Determine which of the indicated column vectors are eigenvectors of the given matrix
a. give the corresponding eigenvalue.
a.(6 pts) a = [ 4 2 5 1 ]; ????1 = [ 5 −2 ], ????2 = [ 2 5 ], ????3 = [ −2 5 ]
b.(9 pts) a = [ 1 −2 2 −2 1 −2 2 2

Respuesta :

]Eigenvectors are found by the equation [tex](A-\lambda I) \vec{v} = 0$[/tex] implying that [tex]\det(A-\lambda I) = 0[/tex]. We then can write:

[tex] A-\lambda I = \left [ \begin{array}{cc} 4-\lambda & 2 \\ 5 & 1-\lambda \end{array}\right ] [/tex] 

And:

[tex] \det(A-\lambda I) = (4-\lambda)(1-\lambda) - 10 = 0 [/tex] 

Gives us the characteristic polynomial:

[tex] \lambda^2 - 5 \lambda -6 = 0 \implies \lambda_1 = -1, \lambda_2 = 6[/tex]

So, solving for each eigenvector subspace:

[tex]\left [ \begin{array}{cc} 4 & 2 \\ 5 & 1 \end{array} \right ] \left [ \begin{array}{c} x \\ y \end{array} \right ] = \left [ \begin{array}{c} -x \\ -y \end{array} \right ] [/tex]

Gives us the system of equations:

[tex] 4x + 2y = -x \newline 5x + y = - y [/tex] 

Producing the subspace along the line [tex] y = -\frac{5}{2} x [/tex]

We can see then that 3 is the answer. 



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