18.06 Problem Set #4 Solutions

1. C(A^{T}): Denote the matrices in the problem as A = LU. The row space of A is the same as the row space of its row echelon matrix U. So a basis for C(A^{T}) is {(0, 1, 2, 3, 4), (0, 0, 0, 1, −1)}.

N(A): Since L is invertible, the null-space of A is equal to the nullspace of U. Look at U, columns 1, 3, 5 give a free variable each. So a basis for N(A) is {(1, 0, 0, 0, 0), (0, −2, 1, 0, 0), (0, −7, 0, 1, 1)}.

C(A): Each column of A is a linear combination of column vectors of L with coefficients given by the corresponding column of U. For example, the fourth column of A is 3 times first column of L plus 1 times second column of L. But the third row of U is a zero row. When expressing column vectors of A as linear combinations of column vectors of L, the coefficient of the third column vector of L is always zero. Hence, it is enough to use the first two columns of L to express any column of A. Also considering the rank of A is two, a basis of C(A) would be first two column vectors of L. That is {(1, −4, 8), (0, 1, 3)}.

N(A^{T}): N(A^{T}) is the orthogonal complement of C(A), so a basis of N(A^{T}) is {(−20, −3, 1)}.

2. (a) Denote the i-th column vector of A as v_{i}, then the entry of A^{T} A in the i-th row, j-th column is the inner product of v_{i} and v_{j}. v_{i} · v_{i} = |v_{i}|^{2} = 1. v_{i} · v_{j} = 0 if i ≠ j because v_{i} and v_{j} are orthogonal. Hence A^{T} A = I the identity matrix.

(b) Just check the conditions.

(c) For example, [ \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ] is an orthogonal matrix, but two times it, [ \begin{matrix}2 & 0 \\ 0 & 2 \end{matrix} ] is not orthogonal

because the first column vector (2, 0) has norm 2 so not a unit vector.

(d) A^{T} = A^{−1} ⇔ A^{T} A = I and the argument in part (a) is reversible.

3. S^{⊥}: Put those vectors spanning S as row vectors of a matrix A:

A = [ \begin{matrix}1 & 0 & −2 & 0 \\ 0 & 2 & 4 & −1 \\ 2 & 2 & 0 & −1 \end{matrix} ]

Now use the row echelon form to find S^{⊥} = C(A^{T})^{⊥} = N(A) as the span of {(2, −2, 1, 0), (0, 1/2, 0, 1)}.

(S^{⊥})^{⊥}: (S^{⊥})^{⊥} = null-space of [ \begin{matrix}2 & −2 & 1 & 0 \\ 0 & 2 & 0 & 1 \end{matrix} ], is the span of \{(−1/2, 0, 1, 0), (−2, −2, 0, 1)\}.

(S^{⊥})^{⊥} = S: It is a simple calculation to express the spanning vectors from S as linear combinations of basis vectors of (S^{⊥})^{⊥}, or vice versa. So the two vector spaces are the same. (S^{⊥})^{⊥} = S is generally true for any vector space S.

4. U^{⊥}: U^{⊥} is the null-space of the matrix with row vectors the spanning vectors of U, i.e. [ \begin{matrix}1 & −2 & 0 & 3 \\ 0 & 1 & 0 & −1 \end{matrix} ].

A basis for U^{⊥} is {(0, 0, 1, 0), (−1, 1, 0, 1)}.

V^{⊥}: V^{⊥} is the null-space of the matrix with row vectors the spanning vectors of V, i.e. [ \begin{matrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{matrix} ].

A basis for V^{⊥} is {(0, 0, 0, 1)}.

U ∩ V: U is the nullspace of the matrix whose rows are the basis vectors of U^{⊥} from above. It follows that any vector (x_{1}, x_{2}, x_{3}, x_{4}) in U will satisfy x_{3} = 0, and −x_{1} + x_{2} + x_{4} = 0. Similarly, a vector (x_{1}, x_{2}, x_{3}, x_{4}) in V has x_{4} = 0. For a vector to be in both U and V, it must satisfy all these equations, that is, it must be in the nullspace of

A = [ \begin{matrix}0 & 0 & 1 & 0 \\ −1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 \end{matrix} ]

The usual computation shows that N(A) is span of {(1, 1, 0, 0)}.

Alternatively: Suppose v is in U ∩ V. Since v is in U we have v = a(1, −2, 0, 3) + b(0, 1, 0, 1) = (a, −2a + b, 0, 3a − d) for constants a, b. But v is in V so we must the last component equal to zero, that is, 3a − d = 0 so 3a = d. Hence, v = (c, −2c + 3c, 0, 0) = c(1, 1, 0, 0), so U ∩ V is a subset of span{(1, 1, 0, 0)}. But any vector in the span of (1, 1, 0, 0) is in U ∩V, Hence U ∩V = span{(1, 1, 0, 0)}.

5. (a) Just follow the procedures, sorry for the poor numbers provided by Wei :).

A^{T} b = [ \begin{matrix}−4 \\ 3 \end{matrix} ] , A^{T} A = [ \begin{matrix}6 & 1 \\ 1 & 10 \end{matrix} ] , x̂ = [ −\frac{43}{59}_{\frac{22}{59}} ] .p = [ −\frac{43}{59} ; \frac{108}{59} ; \frac{23}{59} ]

(b)

A^{T} b = [ \begin{matrix}0 \\ 2 \end{matrix} ] , A^{T} A = [ \begin{matrix}5 & −5 \\ −5 & 15 \end{matrix} ] , x̂ = [ \frac{1}{5} \frac{1}{5} ] .p = [ \frac{2}{5} \frac{2}{5} \frac{1}{5} \frac{1}{5} ]

6. P multiplies a vector in ℝ^{m} to give another vector in ℝ^{m}. We know P has to be an m × m matrix. For all v ∈ ℝ^{m}, P v ∈ C(A^{T}), so C(P^{T}) is contained in C(A^{T}). On the other hand, P v = v for all v ∈ C(A^{T}). So C(P^{T}) contains all vectors in C(A^{T}). Thus C(P^{T}) = C(A^{T}). So r(P) = r(A) = n.