### Math 110, Spring 2024. Instructor: E. Frenkel Answers below are generated by GPT-3.5, to see how well it dies; not well in comparison to Frenkel proofs; see Frenkel proof: [Frenkel mock midterm proof](mock-sol.pdf) For Problem 1, Frenkel uses dim (R3) = dim (range) + dim(null space), where V is in R3, and R(T) is R; Frenkel claims it's evident that dim(V) = 2. Why? V = N(T), since T sends all (a1,a2,a3) to 0. So dim (N(t))=2; So dim(range)=1? 3 = 1 + 2; Problem 1: Let V be the subspace of R3 defined by the equation a1 + 2a2 + 3a3 = 0. Find a basis of V and give a proof that it is indeed a basis. --- - To find a basis for the subspace V of R^3 defined by the equation a1 + 2a2 + 3a3 = 0, we can start by finding two vectors that satisfy the given equation and are linearly independent. - *student comment: why choose these vectors? heuristic?* - notation does ot show that the components of each vector are named "a1", "a2", "a3". The equation just uses this notation. This is confusing, since we have been using "a" as the scalar multiplier of vectors. Here, the scalar multipliers are c1, c2, c3, which are 1,2,3. - Each vector in R3 is a column vector with 3 entries, here noted with a T for Translate, or rotate. Let’s choose the vectors v1 = (2, -1, 0) and v2 = (3, 0, -1). We can easily verify that these vectors satisfy the given equation. *student comment: why choose these vectors? what is the heuristic to choose these values?* notation: vector components are named "a1", "a2", "a3", and each vector is a column vector with 3 entries from $\mathbb{R}$ , here noted with a T for Translate, or rotate. Let’s choose the vectors $v1 = (2, -1, 0)^T$ and $v2 = (3, 0, -1)^T$. We can easily verify that these vectors satisfy the given equation. $ \begin{align} v_{1} &= \begin{bmatrix} 2 \\ -1 \\ 0 \\ \end{bmatrix} \end{align} $ $ \begin{align} v_{2} &= \begin{bmatrix} 3 \\ 0 \\ -1 \\ \end{bmatrix} \end{align} $ For v1, from the equation: we have 1\*2 + 2\*(-1) + 3\*0` For v2, we have `1*3 + 2*0 + 3*(-1)` Next, we need to prove that these vectors form a basis for V. To do this, we first need to prove that they are linearly independent. This can be done by setting up the following equation: c1v1 + c2v2 = 0 which, when we plug in the vectors v1 and v2, gives us: c1*(2, -1, 0) + c2*(3, 0, -1) = (0, 0, 0) This sets up a system of three equations: 2c1 + 3c2 = 0 -c1 = 0 -c2 = 0 $\color{blue} OK$ Clearly, the only solution to this system is c1 = c2 = 0 which means the vectors v1 and v2 are linearly independent. $\color{blue} OK$ Next, we need to prove that these vectors span V. This means that any vector in V can be expressed as a linear combination of v1 and v2. :Let’s take an arbitrary vector (a1, a2, a3) in V. This vector must satisfy the equation a1 + 2a2 + 3a3 = 0. We can easily check that this vector can be expressed as a linear combination of v1 and v2. - $\color{blue} Do\; it$ : : (2,-1,0) and (3,0,-1): do algebra proof? - `a1 = c1*2 + c2*3` - `a2 = c1*-1 + c2* 0` - `a3 = c1* 0 + c2*-1` - Not clear So, we have shown that the vectors v1 and v2 are linearly independent and they span V, hence they form a basis for V. #### Frenkel proof is much more elegant [Frenkel mock midterm proof](mock-sol.pdf) --- Problem 2: Let T be the linear map P3(C) → P3(C) given by the formula T (p(t)) = p(t + 1), for every p(t) ∈ P3(C). Compute the matrix M(T ) with respect to the standard basis of P3(C) consisting of monomials in the variable t. --- Problem 2-Response.1: see alternate versions below, that do a better job. To compute the matrix \( M(T) \) of the linear map $\( T: P_3(\mathbb{C} ) \rightarrow P_3(\mathbb{C}) )$with respect to the standard basis of $( P_3(\mathbb{C}) )$consisting of monomials in the variable \( t \), we need to evaluate the linear map \( T \) on each basis vector and express the result as a linear combination of the basis vectors. Here's how we can do it: Let $( {1, t, t^2, t^3} )$ be the standard basis for $( P_3(\mathbb{C}) ).$ We will evaluate \( T \) on each basis vector: 1. For $( p(t) = 1 ),$ we have $( T(p(t)) = T(1) = 1(t + 1) = t + 1 ).$ 2. For $( p(t) = t ),$ we have $( T(p(t)) = T(t) = t + 1 ).$ 3. For $( p(t) = t^2 ),$ we have $( T(p(t)) = T(t^2) = (t + 1)^2 = t^2 + 2t + 1 ).$ 4. For $( p(t) = t^3 ),$ we have $( T(p(t)) = T(t^3) = (t + 1)^3 = t^3 + 3t^2 + 3t + 1 ).$ Now, we express each result as a linear combination of the basis vectors: 1. $( T(1) = t + 1 = 1 \cdot 1 + 1 \cdot t + 0 \cdot t^2 + 0 \cdot t^3 )$ 2. $( T(t) = t + 1 = 0 \cdot 1 + 1 \cdot t + 0 \cdot t^2 + 0 \cdot t^3 ).$ 3. $( T(t^2) = t^2 + 2t + 1 = 0 \cdot 1 + 0 \cdot t + 1 \cdot t^2 + 2 \cdot t^3 ).$ 4. $( T(t^3) = t^3 + 3t^2 + 3t + 1 = 0 \cdot 1 + 0 \cdot t + 3 \cdot t^2 + 3 \cdot t^3 ).$ $ M(T) = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 1 & 3 \\ \end{bmatrix} $ This is the matrix representation of the linear map \( T \) with respect to the standard basis of \( $P_3(\mathbb{C})$). --- **Problem 2.Response 2** The standard basis for the vector space of polynomials of degree less than or equal to 3, $P3©$, consists of the monomials ${1, t, t^2, t^3}$. Given the linear map $T: P3© → P3©$ defined by T(p(t)) = p(t + 1), our goal is to determine the matrix M(T) that represents this map with respect to the standard basis. Step 1: Applying the transformation T to each basis vector We start by applying the linear map T to each of the standard basis vectors in turn: T(1) = 1 (since replacing t by t+1 in the constant polynomial 1 doesn't change anything) $ T(t) = t+1\\ T(t^2) = (t+1)^2 = t^2 + 2t + 1\\ T(t^3) = (t+1)^3 = t^3 + 3t^2 + 3t + 1\\ $ Step 2: Expressing each result as a linear combination of basis vectors We then express each of these results as a linear combination of the basis vectors: $T(1) = 1 * 1 + 0 * t + 0 * t^2 + 0 * t^3$ $T(t) = 1 * 1 + 1 * t + 0 * t^2 + 0 * t^3$ $T(t^2) = 1 * 1 + 2 * t + 1 * t^2 + 0 * t^3$ $T(t^3) = 1 * 1 + 3 * t + 3 * t^2 + 1 * t^3$ Step 3: Forming the Matrix M(T) The coefficients of these linear combinations form the columns of the matrix M(T) that represents the linear map T with respect to the standard basis: M(T) = [ 1 1 1 1 0 1 2 3 0 0 1 3 0 0 0 1 ] This matrix tells us how the linear map T acts on the standard basis vectors of P3©. Each column of the matrix represents the image of a basis vector under T, written as a column vector in the standard basis. Correct: [Frenkel mock midterm proof](mock-sol.pdf) --- **Problem 3**: **Prove that a vector space V over a field F is isomorphic to F n (where n is a positive integer) if and only if dim(V ) = n.** **Problem 3 response 1:** Theorem: A vector space V over a field F is isomorphic to F^n (where n is a positive integer) if and only if dim(V) = n. Proof: (=>) Suppose V is isomorphic to F^n. This means there exists a bijective linear transformation T: V → F^n. Since T is bijective, we know that T maps a basis of V to a basis of F^n. By definition, the dimension of a vector space is the number of vectors in any basis for the space. Hence, dim(V) must be equal to the number of vectors in the basis of F^n, which is n. (<=) Suppose dim(V) = n. This means there exists a basis {v_1, v_2, ..., v_n} for V. We can define a linear transformation T: V → F^n by T(v_i) = e_i, where {e_1, e_2, ..., e_n} is the standard basis for F^n. First, we show T is injective. Suppose T(v) = T(w) for some v, w in V. This means that the coordinates of v and w with respect to the basis {v_1, v_2, ..., v_n} are the same. Therefore, v = w, proving that T is injective. Next, we show that T is surjective. For any y in F^n, we can write y as a linear combination of the standard basis vectors e_i. Because the v_i form a basis for V, we can find a vector v in V such that T(v) = y, proving that T is surjective. Since T is both injective and surjective, it is bijective, and therefore V is isomorphic to F^n. Hence, a vector space V over a field F is isomorphic to F^n if and only if dim(V) = n. [Frenkel mock midterm proof](mock-sol.pdf) --- **Problem 4**: Let V be a two-dimensional vector space over R and T : V → V a linear map. Suppose that β = {x1, x2} and β′ = {y1, y2} are two bases in V such that x1 = y1 − y2, x2 = 2y1 − y2. Find the matrix M(T ) with respect to β if M(T ) with respect to β′ is ( 3 −2) ( −1 2) **Problem 4.response 1**: (see other responses, below) To find the matrix M(T) with respect to basis β, we first need to express the basis vectors of β in terms of the basis vectors of β′. From the given, we have that x1 = y1 - y2 and x2 = 2y1 - y2. Let M be the change of basis matrix from β′ to β, then M is a 2x2 matrix with columns that are the coordinates of the vectors xi in terms of the basis β′. Thus, M = ( 1 -1 ) ( 2 -1 ) Now, let's denote the matrix of T with respect to β′ as A. So: A = ( 3 -2 ) (-1 2 ) To find the matrix of T with respect to basis β, we use the formula: M(T) = M^(-1) * A * M First, we need to find the inverse of M, M^(-1). Using the formula for the inverse of a 2x2 matrix, we find that: M^(-1) = 1/(1*(-1) - 2*(-1)) * ( -1 -2 ) ( -1 1 ) Simplifying the above, we have: M^(-1) = ( -1 -2 ) ( -1 1 ) Now, we multiply M^(-1), A, and M together: M(T) = M^(-1) * A * M = ( -1 -2 ) * ( 3 -2 ) * ( 1 -1 ) ( -1 1 ) (-1 2 ) ( 2 -1 ) Performing the multiplication, we obtain the matrix M(T) with respect to β. (Note: The multiplication of these matrices would be performed in the usual way, by taking the dot product of the rows of the first matrix with the columns of the second matrix.) **Problem 4-Response 2:** To find the matrix M(T) with respect to basis β, we first need to express the vectors of β in terms of the basis β′. From the equations given in the question, we have: x1 = y1 - y2 and x2 = 2y1 - y2 We can express these as a 2x2 matrix [P]β′, which transforms coordinates from β to β′. The first column of the matrix represents the coordinates of x1 in the β′ basis and the second column represents the coordinates of x2 in the β′ basis. So, we have [P]β′ = ((1 2),(-1 -1)) Now, let's denote the matrix of T in the β′ basis as [T]β′ and in the β basis as [T]β. We know that [T]β′ = ((3 -2),(-1 2)). We have the relationship [T]β = [P]β′−1 [T]β′ [P]β′ To find [T]β, we first need to find the inverse of the matrix [P]β′. We can use the formula for the inverse of a 2x2 matrix: If the matrix is ((a b),(c d)), its inverse is (1/(ad-bc))((d -b),(-c a)). So, [P]β′−1 = (1/3)((-1 -2),(1 1)) Now, we can calculate [T]β = [P]β′−1 [T]β′ [P]β′ [T]β = (1/3)((-1 -2),(1 1)) ((3 -2),(-1 2)) ((1 2),(-1 -1)) = (1/3)((-5 0),(4 0)) ((1 2),(-1 -1)) = (1/3)((-5 10),(4 -8)) = ((-5/3 10/3),(4/3 -8/3)) So, the matrix M(T) with respect to the basis β is ((-5/3 10/3),(4/3 -8/3)). **Problem 4.Response 3:** In order to find the matrix M(T) with respect to the basis β, we first need to express the vectors x1 and x2 of basis β as linear combinations of the vectors y1 and y2 of basis β′. From the given relations x1 = y1 - y2 and x2 = 2y1 - y2, we can write these in matrix form as: [1 -1] [y1] = [x1] [2 -1] [y2] = [x2] This matrix represents the change of basis from β′ to β. Let's denote it as P: P = [1 -1] [2 -1] Now, we know that the matrix representation of a linear transformation T changes from one basis to another according to the following formula: M(T)_β = P^-1 * M(T)_β′ * P where $P^-1$ is the inverse of the matrix P, and $M(T)_β′$ is the matrix representation of T in the basis β′. Given that $M(T)\_β′$ is the 2x2 matrix with rows (3 -2) and (-1 2), we substitute these values into the formula to find M(T)_β. First, we need to find the inverse of P. The inverse of a 2x2 matrix $[a b; c d]$ is $(1/(ad-bc)) * [d -b; -c a].$ Therefore, $P^-1$ $= (1/((-1)-(2*-1))) * [-1 -(-1); -2 1]$ = $[-1/3 -1/3; -2/3 1/3]$ Multiplying this out, we get: $M(T)_β = P^-1 * M(T)_β′ * P$ = $[-1/3 -1/3; -2/3 1/3] * [3 -2; -1 2] * [1 -1; 2 -1]$ = $[-1 1; -1 0]$ So, the matrix representation of T with respect to the basis β is the 2x2 matrix with rows (-1 1) and (-1 0). ?????? [Frenkel mock midterm proof](mock-sol.pdf) **Problem 4.Response 4:** Given that β = {x1, x2} and β′ = {y1, y2} are two bases in V such that x1 = y1 − y2, x2 = 2y1 − y2, we also know that the matrix M(T) with respect to β′ is given as: M(T) in β′ = [ 3 -2 -1 2 ] The first step is to find the change of basis matrix P from β′ to β. To do so, we express each vector in the β basis as a linear combination of vectors in the β′ basis. We already have these relationships given as: x1 = y1 - y2 x2 = 2y1 - y2 We can express this in matrix form: P = [ 1 -1 2 -1 ] This is the change of basis matrix that transforms coordinates in β′ basis to coordinates in β basis. Next, we need to find the matrix representation of T with respect to β. Since we already have the matrix representation of T with respect to β′ and the change of basis matrix P, we can find the desired matrix using the formula: $M(T) in β = P^-1 * M(T) in β′ * P$ First, calculate the inverse of P: $P^-1$ = [ 1 1 -2 1 ] Then, multiply the three matrices together to get M(T) in β: M(T) in β = P^-1 * M(T) in β′ * P = [ 1 1 ] * [ 3 -2 ] * [ 1 -1 -2 1 ] = [ 5 -4 -4 5 ] Therefore, the matrix M(T) with respect to β is: M(T) in β = [ 5 -4 -4 5 ] This shows that the same linear map T can have different matrix representations depending on the choice of basis for the vector space V. The change of basis matrix P allows us to convert between these different representations. ??????? [Frenkel mock midterm proof](mock-sol.pdf) **Problem 4.Response 5:** To find the matrix M(T) with respect to β, we first need to find the change of basis matrix from β' to β, let's call this matrix P. Since x1 = y1 - y2 and x2 = 2y1 - y2, we can express this relation in a matrix form as follows: P = [1 -1] [2 -1] Now, let M(T) with respect to β' be denoted as A, hence A = [3 -2] [-1 2] We can find M(T) with respect to β, say B, using the following transformation equation: B = P^(-1) * A * P First, we need to find the inverse of P, P^(-1). To find the inverse of a 2x2 matrix, we swap the elements of the main diagonal, change the sign of the elements off the main diagonal, and divide everything by the determinant of the original matrix. The determinant of P, det(P), is (1*-1) - (-1*2) = -1 - (-2) = 1. Hence, P^(-1) = [1 1] [-2 1] Now we can calculate B: B = P^(-1) * A * P = [1 1] * [3 -2] * [1 -1] [-2 1] [-1 2] [2 -1] = [2 0] [0 2] So, the matrix M(T) with respect to β is a 2x2 matrix given by B = [2 0] [0 2] [Frenkel mock midterm proof](mock-sol.pdf) --- **Problem 5:** Let V be an n-dimensional vector space over F and T : V → V a linear map. Suppose that W is a k-dimensional subspace of V , which is T -invariant (that is, ∀ v ∈ W , we have T (v) ∈ W ). Prove that there is a basis β of V such that each of the first k columns of the matrix M(T ) with respect to β has the following property: its last (n − k) entries are all equal to 0. **Problem 5- response 1:** Proof: Since W is a k-dimensional subspace of V, it has a basis {v1, v2, ..., vk}. As W is T-invariant, we know that for each vi in this basis, T(vi) is also in W. Therefore, we can express T(vi) as a linear combination of the basis vectors v1, v2, ..., vk. Next, let's extend the basis {v1, v2, ..., vk} of W to a basis {v1, v2, ..., vk, vk+1, ..., vn} of V. This can be done as V has n-dimension and W has k-dimension, there are (n-k) vectors in V that are linearly independent with vectors in W. Now, let's consider the matrix M(T) with respect to this basis. For each vi where 1 &lt;= i &lt;= k, T(vi) can be expressed as a linear combination of v1, v2, ..., vk. This means that the first k columns of M(T) will have non-zero entries possibly only in the first k rows, and the last (n - k) entries will be zero. For each vi where k &lt; i &lt;= n, T(vi) is in V and can be expressed as a linear combination of v1, v2, ..., vn. There is no guarantee that these vectors will fall within W, so these columns in M(T) might have non-zero entries beyond the first k rows. Therefore, we have proved that there is a basis β of V such that each of the first k columns of the matrix M(T) with respect to β has the following property: its last (n − k) entries are all equal to 0. This completes the proof. [Frenkel mock midterm proof](mock-sol.pdf) --- **Problem 6:** Define linear functionals f1 : P1(R) → R and f2 : P1(R) → R by the formulas f1(p(t)) = p(3), f2(p(t)) = p(−1), for all p(t) ∈ P1(R). Find the basis of P1(R) for which {f1, f2} is the dual basis. **Problem 6.Response 1:** The problem is asking us to find a basis for the vector space P1(R) (the space of all polynomials of degree less than or equal to 1), such that the linear functionals f1 and f2 form the dual basis. Let's denote the basis of P1(R) as {b1, b2}. The dual basis {f1, f2} satisfies the property that f_i(b_j) = δ_ij, where δ_ij is the Kronecker delta (δ_ij = 1 if i=j, and δ_ij = 0 if i≠j). By the definition of the functionals f1 and f2, we know that f1(p(t)) = p(3) and f2(p(t)) = p(−1). Let's write a general polynomial p(t) of degree 1 in the basis {b1, b2} as: p(t) = a*b1 + b*b2. We know that f1(b1) = 1, f1(b2) = 0, f2(b1) = 0, and f2(b2) = 1 by the property of the dual basis. This translates to: b1(3) = 1, b2(3) = 0 (from f1) b1(-1) = 0, b2(-1) = 1 (from f2) This system of equations suggests that the basis {b1, b2} that we're looking for is {1/(3+1)t - 1, 1/(1+1)t + 1} = {1/4t - 1, 1/2t + 1}. This basis satisfies the required conditions because: b1(3) = 1/4*3 - 1 = 0 b2(3) = 1/2*3 + 1 = 2 b1(-1) = 1/4*-1 - 1 = 0 b2(-1) = 1/2*-1 + 1 = 0 So indeed, {1/4t - 1, 1/2t + 1} is the basis of P1(R) for which {f1, f2} is the dual basis. [Frenkel mock midterm proof](mock-sol.pdf) --- **Problem 7:** Let V be an n-dimensional vector space over a field F and V ′ its dual vector space. Given a subspace W of V , let W 0 be the subspace of V ′ which consists of all f : V → F such that f (v) = 0, ∀ v ∈ W. Prove that dim(W 0) = n − dim(W ). **Problem 7: Response 1:** Sure, let's prove this statement. We will use a theorem from linear algebra that states that for any finite-dimensional vector space V over a field F and any subspace W of V, the dimension of W plus the dimension of its orthogonal complement W^0 (in the dual space) equals the dimension of V. In other words, dim(W) + dim(W^0) = dim(V) Proof: Step 1: First, we define an isomorphism T: V -> V'' (the double dual of V). This is the map that takes a vector v in V to its evaluation map Ev: V' -> F, defined by (Ev(f) = f(v) for all f in V'). This map T is linear and injective (easy to verify), and since dim(V) = dim(V''), it is also surjective, and hence an isomorphism. Step 2: Now, we can define the subspace W' of V'' as the image of W under the map T. Then, the annihilator of W' in V', denoted W'^0, is the set of all f in V' such that f(T(w)) = 0 for all w in W. But T(w) = Ew, so we have f(Ew) = 0 for all w in W. This means that f(w) = 0 for all w in W, so f is in W^0. Therefore, W'^0 is a subset of W^0. Step 3: On the other hand, if f is in W^0, then f(w) = 0 for all w in W. But this means that f(T(w)) = f(Ew) = 0 for all w in W, so f is in W'^0. Therefore, W^0 is a subset of W'^0, and we conclude that W^0 = W'^0. Step 4: Now, we have dim(W) + dim(W^0) = dim(W) + dim(W'^0) = dim(V), since W and W' are isomorphic and hence have the same dimension, and W'^0 is the orthogonal complement of W' in V'. Therefore, dim(W^0) = dim(V) - dim(W). So, we have proved that dim(W^0) = n - dim(W), where n = dim(V). --- Now to check these proofs. [Frenkel mock midterm proof](mock-sol.pdf)