Systems of Equations
List the three elementary row operations
1. Interchanging two rows, e.g. r_2 <-> r_3
2. Scaling a row e.g. r_1 -> 3r_1
3. Adding a multiple of one row to another row, e.g. r_1 -> r_1 + 2r_2
det(AB) = ?
det(A) det(B)
If A is an nxn matrix, then Row(A) = Col(A).
False! Their dimensions are the same, but they could be very different! Example:
1 0
2 0
The row space is all vectors of the form (x,0). The column space is all vectors of the form (x,2x).
If A is a matrix with A^T = A, then A is called?
Symmetric
Is (1,1,1) an eigenvector for
2 -1 2
-1 2 2
-1 -2 6
If so, what is its eigenvalue?
Yes! If you take the product you get
(3,3,3) so its an eigenvector with eigenvalue 3.
Write down the augmented matrix corresponding to the following system:
2x_1 + 3x_3 = -1
-x_1 - 2x_2 = 0
x_1 + x_2 - x_3 = 1
Row 1: 2 0 3 -1
Row 2: -1 -2 0 0
Row 3: 1 1 -1 1
A is invertible!
If a set of vectors span R^n, then they are also linearly independent in R^n.
False! You can span but not be linearly independent. For example (1,0),(0,1), (1,1) span R^2, but are not linearly independent.
There exists a 5x7 matrix with Rank(A) = Nullity(A).
False! Rank(A) + Nullity(A) = 7 in this case. But if they're equal, their sum is even. So this is impossible.
Find the characteristic equation and eigenvalues of
2 -3
0 2
What are their algebraic multiplicities? Can you decide if the matrix is diagonalizable?
2 is the only eigenvalue, with algebraic multiplicity 2.
Can't decide if it's diagonalizable yet, have to check gemu(2).
Show that there do not exist scalars c_1,c_2,c_3 such that
c_1 (-2,9,6) + c_2 (-3,2,1) + c_3 (1,7,5) = (0,5,4)
Show that the matrix
-2 -3 1 0
9 2 7 5
6 1 5 4
Is inconsistent.
Given a matrix A with det(A) = 2. If I form B by doing the following row operations to A, what is det(B)?
r_1 -> r_1 + 2r_2
r_2 <-> r_3
r_3 -> -2 r_3
2 det(A) = 4.
First row operation doesn't do anything. Second negates so -2. Third scales by -2, so get 4.
If n vectors in R^n are linearly independent, they also span.
True!
If A is 9 x 5 with rank(A) = 5, does Ax = b have a solution for every b in R^9?
No! The column space is a subspace of R^9 in this case, and it has dimension 5. So there are many b that are not in the span of the columns!
So full rank does not always imply there is a solution to Ax = b.
Find the E_5 eigenspace of
-3 0 7
2 5 1
-1 0 5
It's one dimensional with basis given by the vector
(0,1,0)
The most consistent way to do this is to find Null(A - 5I). But... you could notice that (0,1,0) gets scaled by 5 because the second column is (0,1,0) times 5.
Solve for z in the system using Cramer's rule:
3x+2y+6z = 4
x + z = 0
7x + 3y + 9z = 1
Find the inverse of
1 -1 1
2 3 0
0 2 -1
3 -1 3
-2 1 -2
-4 2 -5
Is the subset of R^2 given by (x,y) such that xy >= 0 a subspace of R^2? If so, find a basis for the susbspace.
No it's not! It is not closed under addition.
(-1,-1) and (0,2) are both in the subset, but their sum (-1,1) is not!
If a square matrix has characteristic polynomial
(L-1)^3 (L-2)^2 (L-1)(L-5)
What is the size of the matrix? What needs to be true about gemu(1) and gemu(2) for this matrix to be diagonalizable? What's the determinant of the matrix?
The size is the degree of the characteristic polynomial, so it's 7x7.
gemu(1) = 3 and gemu(2) = 2 implies it's diagonalizable. Since these are the almu's.
Determinant is 1^3 2^2 *1*5 = 20. Product of e-values.
0 3
2 -1
diagonalizable? If so, find the matrix P that diagonalizes it.
Yes! Two distinct eigenvalues L = -3,2.
P = -1 3/2
1 1
works!
Solve the system corresponding to the following augmented matrix using Gauss-Jordan elimination:
1 0 3 1
0 1 2 4
x_3 = t_1 is free
x_2 = 4 - 2t_1
x_1 = 1 - 3 t_1
So solution is
(0,4,1) + t_1 (-3, -2, 1).
If
P_{B_1 ->B_2} = 3 1 P_{B_2 -> B_3} = 7 2
5 2 4 -1
Find P_{B_3 -> B_1}
It's the inverse of P_{B_1 -> B_3} = P_{B_2 -> B_3} P_{B_1 ->B_2}
So
1/15 times -2 11
7 -31
Is the set of matrices A in M_{22} such that AC = CA for all other matrices C a subspace of M_{22}? If so, find a basis for this subspace.
Yes it is! Given two such matrices A,B, and any matrix C
The zero matrix satisfies this! 0 C = C 0
(A+B)C = AC + BC = CA + CB = C(A+B)
(kA)C = kCA = C(kA)
How to find a basis...
If A is a 10x27 matrix with Rank(A) = 9, what is Nullity(A)? What is Rank(A^T)? What is Nullity(A^T)?
Nullity(A) = 18
Rank(A^T) = Rank(A) = 9
Nullity(A^T) = 10 - Rank(A^T) = 10-9 = 1
If A is a 3x3 matrix with eigenvalues L = -1,2 with gemu(-1) = 1 and gemu(2) = 2. Is A^{-1} diagonalizable? What does the diagonalization of A^{-1} look like?
Yes it is! It's eigenvalues are -1 and 1/2.
gemu(-1) = 1 still and gemu(1/2) = 2 still.
One option for the diagonalization is
1/2 0 0
0 1/2 0
0 0 -1