ma1513 2 Flashcards
p is a projection of vector v onto the plane
v - p is orthogonal to the plane
projection p of a vector u onto V where V is a subspace of Rn</sub>
u-p is orthogonal to V i.e. dot product of (u-p) with every vector in V is 0
T(0)
linear transformation
0
T(u + v)
T(u) + T(v)
T(cu)
cT(u)
If x is an eigenvector associated with the eigenvalue λ, the eigenvalue of kx will be
λ
Zero vector can be an eigenvector T/F
F
0 can be an eigenvalue T/F
T
Eigenvalues of a triangular matrix
All diagonal entries
Eigenspace
Solution space of the homogeneous system (λI - A)x = 0
Finding the basis for eigenspace
- Solve the system (λI - A)x = 0 using GE
- Get the general soln and separate the parameters
- Write the associated vectors in the span
Eigenvalue of I3
Only one eigenvalue 1
Eigenspace of I3
R3
If A is an n x n matrix, sum of multiplicities of eigenvalues
n
relation bw dimension of eigenspace and multiplicity
dim Eλi <= ri
Condition for n x n matrix A to be diagonalizable
A has n linearly independent eigenvectors
BD
where D is a diagonal matrix
BD = (b1d1 b2d2 … bndn)
A is an n x n matrix
|S| < n
where S is the total number of basis vectors from all eigenspaces
A is not diagonalizable
A is an n x n matrix
|S| = n
where S is the total number of basis vectors from all eigenspaces
A is diagonalizable
Algorithm for diagonalization
- Solve the characteristic polynomial det(𝜆𝜆I – A) to find all distinct eigenvalues 𝜆1, 𝜆2, …, 𝜆k.
- For each 𝜆i, find a basis S𝜆𝑖 for the eigenspace E𝜆𝑖 by solving (𝜆iI – A)x = 0
- Let S = S𝜆1 ∪ S𝜆2 ∪ … ∪ S𝜆k . (Then |S| is the total number of basis vectors from all the
eigespaces)
relation bw dimension of eigenspace and multiplicities
dim Eλi = |Sλi|
dim Eλi <= ri
Condition when A is diagonalizable
in terms of Eλi and multiplicity
dim Eλi = ri
then |S| = r1 + r2 + .. + rk = n
Condition when A is not diagonalizable
in terms of Eλi and multiplicity
dim Eλi < ri
then |S| < r1 + r2 + .. + rk = n
remember this
when we carry out the diagonalization algorithm, after step 1, if we obtain n distinct eigenvalues, then we can straight away conclude that the matrix is
diagonalizable without going through the remaining steps