linear algebra final (5.1-5.3, 6.1-6.3, 6.5, 6.6)
orthogonality
2 vectors u and v in Rₙ are orthogonal (to each other) if u∙v=0
for a square matrix A, vectors in Col A are orthogonal to vectors in Nul A
FALSE
the determinant of A is the product of the diagonal entries in A
FALSE in general TRUE if A is triangular
If λ+5 is a factor of the characteristic polynomial of A, then 5 is an eigenvalue of A
FALSE; -5 is an eigenvalue λ+5=0 -> λ=-5
A is diagonalizable if A=PDP⁻¹ for some matrix D and some invertible matrix P
FALSE; D must be a diagonal matrix
det Aᵀ = (-1) det A
FALSE; det Aᵀ = det A Theorem 3: Properties of Determinants
an elementary row operation on A does not change the determinant
FALSE; interchanging rows and multiplying a row by a constant changes the determinant
A is diagonalizable if and only if A has n eigenvalues, counting multiplicites
FALSE; it ALWAYS has n eigenvalues, counting multiplicity
if A is diagonalizable, then A has n distinct eigenvalues
FALSE; it could have repeated eigenvalues as long as the basis of each eigenspace is equal to the multiplicity of that eigenvalue
the orthogonal projection ŷ of y onto a subspace W can sometimes depend on the orthogonal basis for W used to compute ŷ
FALSE; it is ALWAYS independent of basis
if A is diagonalizable, then A is invertible
FALSE; it's invertible if it doesn't have a zero eigenvector, but this doesn't affect diagonalizability
the best approximation to y by elements of a subspace W is given by the vector y - proj𝓌y
FALSE; the best approx. is proj𝓌y
A is diagonalizable if A has n eigenvectors
FALSE; the eigenvectors must be linearly independent
If A is invertible, then A is diagonalizable
FALSE; these are not directly related
If Ax=λx for some vector x, then λ is an eigenvalue of A
FALSE; this is true as long as the vector is not the 0 vector
if an nxp matrix U has orthonormal columns, then UUᵀx=x for all x in Rⁿ
FALSE; this only holds true if U is a square matrix
thm 3 (275) [props of determinants]
Let A and B be nxn matrices: 1. A is invertible if and only if detA!=0 2. detAB = (det A)(det B) 3. det Aᵀ= det A 4. if A is triangular, then det A is the product of the entries on the main diagonal of A 5. a row replacement operation on A does NOT change the determinant a row interchange changes the sign of the determinant a row scaling also scales the determinant by the same scalar factor
A is a 3x3 matrix with 2 eigenvalues. each eigenspace is 1D. is A diagonalizable; why?
NO; the sum of the dimensions of the eigenspaces do NOT equal 3 1+1 != 3
(det A)(det B)= detAB
TRUE
If A is 3x3 with columns a₁, a₂, a₃, then detA equals the volume of the parallelepiped determined by a₁, a₂, a₃
TRUE
a row replacement operation on A does not change the eigenvalues
TRUE
for any scalar c, u.(cv) = c(u.v)
TRUE
for each y and each subspace W, the vector y - proj𝓌y is orthogonal to W
TRUE
if W is a subspace of Rⁿ and if v is in both W and W⟂, then v must be the zero vector
TRUE
if the columns of an nxp matrix U are orthonormal, then UUᵀy is the orthogonal projection of y onto the column space of U
TRUE
if the distance from u to v equals the distance from u to -v, then u and v are orthogonal
TRUE
if y = z₁ + z₂, where z₁ is in a subspace W and z₂ is in W⟂, then z₁ must be the orthogonal projection of y onto W
TRUE
if y is in a subspace W, then the orthogonal projection of y onto W is y itself
TRUE
in the Orthogonal Decomp Thm, each term in formula(2) for ŷ is itself an orthogonal projection of y onto a subspace of W
TRUE
the multiplicity of a root r of the characteristic equation of A is called the algebraic multiplicity of r as an eigenvalue of A
TRUE
v.v = ||v||^2
TRUE
if Rⁿ has a basis of eigenvectors of A, then A is diagonalizable
TRUE;
a matrix A is not invertible if and only if 0 is an eigenvalue of A
TRUE; Invertible Matrix Theorem
if vectors v1,....vp span a subspace W and if x is orthogonal to each vj for j=1,....p, then x is in Wperp
TRUE; any vector in W can be written as linear combos of basis vectors,
If AP=PD, with D diagonal, then the nonzero columns of P must be eigenvectors of A
TRUE; each column of PD is a column of P*A, and is equal to the corresponding entry in D*vector P. This satisfies the eigenvector definition as long as the column is nonzero
Finding an eigenvector of A may be difficult, but checking whether a given vector is in fact an eigenvector is easy
TRUE; just see if Ax is a scalar multiple of x
a number C is an eigenvalue of A if and only if the equation (A-cI)x=0 has a nontrivial solution
TRUE; this is a rearrangement of the equation Ax=λx
If z is orthogonal to u₁ and to u₂ and if W=Span{u₁, u₂} then z must be in W⟂
TRUE; z will be orthogonal to any linear combo of u₁ and u₂
A is a 5x5 matrix with 2 eigenvalues. 1 eigenspace is 3D, and the other is 2D. is A diagonalizable; why?
YES; the sum of the dimensions of the eigenspaces equal 5 3+2=5
inner product
a matrix product uᵀv or u·v where u and v are vectors; if u·v = 0, u and v are orthogonal
eigenvector; basis
a nonzero vector x such that Ax=λx for some scalar λ a basis consisting entirely of eigenvectors of a given matrix
eigenvalue
a scalar λ such that Ax=λx has a solution for some nonzero(nontrivial) vector x
diagonalization thm (282)
an nxn matrix A is diagonalizable if and only if A has n linearly independent eigenvectors A=PDP⁻¹, with D a diagonal matrix, if and only if the columns of P are n linearly independent eigenvectors of A. in this case, the diagonal entries of D are eigenvalues of A that correspond, respectively, to the eigenvectors in P
characteristic equation
det(A-λI)=0
explain why a 2x2 matrix can have at most 2 distinct eigenvalues explain why an nxn matrix can have at most n distinct eigenvalues
eigenvectors corresponding to distinct eigenvalues are linearly independent; 2x2 matrices must fit in R² space, meaning it can have, at most, 2 linearly independent vectors
distance; dist(u,v)
for u and v in Rₙ; the length of the vector u-v dist(u, v) = ||u-v||
thm 4 (338)
if S = {u₁,...uₚ} is an orthogonal set of nonzero vectors in Rₙ, then S is linearly independent and hence is a basis for the subspace spanned by S
A is a 4x4 matrix with 3 eigenvalues. 1 eigenspace is 1D and 1 other is 2D. Is it possible that A is not diagonalizable; why?
if the remaining
thm 2 (270)
if v₁,...vᵣ are eigenvectors that correspond to distinct eigenvalues λ₁,...λᵣ of an nxn matrix A, then the set {v₁,...vᵣ} is linearly independent
invertible matrix theorem (275)
let A be an nxn matrix. Then A is invertible if and only if: 1. The number 0 is NOT an eigenvalue of A 2. the determinant of A is NOT 0
thm 5 (339)
let {u₁,...uₚ} be an orthogonal basis for a subspace W of Rₙ. for each y in W, the weights in the linear combo: y=c₁u₁ + ... + cₚuₚ are given by cⱼ = (y ∙ uⱼ)/(uⱼ ∙ uⱼ) (j=1,....p)
let λ be an eigenvalue of an invertible matrix A. Show that λ⁻¹ is an eigenvalue of A⁻¹
since A is invertible: λ!=0 there is a non-zero vector x such that Ax=λx Ax=λx -> [A⁻¹](Ax)=[A⁻¹](λx) -> x=λ(A⁻¹x) -> [λ⁻¹]x=[λ⁻¹]λ(A⁻¹x) -> A⁻¹x=λ⁻¹x
length (or norm)
the scalar ||v|| = √v·v = √{v, v}
formula 2 (340)
ŷ = projₗy = [(y∙u)/(u∙u)] u
