— David Wagner 2008/01/22 14:04

[Please note. My handwriting has never been very neat and has deteriorated over the years from lack of use. The software I prefer to use for most assignments does not do any kind of calculation for me, it just makes my work easier to read.–dw]

(Final Exam MAT 2233 Spring 1997)

1. Define (a) subspace of a vector space V,

The vector space W is a subspace of V when it is a subset of V and the same vector space operations as V.

(b) nullspace of an m × n matrix A,

The nullspace of A is the set of all vectors X such that AX = 0.

(c ) basis of a vector space,

A basis is a linearly independent spanning set. That is, a basis of a vector space V is a set of linearly independent vectors which can be linearly combined to produce every element in V.

(d) inverse of an n × n matrix,

The inverse A^-1 of n x n matrix A is an n x n matrix such that A^-1A = A^-1A = I_n

(e) column space,

The column space of a matrix is the set of all linear combinations of its column vectors.

(f) dimension of a vector space,

The dimension of a vector space V is the number of vectors in a basis of V.

(g) elementary permutation matrix,

An elementary permutation matrix has the effect of exchanging only two rows of a matrix.

(h) A^T,

For the m x n matrix A, with i and j integers such that 1 ≤ i ≤ n and 1 ≤ j ≤ m, A_ij^T = A_ji.

In other words, the rows of A^T are the columns of A.

(i) linear independence,

A set of vectors is linearly independent if none of them is a (finite) linear combination of any of the others.

(j) inner product of two n-vectors x and y.

An inner product is a functional <⋅,⋅> on vector space HxH such that if f,g,h are elements of H and α is an element of ℜ, then

1. <f,f> ≥ 0 and <f,f> =0 iff f=0 (The functional is positive definite.)
2. <f+g,h> = <f,h> + <g,h> (The functional is bilinear.)
3. <αf,g> = α<f,g> (The functional is homogenous.)
4. <g,f> = <f,g> (The functional is symmetric.)

2. Decide whether the following vectors are independent in ℜ³.

Try the old brute force method.

1. →
   • ;
   • ;
   • .
2. ✔.

Since v₂ may be expressed as a linear combination of v₁ and v₃, they must not be linearly independent.

3. State the Fundamental Theorem of Linear Algebra Parts 1 and 2. Draw a schematic diagram to illustrate Part 2.

1. PA = LDU
2. Every m x n matrix A can be uniquely expressed as the sum of a vector in its nullspace (kernel) and one in its column space (range).

4. Find the dimension and a basis for each of the four fundamental subspaces of .

• By inspection, the dimension appears to be 2.
  • The fourth column simply repeats the first.
  • The third column may be obtained by adding twice the first from the second.
  • The first two columns are linearly independent.

1. A basis of the column space would therefore be
2. A nullspace basis is .
   • 
   • 
   • (repeated, but used to find a second linearly independent vector.)
3. Row three is a repetition of row 1 but rows one and two are linearly independent, so A basis for the rowspace would be .
4. The remaining basis is the nullspace of : .
   • 
   • 
   • 
   • (repeated, but used to find a second linearly independent vector.)

[Here, my lousy time-management skills become apparent…]

5. Solve the linear system

(a) by performing an LU factorization of A, then solving Ax = b using forward and backward substitution, and

(b) by finding A⁻¹ (checking that you have indeed found the inverse) and multiplying both sides by that matrix.

6. Decide whether the following statements are true or false.

(a) Suppose V is a vector space of dimension 5 and W is a subspace of dimension 4. Every basis for V can be reduced to a basis of W by removing 1 vector.

First, I'll assume W is meant to be given as a subspace of V. Will removing any defining vector of V leave a definition of W? No, you have to take away the one pointing in the direction W does not include. However, the question asks if a basis of V can be reduced to a basis W in this way, which it can.

(b) If A is a matrix and Ax = Ay, then x − y belongs to the nullspace of A.

Matrix pre-multiplication distributes over subtraction: Ax = Ay ⇒ Ax - Ay = 0 = A(x - y) = 0✔

(c ) A 3 × 5 matrix never has linearly independent columns.

(d) If A and B commute, then <m>(A − B)^2 = A^2 − 2BA + B^2</m>.

(A − B)^2 = (A − B)(A − B) = AA - BA -AB + BB = A^2 − BA - BA + B^2 = A^2 − 2BA + B^2.✔

(e) The set of vectors b in ℜ³ with b₁ · b₃ = 0 is a subspace of ℜ³.

b₁ · b₃ = 0 implies b spans ℜ², a two-dimensional subspace of ℜ³. I'm not certain what happens for the cases where b₂ is zero or not linearly independent of b₁ and b₃, but I think these just define smaller subspaces.

(f) If A and B are invertible then A + B is invertible.

• Certainly not true if A + B = 0.
• Also not true if A cannot be added to B. They may not be the same size. Each matrix may be composed of incompatible elements which cannot be summed (apples and oranges). These could even be matrices of some bizarre type for which inversion is defined but addition is not.

(g) A scalar multiple of the identity matrix I_n commutes with every n × n matrix.

(bI)A = A(bI) is true if scalar multiplication of a matrix distributes since AI = IA.

1. b I A = AbI;
2. b I A = b A I;
3. b I A = b I A.✔

(h) If the vectors x₀, x₁, x₂, …, x_m span a subspace S, then dim(S) = m + 1.

There are m+1 linearly independent vectors since the index m starts at zero instead of one.

(i) Two nonzero vectors which are orthogonal, are necessarily linearly independent.

Isn't this more-or-less by definition?

(j) A set of three vectors in ℜ⁴ cannot span ℜ⁴.

A set of three vectors in ℜ⁴ can at most span ℜ³.