Input interpretation
![palladium(II) ions | formal charges](../image_source/a4341372ada2eef01070a0f72d4b9e46.png)
palladium(II) ions | formal charges
Result
![(2)](../image_source/800237712fc4b98b1d7a68152a27e6f3.png)
(2)
Table
![2](../image_source/74cdae1332d9292a5089fe82fe3b6402.png)
2
Characteristic polynomial
![2 - λ](../image_source/0fae2ee3c83ab3862f13174b0f10a7c0.png)
2 - λ
Possible intermediate steps
![Find the characteristic polynomial of the matrix M with respect to the variable λ: M = (2) To find the characteristic polynomial of a matrix, subtract a variable multiplied by the identity matrix and take the determinant: left bracketing bar M - λ I right bracketing bar left bracketing bar M - λ I right bracketing bar | = | left bracketing bar 2 - λ 1 right bracketing bar | = | left bracketing bar 2 - λ right bracketing bar invisible comma = left bracketing bar 2 - λ right bracketing bar The determinant of a diagonal matrix is the product of its diagonal elements: left bracketing bar 2 - λ right bracketing bar 2 - λ = 2 - λ: Answer: | | 2 - λ](../image_source/41a36e71d921e021ef3d41bc2a32cdb8.png)
Find the characteristic polynomial of the matrix M with respect to the variable λ: M = (2) To find the characteristic polynomial of a matrix, subtract a variable multiplied by the identity matrix and take the determinant: left bracketing bar M - λ I right bracketing bar left bracketing bar M - λ I right bracketing bar | = | left bracketing bar 2 - λ 1 right bracketing bar | = | left bracketing bar 2 - λ right bracketing bar invisible comma = left bracketing bar 2 - λ right bracketing bar The determinant of a diagonal matrix is the product of its diagonal elements: left bracketing bar 2 - λ right bracketing bar 2 - λ = 2 - λ: Answer: | | 2 - λ
Eigenvalues
![λ_1 = 2](../image_source/5a20bd8a9de9a0426957bde56a323db2.png)
λ_1 = 2
Possible intermediate steps
![Find all the eigenvalues of the matrix M: M = (2) Find λ element C such that M v = λ v for some nonzero vector v: M v = λ v The only value of λ for which M v = λ v for any nonzero v is 2: Answer: | | 2](../image_source/f225e2ce1153b7c5981f0001fd933bc4.png)
Find all the eigenvalues of the matrix M: M = (2) Find λ element C such that M v = λ v for some nonzero vector v: M v = λ v The only value of λ for which M v = λ v for any nonzero v is 2: Answer: | | 2
Eigenvectors
![v_1 = (1)](../image_source/816ca988316a1b16fc9939c365830598.png)
v_1 = (1)
Possible intermediate steps
![Find all the eigenvalues and eigenvectors of the matrix M: M = (2) Find λ element C such that M v = λ v for some nonzero vector v: M v = λ v The only value of λ for which M v = λ v for any nonzero v is 2: 2 The equation M v = λ v is satisfied by each v element C^1, which means a suitable eigenvalue/eigenvector pair is: Answer: | | Eigenvalue | Eigenvector 2 | (1)](../image_source/3b25f9e8102859d5a4873dce4621576f.png)
Find all the eigenvalues and eigenvectors of the matrix M: M = (2) Find λ element C such that M v = λ v for some nonzero vector v: M v = λ v The only value of λ for which M v = λ v for any nonzero v is 2: 2 The equation M v = λ v is satisfied by each v element C^1, which means a suitable eigenvalue/eigenvector pair is: Answer: | | Eigenvalue | Eigenvector 2 | (1)