tex]
Now we have a = b
[tex]\frac{1}{\sqrt{2} } \left[\begin{array}{ccc}1\\1\end{array}\right][/tex]
If you want to learn more, you can read:
https://brainly.com/question/14415835
