Mathematics 244, section 1 – Linear Algebra Solutions for Exam 2 – March 23, 2007

I. (15) Let V be a vector space and let S be a finite subset of V with Span(S) = V. If T is a linearly independent subset of V, show that |T | ≤ |S|. (Here |S| denotes the number of elements in S, and similarly for |T |.)

Solution: Say S = {v_{1}, …, v_{m}} and T = {w_{1}, …, w_{n}} in V, where Span(S) = V. We will show that if n > m, then T must be linearly dependent. Since Span(S) = V, for each j, 1 ≤ j ≤ n, there exist scalars a_{ij} such that

(1)  w_{j} = a_{1j}v_{1} + ⋯ + a_{mj}v_{m}.

Suppose we have scalars c_{1}, …, c_{n} such that

(2)  0 = c_{1}w_{1} + c_{2}w_{2} + ⋯ + c_{n}w_{n}.

Then substituting from the equations (1), we have

0 = c_{1}(a_{11}v_{1} + ⋯ + a_{m1}v_{m}) + c_{2}(a_{12}v_{1} + ⋯ + a_{m2}v_{m}) + ⋯ + c_{n}(a_{1n}v_{1} + ⋯ + a_{mn}v_{m}).

Rearranging this equation gives:

0 = (a_{11}c_{1} + ⋯ + a_{1n}c_{n})v_{1} + ⋯ + (a_{m1}c_{1} + ⋯ + a_{mn}c_{n})v_{n}.

This will be satisfied if

\begin{aligned}a_{11}c_{1} + ⋯ + a_{1n}c_{n} & = 0 \\ ⋮ & ⋮ \\ a_{m1}c_{1} + ⋯ + a_{mn}c_{n} & = 0 \end{aligned}

But this is a homogeneous system of m linear equations in the variables c_{1}, …, c_{n}. Since we assume that n > m, there are free variables, hence nontrivial solutions. This implies that T is linearly dependent if n > m from (2). Hence if T is linearly independent, then n = |T | ≤ |S| = m.

II. All parts of this question refer to the matrix

A = ( \begin{matrix}1 & −2 & 0 & s & 2 \\ −2 & −1 & −1 & 2s & −1 \\ 1 & 3 & 1 & s^{2} & −1 \end{matrix} )

A) (15) For which value(s) of the scalar s does A satisfy dim Nul(A) = 2?

Solution: We begin by applying row operations R_{2} ↦ R_{2} + 2R_{1}, R_{3} ↦ R_{3} − R_{1}, and then R_{3} ↦ R_{3} + R_{2} the matrix A.

\begin{aligned}( \begin{matrix}1 & −2 & 0 & s & 2 \\ −2 & −1 & −1 & 2s & −1 \\ 1 & 3 & 1 & s^{2} & −1 \end{matrix} ) & → ( \begin{matrix}1 & −2 & 0 & s & 2 \\ 0 & −5 & −1 & 4s & 3 \\ 0 & 5 & 1 & s^{2} − s & −3 \end{matrix} ) \\ & → ( \begin{matrix}1 & −2 & 0 & s & 2 \\ 0 & −5 & −1 & 4s & 3 \\ 0 & 0 & 0 & s^{2} + 3s & 0 \end{matrix} ) \end{aligned}

From this, we can see that there are always pivots in columns 1 and 2. Moreover, if s^{2} + 3s ≠ 0, then there is a pivot in column 4 as well. To get dim Nul(A) = 2 (not 3), we need three pivot columns (2 free variables), so the condition is s^{2} + 3s ≠ 0. dim Nul(A) = 2 for all real s other than s = 0, s = −3.

B) (10) Say s = 1. Give bases of Col(A) and Nul(A).

Solution: We substitute s = 1 in the last matrix above and continue the reduction to row-reduced echelon form. The result is:

( \begin{matrix}1 & 0 & 2/5 & 0 & 4/5 \\ 0 & 1 & 1/5 & 0 & −3/5 \\ 0 & 0 & 0 & 1 & 0 \end{matrix} )

Thus, a basis for Col(A) is given by columns 1,2,4 of the original matrix A (taking s = 1 of course):

\{( \begin{matrix}1 \\ −2 \\ 1 \end{matrix} ) , ( \begin{matrix}−2 \\ −1 \\ 3 \end{matrix} ) , ( \begin{matrix}1 \\ 2 \\ 1 \end{matrix} )\} .

A basis for Nul(A) comes from parametrizing the solutions of Ax = 0 with the free variables x_{3}, x_{5} as usual:

\{( \begin{matrix}−2/5 \\ −1/5 \\ 1 \\ 0 \\ 0 \end{matrix} ) , ( \begin{matrix}−4/5 \\ 3/5 \\ 0 \\ 0 \\ 1 \end{matrix} )\}

III. (10) Use Cramer’s Rule to solve the system

\begin{aligned}3x_{1} + 2x_{2} & = 9 \\ −8x_{1} + 3x_{2} & = −12 \end{aligned}

Solution: We have

x_{1} = \begin{matrix}det & ( \begin{matrix}9 & 2 \\ −12 & 3 \end{matrix} ) \\ det & ( \begin{matrix}3 & 2 \\ −8 & 3 \end{matrix} ) \end{matrix} = 51/25

and

x_{2} = \begin{matrix}det & ( \begin{matrix}3 & 9 \\ −8 & −12 \end{matrix} ) \\ det & ( \begin{matrix}3 & 2 \\ −8 & 3 \end{matrix} ) \end{matrix} = 36/25

IV. Let V = M_{2×2}(R), the vector space of all 2×2 matrices with real entries. Let W = R^{2}.

A) (10) Is T : V → W defined by T (A) = ( \begin{matrix}det(A) \\ det(A^{2}) \end{matrix} ) a linear mapping? Why or why not?

Solution: No, T is not a linear mapping. The easiest way to see this is to note that if we multiply the matrix A by a scalar c, then

T (cA) = ( \begin{matrix}det(cA) \\ det((cA)^{2}) \end{matrix} ) = ( \begin{matrix}c^{2} det(A) \\ c^{4} det(A^{2}) \end{matrix} ) .

If c ≠ 1, then this is not the same as cT (A). For instance if A = I and c = 2, we get

T (2I) = ( \begin{matrix}4 \\ 16 \end{matrix} ) ≠ 2T (I) = ( \begin{matrix}2 \\ 2 \end{matrix} ) .

B) (10) Show that

H = \{A = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ V : a + b = c + d\}

is a vector subspace of V.

Solution: The zero matrix ( \begin{matrix}0 & 0 \\ 0 & 0 \end{matrix} ) is in H since 0 + 0 = 0 + 0. If A = ( \begin{matrix}a & b \\ c & d \end{matrix} )

and B = ( \begin{matrix}e & f \\ g & h \end{matrix} ) are in H, then a + b = c + d = 0 and e + f = g + h = 0. Then

A + B = ( \begin{matrix}a + e & b + f \\ c + g & d + h \end{matrix} ) and

(a + e) + (b + f) = (a + b) + (e + f) = (c + d) + (g + h) = (c + g) + (d + h)

(using commutativity and associativity of addition in R). Hence A + B ∈ H. Finally,

if A ∈ H and r ∈ R, then rA = ( \begin{matrix}ra & rb \\ rc & rd \end{matrix} ) and ra+rb = r(a+b) = r(c+d) = rc+rd.

Therefore rA ∈ H.

C) (5 Extra Credit) Find a basis for H from part B and determine its dimension.

Solution: In the equation a + b − c − d = 0, b, c, d are free variables. Hence dim H = 3, and a basis consists of

( \begin{matrix}−1 & 1 \\ 0 & 0 \end{matrix} ) , ( \begin{matrix}1 & 0 \\ 1 & 0 \end{matrix} ) , ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) .

V. (True-False) For each true statement, give a short proof. For each false statement, give a counterexample. (Note: As always, “true” means “true in every case.” Partial credit will be given for definitions or theorems that apply even if you don’t find a complete solution.)

A) (10) Let A be a 4 × 4 matrix in which R_{1} + 2R_{2} = R_{3} + R_{4} (R_{i} is row i of the matrix). Then det(A) = 0.

Solution: This is True. For a proof, we will use the fact that “replacement operations” R_{i} ↦ R_{i} +cR_{j} do not change the value of the determinant. Hence if we apply the row operations R_{1} ↦ R_{1} + 2R_{2}, and R_{4} ↦ R_{4} + R_{3}, we will obtain matrix with two equal rows (rows 1 and 4). Hence if we apply another operation such as R_{4} ↦ R_{4} − R_{1}, we obtain a row of zeroes. Expanding the determinant along row 4, we obtain det(A) = 0.

B) (10) There exist linear mappings T : R^{2} → R^{2} with standard matrices A satisfying Nul(A) = Col(A).

Solution: This is True. Since the statement says “there exists such a matrix,” it suffices to produce one. From the equation dim Nul(A) + dim Col(A) = 2, we can see that to get equality, we want both terms = 1. This means that we want a 2 × 2 matrix A whose columns are scalar multiples, and such that Ax = 0 for x = both of its columns. The condition dim Col(A) = 1 means that

A = ( \begin{matrix}a & ca \\ b & cb \end{matrix} )

for some a, b, c. Then

( \begin{matrix}a & ca \\ b & cb \end{matrix} ) ( \begin{matrix}a \\ b \end{matrix} ) = ( \begin{matrix}0 \\ 0 \end{matrix} )

implies that a^{2}+cab = 0 and ab+cb^{2} = 0. There are infinitely many different solutions of these equations. If we take c = −1, for instance, then a = 1, b = 1 gives one:

A = ( \begin{matrix}1 & −1 \\ 1 & −1 \end{matrix} )

is such a matrix, with

Col(A) = Span \{( \begin{matrix}1 \\ 1 \end{matrix} )\} = Nul(A).

(There are infinitely many others too.) C) (10) Let A, B, C be n×n matrices. If det(A^{t}BC^{5}) = 23, then A, B, C are all invertible

matrices. Solution: This is True. From properties of determinants we know

23 = det(A^{t}BC^{5}) = det(A) det(B)(det(C))^{5}.

Since the product is 23 ≠ 0, det(A), det(B), det(C) must all be ≠ 0. This means that A, B, C are all invertible.