Show that&nbsp;λ&nbsp;is an eigenvalue of A and Identify one eigenvector that matches this eigenvalue..A=[4−25−7],&nbsp;λ=−6

Question

Show that&amp;nbsp;λ&amp;nbsp;is an eigenvalue of A and Identify one eigenvector that matches this eigenvalue..A=[4−25−7],&amp;nbsp;λ=−6

Muspee · Accepted Answer

Step 1Solution:In order to show that&amp;nbsp;λ=−6&amp;nbsp;is eigenvalue for the matrixA=[5−25−7]We need demonstrate that there is at least one vector.x=[x1x2]such thatAx=λxConsider drawing some conclusions from this situation.Ax=[4−25−7]⋅[x1x2]=[4x1−2x25x1−7x2]λx=λ⋅[x1x2]=[−6x1−6x2]If we equalize thise two equations we get:[4x1−2x25x1−7x2]=[−6x1−6x2]and from here we get the following system:1)&amp;nbsp;4x1−2x2=−6x12)&amp;nbsp;5x1−7x2=−6x2this indicates that3)&amp;nbsp;10x1−2x2=0and 4)&amp;nbsp;5x1−x2=0but equations (3) and (4) are equivalent so we only take 4th into consideration and conclude thatx2=5x1This means that all eigenvectors have this formx=[x15x1]where&amp;nbsp;x1&amp;nbsp;is random scalar different from zero. For example if we take&amp;nbsp;x1=10&amp;nbsp;we get the eigenvectorx=[1050]

Aretha Frazier · Accepted Answer

Step 1 Let A be an n×n matrix. A scalar λ is called an eigenvalue of A if there is a nonzero vector x such that Ax=λx. Such a vector x is called an eigenvector of A corresponding to λ Step 2 We must show that there is nonzero vector x such that Ax=λx In other words, that equation Ax−λx=0 (A−λI)x=0 We have to compute the Null space of matrix A−λI Step 3 A−λI=[4−25−7]−[−600−6] =[10−25−1] [10−2|05−1|0]R2−12R1→[10−2|000|0] 10x1−2x2=0 x2=5x1 Arbitrary vector from Null space is: x=[x1x2]=[x15x1]=x1[15] Choose any value for x1 to obtain one eigenvector. x1=1 gives us eigenvector v=[15]

Show that \lambda is an eigenvalue of A and find one eigenvector correspon

Answered question

Answer & Explanation

New Questions in Other