Linear Algebra Review Session 4

Linear Algebra: Review Session 4 Eigenvalues and Eigenvectors Computations and Diagonalizability Ex. 1. Let A be the matrix A01 21-2 0 11 (a) Define what it means for a real number AER to be an eigenvalue of A. (b) Find all eigenvectors of A. (e) Find a diagonal matrix D and an invertible matrix P such that A PDP-1, or show that no such matrices exist. 20 Ex. 2. Consider the matrix A-30-5 10 (a) Find all eigenvalues of A. (b) Find a diagonal matrix D and an invertible matrix P such that A matrices exist.. PDP-1, or show that no such Ex. 3. Let A be the matrix 2 A 0 1 2 1 0 6-1-3 - Find a diagonal matrix D and an invertible matrix P such that D PAP-1, or show that no such matrices exist. Ex. 4. Consider the matrix 10-2 -20 4 Find a basis for R consisting of eigenvectors of A, or else prove that there is no such basis. Ex. 5. Consider the matrix [1-20] A2 5 1 0 22 Is A diagonalizable? Why or why not? Proofs Ex. 6. Let V be a finite-dimensional vector space, and let T: VV be a linear transformation. Prove that 0 is an eigenvalue of T if and only if the image (ie., range) of T is not equal to V. Ex. 7. Let A, B be nxn matrices that commute, Le. ABBA. Let ER" be an eigenvector of A such that Bu0. Prove that Bu is also an eigenvector of A. Ex. 8. Suppose that a 3x3 matrix A has 0 as an eigenvalue. (a) What are the possible values of the rank of A? Justify your answer. (b) Let T: RR be the linear transformation given by T(z) Ar. Can T possibly be one-to-one? Can T be onto? Justify your answer. Ex. 9. Let A be annxn matrix and R be a scalar that is NOT an eigenvalue of A. Suppose that is an eigenvalue for the matrix B(Aol) with corresponding eigenvector v. Prove that u is also an eigenvector for A and find a formula for the corresponding eigenvalue of A in terms of u and a. Ex. 10. Prove that if T: VV is a linear map then the eigenspace Es corresponding to the eigenvalue X = 0 is equal to Ker(T). Linear Algebra Review Session 4: Eigenvalues & Eigenvectors Ex. 1 [A=[\begin{matrix}2&1&-2\ 0&1&0\ 1&1&-1\end{matrix}]] a) A real number λ is an eigenvalue of A if there exists a nonzero vector 𝑣 ∈ R³ such that A𝑣 = λ𝑣. b) det(A-λI) = det([\begin{matrix}2-λ&1&-2\ 0&1-λ&0\ 1&1&-1-λ\end{matrix}]) = 0 |1-λ -2| |1 -1-λ| |1 -1-λ| = (1-λ)[(2-λ)(-1-λ)+2] = (1-λ)[λ²+λ-2λ-2+2] = (1-λ)(λ²-λ) = (1-λ)λ(λ-1) Thus, the eigenvalues of A are 0 and 1. => λ=0,1,1 c) Find eigenvectors (A-0I)x=0 => [\begin{matrix}2&1&-2\ 0&1&0\ 1&1&-1\end{matrix}]x=[\begin{matrix}0\ 0\ 0\end{matrix}] => 2x+y-2z=0 y=-2x+2z x+y-z=0 Since det(A-0I)≠0 => A is diagonalizable, so we find D and P. x, z free v₁=[\begin{matrix}1\ -2\ 0\end{matrix}], v₂=[\begin{matrix}0\ 2\ 1\end{matrix}] (A-1I)x=0 => [\begin{matrix}1&1&-2\ 0&0&0\ 1&1&-2\end{matrix}][\begin{matrix}x\ y\ z\end{matrix}]=[\begin{matrix}0\ 0\ 0\end{matrix}] => 2x+y-2z=0 2z=0 2x-2z=0 y=z 0 y=0 =0 x-z=0 y=0 y=0 x+y-z=0 v₃=[\begin{matrix}0\ 0\ 1\end{matrix}] Then if D=[\begin{matrix}1&0&0\ 0&1&0\ 0&0&0\end{matrix}] and P=[\begin{matrix}1&0&1\ -2&2&0\ 0&1&1\end{matrix}] we have A=PDP^(-1) Ex.2 A=[\begin{matrix}2&0&-1\ 3&0&-1\ 1&0&0\end{matrix}] a) det(A-λI) = det([\begin{matrix}2-λ&0&-1\ 3&-λ&-1\ 1&0&-λ\end{matrix}]) = -λ(λ²-2λ+1) = -λ(λ-1)(λ-1) eigenvalues are 1 and 0, b) (A-1I)x=0 => [\begin{matrix}1&0&-1\ 3&-1&-5\ 1&0&-1\end{matrix}][\begin{matrix}x\ y\ z\end{matrix}]=[\begin{matrix}0\ 0\ 0\ 0\end{matrix}] => x-z=0 x=z x=z 3x-y-5z+0 => x=z 3x-z=0 3z-y-5z=0 y=-2z. Since there is only one free variable, dim(ker(A-1I)) = dim(E₁) = 1. Thus, the geometric multiplicity of the eigenvalue 1 is 1. Since the algebraic multiplicity of λ=1 is 2, A is not diagonalizable, Hence, there do net exist matrices P and D (D diagonal) s.t. A=PDP^(-1) Image Analysis Textual Content: Ex. 3 Eigenvalues: λ = 0, -1, 2 Eigenvectors: (respectively) [\begin{pmatrix} -1 \ 0 \ 2 \end{pmatrix}], [\begin{pmatrix} -1 \ 0 \ 3 \end{pmatrix}], [\begin{pmatrix} -2 \ -3 \ 3 \end{pmatrix}] (D = \begin{pmatrix} 0 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 2 \end{pmatrix}), (P = \begin{pmatrix} -1 & -1 & -2 \ 0 & 0 & -3 \ 2 & 3 & 3 \end{pmatrix}) Ex. 4 Eigenvalues: λ = 0, 0, 5 Eigenvectors: ... (missing eigenvectors) Ex. 5 Eigenvalues: λ = 2, 3, 3 Eigenvectors: ... (missing eigenvectors) Since the algebraic mult. of λ = 3 is 2 but the geometric multiplicity of 3 is only 1, A is not diagonalizable. Ex. 6 Let V be a finite dim. vector space, and T: V -> V be linear. Let dim V = n. (=>) First suppose that λ = 0 is an eigenvalue of T. Then ∃ a nonzero vector v ∈ V such that Tv = 0. Hence, the dimension of the nullspace of T is at least one. By the rank-nullity theorem, the dimension of the image of T is therefore at most n-1 since dim(Im T) + dim(nullspace T) = n = dim V. Thus, T cannot be surjective if the image of T is not equal to V.   (<=) Now suppose that the image of T is not equal to V. Thus, T is not surjective. Since T: V -> V, T cannot be injective either. Thus, there is the dimension of the nullspace of T is at least one. So, there is a nonzero vector v ∈ V such that Tv = 0. But then Tv = 0 and so v is an eigenvector of T with eigenvalue λ = 0. Ex. 7 Let A, B be nxn matrices s.t. AB = BA. Let v ∈ R^n be an eigenvector of A s.t. Av = λv. Since v is an eigenvector of A, ∃ λ such that Av = λv. Now A(Bv) = (AB)v = (BA)v = B(Av) = B(λv) = λ(Bv). Since Bv ≠ 0, Bv is an eigenvector of A with eigenvalue λ. Ex. 8 Let A be a 3x3 matrix and λ=0 be an eigenvalue of A. a) Since 0 is an eigenvalue of A, ∃ a nonzero vector v ∈ R³ s.t. Av=0v=0. Thus, v ∈ ker(A) and dim(ker A) ≥ 1. By the rank-nullity theorem, null(A) + rank(A)=3. Since null(A) = 1, 2, or 3, the possible values of rank(A) are 2, 1, or 0. b) Let T: R³ -> R³ by T(x) = Ax. From part a), the nullity of A is at least one and hence the dimension of the kernel of T is at least 1. Therefore, T cannot be one-to-one. Since T: R³ -> R³ cannot be one-to-one, it cannot be onto either. Ex. 9 Let A be an nxn matrix and d ∈ R s.t. d is not an eigenvalue of A. Suppose that v is an eigenvector of B=(A-dI) with eigenvalue λ. Since v is an eigenvector of B, note that v ≠ 0, and Bv = (A-dI)v = λv. Applying A-dI to both sides, we find Av = (A-dI)v + dv = λv + dv = (λ+d)v. Now, noting that v ≠ 0, and B is invertible, we must have that v ≠ 0. Thus, we can divide by v and obtain Av = (λ+d)v. Thus, v is an eigenvector of A with eigenvalue λ+d. Ex. 10 Let T: V -> V be linear and E₀ be the eigenspace corresponding to the eigenvalue 0. Let v ∈ E₀. Then Tv = 0v = 0. Hence, v ∈ ker(T). Now let v ∈ ker(T). Then Tv = 0 = 0v, so v ∈ E₀. Thus, E₀ = ker(T).