Last Updated: 29-2-2 9:56 EECS 6A Spring 29 Designing Information Devices and Systems I Discussion 4B Reference Definitions: Matrices and Linear (In)Dependence We ve seen that the following statements are equivalent for an n n matrix A, meaning, if one is true then all are true: A is invertible The equation A, has a unique solution, which is The columns of A are linearly independent For each column vector b R n, A x = b has a unique solution x Null(A) = Conversely, if one of the following is true, then all of the following are true: A is not invertible A for some x The columns of A are linearly dependent There is not be a unique x for every b where A x = b Null(A) contains more than just the zero vector These are part of what is known as the Invertible Matrix Theorem.. Mechanical Eigenvalues and Eigenvectors In each part, find the eigenvalues of the matrix M and the associated eigenvectors. (a) M = 9 Let s begin by finding the eigenvalues: det ( ) () λ λ = det = 9 λ 9 λ The determinant of a diagonal matrix is the product of the entries. ( λ)(9 λ) = From the above equation, we know that the eigenvalues are λ = and λ = 9. UCB EECS 6A, Spring 29, Discussion 4B, All Rights Reserved. This may not be publicly shared without explicit permission.
Last Updated: 29-2-2 9:56 2 For the eigenvalue λ = : which is simply x 2 = or equivalently For the eigenvalue λ = 9: ( 9 (M I) ) 8 x or equivalently span {}. (M 9I) ( ) 9 9 ( ) 9 9 9 8 {} which is simply x = or equivalently or equivalently span. x 2 (b) M = 2 2 Let s begin by finding the eigenvalues: det ( ) () λ λ = det = 2 2 λ 2 2 λ ( λ)(2 λ) 2 = λ 2 3λ = λ(λ 3) = From the above equation, we know that the eigenvalues are λ = and λ = 3. For the eigenvalue λ = : (M I) x = M 2 2 x which is simply x is free and x 2 = x or equivalently x For the eigenvalue λ = 3: (M 3I) ( ) 3 2 2 ( ) 3 2 2 3 2 2 {} or equivalently span. UCB EECS 6A, Spring 29, Discussion 4B, All Rights Reserved. This may not be publicly shared without explicit permission. 2
Last Updated: 29-2-2 9:56 3 which is simply x is free and x 2 = 2x or equivalently (c) (PRACTICE) M = 3 4 9 3 Let s begin by finding the eigenvalues: [ x ] {} or equivalently span. 2x 2 λ λ det 3 4 9 λ = det 3 4 λ 9 = 3 λ 3 λ Without changing the determinant, we can subtract 3 λ of row from row 2. λ λ det 3 4 λ 9 = det 4 λ 9 = 3 λ 3 λ λ(4 λ)(3 λ) = From the above equation, we know that the eigenvalues are λ =, λ = 3, and λ = 4. For the eigenvalue λ = : (M I) x = M 3 4 9 3 which is simply x 3 =, x 2 is free, and x = 4 3 x 2 or equivalently For the eigenvalue λ = 3: 4 3 x 2 x 2 (M 3I) 3 4 9 3 3 3 3 4 9 3 3 3 3 3 9 4 3 or equivalently span. which is simply x =, x 3 is free, and x 2 = 9x 3 or equivalently 9x 3 or equivalently span 9. x 3 UCB EECS 6A, Spring 29, Discussion 4B, All Rights Reserved. This may not be publicly shared without explicit permission. 3
Last Updated: 29-2-2 9:56 4 For the eigenvalue λ = 4: (M 4I) 3 4 9 4 3 4 3 4 9 4 3 4 4 3 9 which is simply x = x 3 = and x 2 = x 2 or equivalently x 2 or equivalently span. (d) (PRACTICE) M = Let s begin by finding the eigenvalues: det ( ) λ = det λ () λ = λ Without changing the determinant, we can add λ of row to row 2. () ([ λ λ det = det λ λ λ ( λ λ ) = λ 2 + = λ ]) = From the above equation, we know that the eigenvalues are λ = i and λ = i. For the eigenvalue λ = i: (M ii) ( ) i ( ) i i [ i i ] {} ix2 i which is simply x = ix 2 and x 2 is free or equivalently or equivalently span. x 2 UCB EECS 6A, Spring 29, Discussion 4B, All Rights Reserved. This may not be publicly shared without explicit permission. 4
.5.4.4.4 Last Updated: 29-2-2 9:56 5 For the eigenvalue λ = i: (M + ii) ( ) + i ( ) i + i [ i i ] {} ix2 i which is simply x = ix 2 and x 2 is free or equivalently or equivalently span. 2. Steady State Reservoir Levels We have 3 reservoirs: A,B and C. The pumps system between the reservoirs is depicted in Figure..2 x 2 A B.3.2.3 C.3 Figure : Reservoir pumps system. (a) Write out the transition matrix representing the pumps system..2.5.4 T =.4.3.3.4.2.3 (b) Assuming that you start the pumps with the water levels of the reservoirs at A = 29,B = 9,C = (in kiloliters), what would be the steady state water levels (in kiloliters) according to the pumps system described above? Hint: If x ss = A ss B ss C ss is a vector describing the steady state levels of water in the reservoirs (in kiloliters), what happens if you fill the reservoirs A,B and C with A ss,b ss and C ss kiloliters of water, respectively, and apply the pumps once? UCB EECS 6A, Spring 29, Discussion 4B, All Rights Reserved. This may not be publicly shared without explicit permission. 5
Last Updated: 29-2-2 9:56 6 Hint II: Note that the pumps system preserves the total amount of water in the reservoirs. That is, no water is lost or gained by applying the pumps. If x ss = A ss B ss C ss is a vector describing the steady state levels of water in the reservoirs, then we know that T x ss = x ss that is, applying the pumps one more time wouldn t change the level of water in any of the reservoirs. This means that x ss is an eigenvector of T associated with the eigenvalue λ =. Therefore,.2.5.4.8.5.4 x ss Null(T I) = Null.4.3.3 = Null.4.7.3.4.2.3.4.2.7.8.5.4 43 We calculate the null space of the matrix.4.7.3, which is simply span 4,.4.2.7 36 43α which means that our steady state reservoirs levels vector is of the form 4α,α R. 36α Furthermore, we know that the pumps system conserves the water, i.e., no water is lost by running the pumps system. Therefore, we know that the total amount of water in the reservoirs at any point in time will be 29 + 9 + = 238 (equal to the original total amount of water in the system). Therefore, we are looking for an eigenvector whose components sum to 238. In other words, we are looking for α such that 43α + 4α + 36α = 238, which yields α = 2. Therefore, the steady state levels of the water 86 in the reservoirs will be 8. 72 3. Eigenvalues and Special Matrices Visualization An eigenvector v belonging to a square matrix A is a nonzero vector that satisfies A v = λ v where λ is a a scalar known as the eigenvalue corresponding to eigenvector v. The following parts don t require knowledge about how to find eigenvalues. Answer each part by reasoning about the matrix at hand. (a) Does the identity matrix in R n have any eigenvalues λ R? What are the corresponding eigenvectors? Multiplying the identity matrix with any vector in R n produces the same vector, that is, I x = x = x. Therefore, λ =. Since x can be any vector in R n, the corresponding eigenvectors are all vectors in R n. d d 2 (b) Does a diagonal matrix d 3 in R n have any eigenvalues λ R? What are the....... d n corresponding eigenvectors? UCB EECS 6A, Spring 29, Discussion 4B, All Rights Reserved. This may not be publicly shared without explicit permission. 6
Last Updated: 29-2-2 9:56 7 Since the matrix is diagonal, multiplying the diagonal matrix with any standard basis vector e i produces d i e i, that is, D e i = d i e i. Therefore, the eigenvalues are the diagonal entries d i of D, and the corresponding eigenvector associated with λ = d i is the standard basis vector e i. (c) Does a rotation matrix in R 2 have any eigenvalues λ R? There are three cases: i. Rotation by (more accurately, any integer multiple of 36 ), which yields a rotation matrix R = I: This will have one eigenvalue of + because it doesn t affect any vector (R x = x). The eigenspace associated with it is R 2. ii. Rotation by 8 (more accurately, any angle of 8 + n 36 for integer n), which yields a rotation matrix R = I: This will have one eigenvalue of because it flips any vector (R x = x). The eigenspace associated with it is R 2. iii. Any other rotation: there aren t any real eigenvalues. The reason is, if there were any real eigenvalue λ R for a non-trivial rotation matrix, it means that we can get R x = λ x for some x, which means that by rotating a vector, we scaled it. This is a contradiction (again, unless R = I). Refer to Figure 2 for a visualization. M x λ x x θ α Figure 2: Rotation will never scale any non-zero vector (by a real number) unless it is rotation by an integer multiple of 36 (identity matrix) or the rotation angle is θ = 8 + n 36 for any integer n ( I). (d) Does a reflection matrix in R 2 have any eigenvalues λ R? Yes, both + and. Why? Reflecting any vector that is on the reflection axis will not affect it (eigenvalue +). Reflecting any vector orthogonal to the reflection axis will just flip it/negate it (eigenvalue ). In other words, the axis of reflection is the eigenspace associated with the eigenvalue + and the orthogonal space to that axis of reflection is the eigenspace associated with the eigenvalue. Refer to Figure 3 for a visualization. UCB EECS 6A, Spring 29, Discussion 4B, All Rights Reserved. This may not be publicly shared without explicit permission. 7
Last Updated: 29-2-2 9:56 8 M x λ x x3 M x 3 = x 3 α M x 2 = x 2 α x Figure 3: Reflection will scale vectors on the reflection axis (by +) and orthogonal to it (by ). (e) If a matrix M has an eigenvalue λ =, what does this say about its null space? What does this say about the solutions of the system of linear equations M x = b? dim(null(m)) > M x = b has no unique solution. (f) Does the matrix have any eigenvalues λ R? What are the corresponding eigenvectors? Hint: What is the rank of the matrix? Note that the matrix is rank-deficient. Therefore, according to part (e), one eigenvalue [ is] λ =. The corresponding eigenvector, which is equivalent to the basis vector for the null space, is eigenvalue is, by inspection, λ = with the corresponding eigenvector because. The other ] =. [ UCB EECS 6A, Spring 29, Discussion 4B, All Rights Reserved. This may not be publicly shared without explicit permission. 8