IIT Madras
Mathematics 2
Graded Assignment 6

Graded Assignment 6

Exercise Questions ❓

alt text alt text alt text alt text

alt text alt text alt text alt text

Exercise Solutions

Here is a detailed breakdown of each question, including the core concepts and step-by-step solutions.

❓ Question 1

  1. A function $T: V \to W$ between two vector spaces $V$ and $W$ is said to be a linear transformation if the following conditions hold:

Condition 1: $T(v_1 + v_2) = T(v_1) + T(v_2)$ for all $v_1, v_2 \in V$. Condition 2: $T(cv) = cT(v)$ for all $v \in V$ and $c \in \mathbb{R}$.

Consider the following function: $T: \mathbb{R}^2 \to \mathbb{R}$ $T(x, y) = \begin{cases} 3x & \text{if } y = 0 \ 4y & \text{if } y \neq 0 \end{cases}$

Which of the following statements is true?

  • Condition 1 holds.
  • Condition 1 does not hold.
  • Condition 2 holds.
  • Condition 2 does not hold.

💡 Concepts Involved

  • Linear Transformation: A function between vector spaces is linear if it preserves the operations of vector addition (additivity, Condition 1) and scalar multiplication (homogeneity, Condition 2).
  • Proving vs. Disproving: To prove a condition holds, you must show it’s true for all possible general cases. To disprove a condition, you only need to find one single counterexample.

✅ Detailed Solution

Let’s test both conditions.

Testing Condition 1 (Additivity): To find a counterexample, we should pick vectors that “cross” the function’s piecewise rule (i.e., where one vector has $y=0$ and the other has $y \neq 0$).

  • Let $v_1 = (1, 0)$. Here, $y=0$, so $T(v_1) = T(1, 0) = 3(1) = 3$.

  • Let $v_2 = (0, 1)$. Here, $y \neq 0$, so $T(v_2) = T(0, 1) = 4(1) = 4$.

  • Now, let’s find the sum of their outputs: $T(v_1) + T(v_2) = 3 + 4 = 7$.

  • Next, let’s add the vectors first and then apply the transformation: $v_1 + v_2 = (1, 0) + (0, 1) = (1, 1)$.

  • For the vector $(1, 1)$, $y \neq 0$, so we use the second rule: $T(v_1 + v_2) = T(1, 1) = 4(1) = 4$.

  • Comparison: We see that $T(v_1 + v_2) = 4$ but $T(v_1) + T(v_2) = 7$.

  • Since $4 \neq 7$, Condition 1 is violated.

  • Conclusion: Condition 1 does not hold.

Testing Condition 2 (Homogeneity): We must check $T(cv) = cT(v)$ for all cases. Let $v = (x, y)$.

  • Case 1: $y = 0$

    • $v = (x, 0)$.
    • LHS: $T(cv) = T(c(x, 0)) = T(cx, 0)$. Since the y-coordinate is 0, $T(cx, 0) = 3(cx) = 3cx$.
    • RHS: $cT(v) = cT(x, 0) = c(3x) = 3cx$.
    • LHS = RHS. The condition holds for this case.

  • Case 2: $y \neq 0$

    • $v = (x, y)$.
    • RHS: $cT(v) = cT(x, y) = c(4y) = 4cy$.
    • LHS: $T(cv) = T(c(x, y)) = T(cx, cy)$.
      • Subcase 2a: $c \neq 0$. If $c \neq 0$, then $cy \neq 0$. So we use the second rule: $T(cx, cy) = 4(cy) = 4cy$. This matches the RHS.
      • Subcase 2b: $c = 0$. If $c = 0$, then $cy = 0$. So we must use the first rule: $T(cx, cy) = T(0, 0) = 3(0) = 0$. The RHS is $cT(v) = 0 \cdot (4y) = 0$. This also matches the RHS.

  • Since the condition $T(cv) = cT(v)$ holds for all possible cases, Condition 2 is satisfied.

  • Conclusion: Condition 2 holds.

Final Answer: The correct statements are “Condition 1 does not hold” and “Condition 2 holds”.

❓ Question 2

  1. Suppose the matrix representation of a linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^3$ with respect to ordered bases $\beta = {(1, 0, 1), (0, 1, 0), (0, 0, 1)}$ for the domain and $\gamma = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}$ for the range, is $I_{3 \times 3}$. Let $A$ denote the matrix representation of $T$ with respect to the standard ordered basis ($E$) of $\mathbb{R}^3$ for both domain and range. Which of the following are true?
  • $A = I_{3 \times 3}$
  • $A$ is a singular matrix
  • $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 1 & 0 & -1 \end{bmatrix}$
  • $\det(A) = 1$
  • $\det(A) = -1$
  • $A = \begin{bmatrix} 0 & 0 & 1 \ 0 & 1 & 0 \ -1 & 0 & 1 \end{bmatrix}$

💡 Concepts Involved

  • Matrix Representation: A matrix $[T]_{\beta}^{\gamma}$ describes how $T$ transforms basis vectors from $\beta$ into coordinates of $\gamma$.
  • Standard Basis: The standard basis for $\mathbb{R}^3$ is $E = {e_1, e_2, e_3} = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}$.
  • Change of Basis: The easiest way to solve this is often to use the definition directly, rather than complex change-of-basis formulas.

✅ Detailed Solution

  1. Understand the Given Information:

    • The basis for the domain is $\beta = {v_1, v_2, v_3} = {(1, 0, 1), (0, 1, 0), (0, 0, 1)}$.
    • The basis for the range is $\gamma = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}$. Notice that $\gamma$ is just the standard basis $E$.
    • The matrix representation is $[T]_{\beta}^{\gamma} = I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$.

  2. Interpret the Matrix: The columns of $[T]_{\beta}^{\gamma}$ tell us what $T$ does to the $\beta$ vectors, expressed in $\gamma$ coordinates.

    • Column 1: $T(v_1) = 1\gamma_1 + 0\gamma_2 + 0\gamma_3 = \gamma_1 = (1, 0, 0)$.
    • Column 2: $T(v_2) = 0\gamma_1 + 1\gamma_2 + 0\gamma_3 = \gamma_2 = (0, 1, 0)$.
    • Column 3: $T(v_3) = 0\gamma_1 + 0\gamma_2 + 1\gamma_3 = \gamma_3 = (0, 0, 1)$.

    So, we know the action of $T$ on the $\beta$ basis vectors:

    • $T(1, 0, 1) = (1, 0, 0)$
    • $T(0, 1, 0) = (0, 1, 0)$
    • $T(0, 0, 1) = (0, 0, 1)$

  3. Find the Standard Matrix A: We want to find $A = [T]_{E}^{E}$. The columns of $A$ are $T(e_1)$, $T(e_2)$, and $T(e_3)$.

    • $e_1 = (1, 0, 0)$

    • $e_2 = (0, 1, 0)$

    • $e_3 = (0, 0, 1)$

    • Column 2: We already know $T(e_2) = T(0, 1, 0) = (0, 1, 0)$.

    • Column 3: We already know $T(e_3) = T(0, 0, 1) = (0, 0, 1)$.

    • Column 1: We need to find $T(e_1) = T(1, 0, 0)$. We can express $e_1$ using the $\beta$ basis vectors.

      • Notice that $v_1 = (1, 0, 1) = (1, 0, 0) + (0, 0, 1) = e_1 + e_3$.
      • Therefore, $e_1 = v_1 - e_3$.
      • Since $T$ is a linear transformation, we can write: $T(e_1) = T(v_1 - e_3) = T(v_1) - T(e_3)$
      • Substitute the values we know: $T(e_1) = (1, 0, 0) - (0, 0, 1) = (1, 0, -1)$.

  4. Assemble the Matrix A:

    • Column 1: $T(e_1) = (1, 0, -1)$
    • Column 2: $T(e_2) = (0, 1, 0)$
    • Column 3: $T(e_3) = (0, 0, 1)$ $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ -1 & 0 & 1 \end{bmatrix}$

  5. Check the Options:

    • $A = I_{3 \times 3}$? False.
    • $A$ is singular? Let’s find the determinant. $\det(A) = 1(1 \cdot 1 - 0 \cdot 0) - 0 + 0 = 1$. Since $\det(A) \neq 0$, $A$ is non-singular. False.
    • $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 1 & 0 & -1 \end{bmatrix}$? False. (Sign error).
    • $\det(A) = 1$? True.
    • $\det(A) = -1$? False.
    • $A = \begin{bmatrix} 0 & 0 & 1 \ 0 & 1 & 0 \ -1 & 0 & 1 \end{bmatrix}$? False.

Final Answer: The only true statement is $\det(A) = 1$.

❓ Question 3

  1. Consider a linear transformation $S: M_2(\mathbb{R}) \to M_2(\mathbb{R})$ such that $S(A) = A^T$. Let $B$ be the matrix representation of $S$ with respect to the ordered bases: ${ \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \ 0 & 1 \end{bmatrix} }$ of $M_2(\mathbb{R})$. Choose the set of correct options:
  • The order of the matrix $B$ is $2 \times 2$.
  • The order of the matrix $B$ is $4 \times 4$.
  • The dimension of the row space of the matrix $B$ is 4.
  • The dimension of the column space of the matrix $B$ is 3.
  • The nullity of the matrix $B$ is 1.
  • The rank of the matrix $B$ is 4.
  • $S$ is surjective.

💡 Concepts Involved

  • Vector Space of Matrices: $M_2(\mathbb{R})$ is the vector space of all $2 \times 2$ matrices. The standard basis is given in the problem.
  • Dimension: The dimension of $M_2(\mathbb{R})$ is 4, as it takes 4 independent numbers to define a $2 \times 2$ matrix.
  • Matrix Representation: Since the domain and codomain are 4-dimensional, the matrix $B$ representing the transformation $S$ must be $4 \times 4$.
  • Rank-Nullity Theorem: For a transformation $T: V \to W$, $\text{rank}(T) + \text{nullity}(T) = \text{dim}(V)$. The rank is the dimension of the image (row space/column space), and the nullity is the dimension of the kernel.
  • Surjective: A transformation is surjective (onto) if its rank equals the dimension of the codomain.

✅ Detailed Solution

  1. Determine the Order of B:

    • The vector space is $M_2(\mathbb{R})$, which has dimension 4.
    • The transformation $S$ maps from a 4-dimensional space to a 4-dimensional space.
    • Therefore, its matrix representation $B$ must be $4 \times 4$.
    • ✅ “The order of the matrix B is $4 \times 4$” is True.
    • ❌ “The order of the matrix B is $2 \times 2$” is False.

  2. Find the Matrix B: Let the basis be $\beta = {v_1, v_2, v_3, v_4}$, where:

    • $v_1 = \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix}$
    • $v_2 = \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix}$
    • $v_3 = \begin{bmatrix} 0 & 0 \ 1 & 0 \end{bmatrix}$
    • $v_4 = \begin{bmatrix} 0 & 0 \ 0 & 1 \end{bmatrix}$

    We find the columns of $B$ by applying $S(A) = A^T$ to each basis vector:

    • Col 1: $S(v_1) = S\left(\begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix}\right) = \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix}^T = \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix} = \mathbf{1}v_1 + \mathbf{0}v_2 + \mathbf{0}v_3 + \mathbf{0}v_4 \implies \text{Col 1} = (1, 0, 0, 0)^T$
    • Col 2: $S(v_2) = S\left(\begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix}\right) = \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix}^T = \begin{bmatrix} 0 & 0 \ 1 & 0 \end{bmatrix} = \mathbf{0}v_1 + \mathbf{0}v_2 + \mathbf{1}v_3 + \mathbf{0}v_4 \implies \text{Col 2} = (0, 0, 1, 0)^T$
    • Col 3: $S(v_3) = S\left(\begin{bmatrix} 0 & 0 \ 1 & 0 \end{bmatrix}\right) = \begin{bmatrix} 0 & 0 \ 1 & 0 \end{bmatrix}^T = \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix} = \mathbf{0}v_1 + \mathbf{1}v_2 + \mathbf{0}v_3 + \mathbf{0}v_4 \implies \text{Col 3} = (0, 1, 0, 0)^T$
    • Col 4: $S(v_4) = S\left(\begin{bmatrix} 0 & 0 \ 0 & 1 \end{bmatrix}\right) = \begin{bmatrix} 0 & 0 \ 0 & 1 \end{bmatrix}^T = \begin{bmatrix} 0 & 0 \ 0 & 1 \end{bmatrix} = \mathbf{0}v_1 + \mathbf{0}v_2 + \mathbf{0}v_3 + \mathbf{1}v_4 \implies \text{Col 4} = (0, 0, 0, 1)^T$

    So, $B = \begin{bmatrix} 1 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \end{bmatrix}$.

  3. Analyze Matrix B and Transformation S:

    • Rank: The columns of $B$ are the standard basis vectors for $\mathbb{R}^4$ (just reordered). They are clearly linearly independent. Therefore, the rank is 4.
      • The rank is the dimension of the column space and the row space.
      • ✅ “The dimension of the row space … is 4” is True.
      • ✅ “The rank of the matrix B is 4” is True.
      • ❌ “The dimension of the column space … is 3” is False.
    • Nullity: By the Rank-Nullity Theorem: $\text{rank}(B) + \text{nullity}(B) = \text{dim}(\text{domain}) = 4$.
      • $4 + \text{nullity}(B) = 4 \implies \text{nullity}(B) = 0$.
      • ❌ “The nullity of the matrix B is 1” is False.
    • Surjectivity: $S$ is surjective if $\text{rank}(S) = \text{dim}(\text{codomain})$.
      • $\text{rank}(S) = \text{rank}(B) = 4$.
      • $\text{dim}(M_2(\mathbb{R})) = 4$.
      • Since $4 = 4$, the transformation $S$ is surjective.
      • ✅ "$S$ is surjective" is True.

Final Answer: The correct options are:

  • The order of the matrix $B$ is $4 \times 4$.
  • The dimension of the row space of the matrix $B$ is 4.
  • The rank of the matrix $B$ is 4.
  • $S$ is surjective.

❓ Question 4

  1. Let $T: \mathbb{R}^n \to \mathbb{R}^m$ be a linear transformation and let ${v_1, v_2, \dots, v_n}$ be a basis of $\mathbb{R}^n$. Suppose ${T(v_1), T(v_2), \dots, T(v_n)}$ is a spanning set for $\mathbb{R}^m$. Choose all the correct statements.
  • $T$ is a surjective linear transformation.
  • $T$ is a linear isomorphism.
  • $m \le n$.
  • For any basis ${w_1, w_2, \dots, w_n}$ of $\mathbb{R}^n$, the set ${T(w_1), T(w_2), \dots, T(w_n)}$ is a spanning set for $\mathbb{R}^m$.
  • The matrix of the linear transformation $T$ has rank equal to $n$.

💡 Concepts Involved

  • Image (Range): The image of $T$, denoted $\text{Im}(T)$, is the set of all possible outputs of $T$. $\text{Im}(T)$ is always spanned by the image of a basis of the domain. So, $\text{Im}(T) = \text{span}{T(v_1), \dots, T(v_n)}$.
  • Surjective (Onto): $T$ is surjective if its image equals its codomain. $\text{Im}(T) = \mathbb{R}^m$.
  • Rank: $\text{rank}(T) = \text{dim}(\text{Im}(T))$.
  • Spanning Set Property: A set of $k$ vectors can only span a space of dimension at most $k$.

✅ Detailed Solution

  1. Analyze the Given Statement:

    • We are given that $\text{span}{T(v_1), \dots, T(v_n)} = \mathbb{R}^m$.
    • From the concept above, we also know $\text{Im}(T) = \text{span}{T(v_1), \dots, T(v_n)}$.
    • Therefore, $\text{Im}(T) = \mathbb{R}^m$.

  2. Check the Options:

    • "$T$ is a surjective linear transformation."
      • This is the definition of surjective: the image equals the codomain.
      • True.
    • "$T$ is a linear isomorphism."
      • An isomorphism must be surjective and injective (one-to-one).
      • $T$ is injective if $\text{nullity}(T) = 0$.
      • We know $\text{rank}(T) = \text{dim}(\text{Im}(T)) = \text{dim}(\mathbb{R}^m) = m$.
      • By Rank-Nullity: $m + \text{nullity}(T) = n$.
      • So, $\text{nullity}(T) = n - m$. This is only 0 if $n=m$.
      • If $T: \mathbb{R}^3 \to \mathbb{R}^2$ (e.g., $T(x,y,z)=(x,y)$), $n=3, m=2$. $T$ is surjective, but $\text{nullity} = 3-2=1$, so it’s not injective.
      • Therefore, $T$ is not necessarily an isomorphism. False.
    • "$m \le n$."
      • We have a set of $n$ vectors ${T(v_1), \dots, T(v_n)}$ that spans $\mathbb{R}^m$.
      • The dimension of $\mathbb{R}^m$ is $m$.
      • A fundamental theorem of linear algebra states that to span a space of dimension $m$, you need at least $m$ vectors.
      • Since our spanning set has $n$ vectors, $n$ must be greater than or equal to $m$.
      • True.
    • “For any basis ${w_1, \dots, w_n}$ of $\mathbb{R}^n$, the set ${T(w_1), \dots, T(w_n)}$ is a spanning set for $\mathbb{R}^m$.”
      • The image of any basis of the domain spans the image of $T$.
      • $\text{span}{T(w_1), \dots, T(w_n)} = \text{Im}(T)$.
      • We have already established from the problem’s premise that $\text{Im}(T) = \mathbb{R}^m$.
      • Therefore, $\text{span}{T(w_1), \dots, T(w_n)} = \mathbb{R}^m$.
      • True.
    • “The matrix of the linear transformation $T$ has rank equal to $n$.”
      • The rank of $T$ is $\text{dim}(\text{Im}(T))$.
      • We found that $\text{Im}(T) = \mathbb{R}^m$, so $\text{rank}(T) = \text{dim}(\mathbb{R}^m) = m$.
      • The rank is $m$, not $n$ (unless $m=n$). False.

Final Answer: The correct statements are:

  • $T$ is a surjective linear transformation.
  • $m \le n$.
  • For any basis ${w_1, w_2, \dots, w_n}$ of $\mathbb{R}^n$, the set ${T(w_1), T(w_2), \dots, T(w_n)}$ is a spanning set for $\mathbb{R}^m$.

❓ Question 5

  1. Consider the following statements: Statement 1: Consider a linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^4$ such that $T$ is not injective. Then $\text{rank}(T) < 3$. Statement 2: if $T: V \to W$ is… (matrix given)… where $\alpha, \beta, \gamma \in \mathbb{R} \setminus {0}$, then the rank of $T$ is 3. Statement 3: If $T: \mathbb{R}^3 \to \mathbb{R}^4$ … $T(x, y, z) = (2x - z, 3y - 2z, z, 0)$, then ${(-3, 1, 1, 0), (1, -5, 1, 0), (3, 5, -1, 0)}$ is a basis of the image space. Statement 4: If $T: M_2(\mathbb{R}) \to M_2(\mathbb{R})$ … $T(A) = PA$, where $P = \begin{bmatrix} -1 & 1 \ 1 & -1 \end{bmatrix}$, then ${…}$ is a basis of the kernel.

Write down the statement numbers corresponding to the correct statements in increasing order.

💡 Concepts Involved

  • Rank-Nullity Theorem: $\text{rank}(T) + \text{nullity}(T) = \text{dim}(\text{domain})$.
  • Injective: $T$ is injective $\iff \text{Kernel}(T) = {0} \iff \text{nullity}(T) = 0$.
  • Rank of a Matrix: The number of linearly independent columns (or rows).
  • Basis of Image Space: A set of vectors $S$ is a basis for $\text{Im}(T)$ if:
    1. Every vector in $S$ is in $\text{Im}(T)$.
    2. $S$ is linearly independent.
    3. $\text{dim}(\text{span}(S)) = \text{rank}(T)$.
  • Kernel (Null Space): The set of all input vectors $v$ such that $T(v) = 0$.

✅ Detailed Solution

  • Statement 1: True

    • $T: \mathbb{R}^3 \to \mathbb{R}^4$. $\text{dim}(\text{domain}) = 3$.
    • By Rank-Nullity: $\text{rank}(T) + \text{nullity}(T) = 3$.
    • $T$ is not injective, which means $\text{nullity}(T) > 0$.
    • Since nullity must be an integer, $\text{nullity}(T) \ge 1$.
    • $\text{rank}(T) = 3 - \text{nullity}(T)$.
    • Since $\text{nullity}(T) \ge 1$, it follows that $\text{rank}(T) \le 3 - 1 = 2$.
    • If $\text{rank}(T) \le 2$, then $\text{rank}(T) < 3$. The statement is True.

  • Statement 2: True

    • We need the rank of $M = \begin{bmatrix} 0 & \alpha & \gamma \ 1 & 0 & \gamma \ 0 & \beta & 0 \ 0 & \beta & \alpha^2 \end{bmatrix}$. We are given $\alpha, \beta, \gamma \neq 0$.
    • Let’s check if the columns $c_1, c_2, c_3$ are linearly independent. Set $k_1 c_1 + k_2 c_2 + k_3 c_3 = 0$:
      1. $\alpha k_2 + \gamma k_3 = 0$
      2. $k_1 + \gamma k_3 = 0$
      3. $\beta k_2 = 0$
      4. $\beta k_2 + \alpha^2 k_3 = 0$
    • From (3), since $\beta \neq 0$, we must have $k_2 = 0$.
    • Substitute $k_2=0$ into (1): $\alpha(0) + \gamma k_3 = 0 \implies \gamma k_3 = 0$. Since $\gamma \neq 0$, $k_3 = 0$.
    • Substitute $k_3=0$ into (2): $k_1 + \gamma(0) = 0 \implies k_1 = 0$.
    • The only solution is $k_1=k_2=k_3=0$. The three columns are linearly independent.
    • The rank of the matrix is 3. The statement is True.

  • Statement 3: True

    • $T(x, y, z) = (2x - z, 3y - 2z, z, 0)$.
    • First, find a basis for the image. The image is spanned by the columns of the standard matrix:
      • $T(1,0,0) = (2, 0, 0, 0)$
      • $T(0,1,0) = (0, 3, 0, 0)$
      • $T(0,0,1) = (-1, -2, 1, 0)$
    • These three vectors are clearly linearly independent (they form an echelon matrix). Thus, $\text{rank}(T) = 3$.
    • The problem gives a set of 3 vectors and claims it’s a basis. Since we know the dimension is 3, we just need to check if they are (1) in the image and (2) linearly independent.
    • (1) All three vectors $(…, …, …, 0)$ are in the correct form. A quick check (or trust) shows they are combinations of the basis we found. E.g., $(-3, 1, 1, 0) = -T(e_1) + T(e_2) + T(e_3)$. All three vectors are in the image.
    • (2) Are they linearly independent? Let’s check the determinant of the matrix formed by their non-zero components: $\det \begin{bmatrix} -3 & 1 & 3 \ 1 & -5 & 5 \ 1 & 1 & -1 \end{bmatrix} = -3(5-5) - 1(-1-5) + 3(1 - (-5)) = -3(0) - 1(-6) + 3(6) = 0 + 6 + 18 = 24$.
    • Since the determinant is $24 \neq 0$, the vectors are linearly independent.
    • We have 3 linearly independent vectors in a 3-dimensional image space. They form a basis. The statement is True.

  • Statement 4: False

    • $T(A) = PA = 0$. We need to find the kernel.
    • Let $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$.
    • $PA = \begin{bmatrix} -1 & 1 \ 1 & -1 \end{bmatrix} \begin{bmatrix} a & b \ c & d \end{bmatrix} = \begin{bmatrix} -a+c & -b+d \ a-c & b-d \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$
    • This gives the conditions $c = a$ and $d = b$.
    • So, any matrix $A$ in the kernel looks like $A = \begin{bmatrix} a & b \ a & b \end{bmatrix}$.
    • A basis for this kernel is $A = a\begin{bmatrix} 1 & 0 \ 1 & 0 \end{bmatrix} + b\begin{bmatrix} 0 & 1 \ 0 & 1 \end{bmatrix}$.
    • A basis is ${ \begin{bmatrix} 1 & 0 \ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \ 0 & 1 \end{bmatrix} }$.
    • The dimension of the kernel (nullity) is 2.
    • The statement provides a set of three vectors. A basis for a 2-dimensional space cannot have three vectors.
    • The statement is False.

Final Answer: Statements 1, 2, and 3 are correct. In increasing order, the answer is 123.

❓ Question 6

  1. Let $T: \mathbb{R}^3 \to \mathbb{R}^2$ be a linear transformation. The matrix of the linear transformation $T$ with respect to the standard bases for the domain and codomain is given by $\begin{bmatrix} 12 & k & -9 \ 8 & -2 & 2k \end{bmatrix}$ where $k \in \mathbb{R}$. Find the value of $k$ for which $T$ is a linear transformation which is not surjective.

💡 Concepts Involved

  • Surjective (Onto): A transformation $T: \mathbb{R}^3 \to \mathbb{R}^2$ is surjective if its image is all of the codomain $\mathbb{R}^2$.
  • Rank: $T$ is surjective if $\text{rank}(T) = \text{dim}(\mathbb{R}^2) = 2$.
  • Not Surjective: The question asks for $T$ to be not surjective. This means $\text{rank}(T) < 2$.
  • Rank of a Matrix: The rank of the matrix $A = \begin{bmatrix} 12 & k & -9 \ 8 & -2 & 2k \end{bmatrix}$ must be less than 2. Since the matrix is not all zeros, the rank must be 1.
  • Rank 1 Condition: A matrix has rank 1 if and only if all its rows are scalar multiples of each other.

✅ Detailed Solution

  1. For $T$ to be not surjective, the rank of its matrix $A$ must be less than 2. $A = \begin{bmatrix} 12 & k & -9 \ 8 & -2 & 2k \end{bmatrix}$

  2. For $\text{rank}(A) < 2$ (i.e., $\text{rank}(A)=1$), the two rows must be linearly dependent. This means Row 1 is a scalar multiple of Row 2. $R_1 = c \cdot R_2$ $(12, k, -9) = c \cdot (8, -2, 2k)$

  3. This gives a system of three equations:

    1. $12 = c \cdot 8$
    2. $k = c \cdot (-2)$
    3. $-9 = c \cdot (2k)$

  4. Solve for $c$ and $k$:

    • From (1): $c = 12/8 = 3/2$.
    • Substitute $c = 3/2$ into (2): $k = (3/2) \cdot (-2) \implies$ $k = -3$

  5. Verify this value of $k$ using equation (3):

    • LHS: $-9$
    • RHS: $c \cdot (2k) = (3/2) \cdot (2 \cdot -3) = (3/2) \cdot (-6) = -18/2 = -9$.
    • LHS = RHS. The value $k = -3$ is correct.

When $k = -3$, the matrix becomes $\begin{bmatrix} 12 & -3 & -9 \ 8 & -2 & -6 \end{bmatrix}$. You can see that Row 1 is $\frac{3}{2}$ times Row 2. Since the rows are dependent, the rank is 1. If the rank is 1, the image is a 1-dimensional line, which is not the entire 2-dimensional codomain $\mathbb{R}^2$. Thus, $T$ is not surjective.

Final Answer: The value of $k$ is -3.

Here is a detailed breakdown of each question from the images you provided.

❓ Comprehension (Questions 7, 8, and 9)

Comprehension Type Question: Suppose a bread-making machine B makes 6 breads from 2 eggs, 3 (in hundreds) grams of wheat, and 1 (in hundred) grams of sugar. B also makes 8 breads from 3 eggs, 4 (in hundreds) grams of wheat, and 2 (in hundreds) grams of sugar, and 10 breads from 5 eggs, 5 (in hundreds) grams of wheat, and 3 (in hundreds) grams of sugar. Suppose the production of breads is a linear function of the amount of eggs, wheat (in hundreds), and sugar (in hundreds) used as raw ingredients.

We can express this as follows: $T: \mathbb{R}^3 \to \mathbb{R}$ $T(x, y, z) = ax + by + cz$ where the co-ordinates in $\mathbb{R}^3$ denote the number of eggs, amount (in grams) of wheat (in hundreds), and amount (in grams) of sugar (in hundreds) used…

💡 Concepts Involved

  • Linear Transformation: The problem states the output (breads) is a linear function $T(x, y, z) = ax + by + cz$ of the inputs (eggs, wheat, sugar).
  • System of Linear Equations: We can use the given data points to create a system of equations to solve for the unknown constants $a$, $b$, and $c$.

Solution for $T(x, y, z)$

First, let’s find the specific formula for $T$ by solving for $a$, $b$, and $c$.

  • Variables:

    • $x$ = number of eggs
    • $y$ = wheat (in hundreds of grams)
    • $z$ = sugar (in hundreds of grams)

  • Data Points:

    1. $T(2, 3, 1) = 6 \implies 2a + 3b + c = 6$
    2. $T(3, 4, 2) = 8 \implies 3a + 4b + 2c = 8$
    3. $T(5, 5, 3) = 10 \implies 5a + 5b + 3c = 10$

  • Solving the System: Let’s use elimination.

    • Subtract (Eq 1) from (Eq 2): $(3a + 4b + 2c) - (2a + 3b + c) = 8 - 6$ $\implies a + b + c = 2$ (Eq 4)
    • Subtract (Eq 2) from (Eq 3): $(5a + 5b + 3c) - (3a + 4b + 2c) = 10 - 8$ $\implies 2a + b + c = 2$ (Eq 5)
    • Now subtract (Eq 4) from (Eq 5): $(2a + b + c) - (a + b + c) = 2 - 2$ $\implies a = 0$
    • Substitute $a = 0$ into (Eq 4): $0 + b + c = 2 \implies c = 2 - b$
    • Substitute $a = 0$ and $c = 2 - b$ into the original (Eq 1): $2(0) + 3b + (2 - b) = 6$ $2b + 2 = 6$ $2b = 4 \implies$ $b = 2$
    • Find $c$: $c = 2 - b = 2 - 2 \implies$ $c = 0$

The solution is $a = 0, b = 2, c = 0$. Therefore, the linear transformation is: $T(x, y, z) = 2y$

This means the number of breads produced is simply twice the amount (in hundreds of grams) of wheat used. The eggs and sugar have no effect.

❓ Question 7

  1. Choose the correct set of options from the the following. (MSQ)
  • $Nullity(T) = 1$
  • $Rank(T) = 1$
  • $Nullity(T) = 2$
  • $Rank(T) = 2$
  • $Nullity(T) = 3$
  • $Rank(T) = 3$
  • $T$ is neither one to one nor onto.
  • $T$ is one to one but not onto.
  • $T$ is onto but not one to one.
  • $T$ is an isomorphism.

💡 Concepts Involved

  • Rank: The dimension of the image (range) of $T$. The image is the set of all possible outputs.
  • Nullity: The dimension of the null space (kernel) of $T$. The null space is the set of all inputs $(x, y, z)$ that result in an output of 0.
  • Rank-Nullity Theorem: $\text{rank}(T) + \text{nullity}(T) = \text{dim}(\text{domain})$.
  • One-to-one (Injective): $T$ is one-to-one if $\text{nullity}(T) = 0$.
  • Onto (Surjective): $T$ is onto if $\text{rank}(T) = \text{dim}(\text{codomain})$.

✅ Detailed Solution

We use the formula $T(x, y, z) = 2y$, with $T: \mathbb{R}^3 \to \mathbb{R}$.

  1. Rank: The image is the set of all possible values of $2y$. Since $y$ can be any real number, $2y$ can also be any real number. The image is $\mathbb{R}$.

    • $\text{dim}(\text{Image}) = \text{dim}(\mathbb{R}) = 1$.
    • $\text{Rank}(T) = 1$.

  2. Nullity: The null space is the set of inputs $(x, y, z)$ where $T(x, y, z) = 0$.

    • $2y = 0 \implies y = 0$.
    • The null space consists of all vectors of the form $(x, 0, z)$, where $x$ and $z$ can be any real numbers.
    • A basis for this space is ${(1, 0, 0), (0, 0, 1)}$.
    • The dimension of this space is 2.
    • $\text{Nullity}(T) = 2$.
    • (Check: $\text{Rank} + \text{Nullity} = 1 + 2 = 3 = \text{dim}(\mathbb{R}^3)$. It’s correct.)

  3. Properties:

    • One-to-one? $\text{Nullity}(T) = 2$, which is not 0. So, $T$ is not one-to-one. (For example, $T(1,0,0) = 0$ and $T(5,0,3) = 0$).
    • Onto? $\text{Rank}(T) = 1$. The codomain is $\mathbb{R}$, and $\text{dim}(\mathbb{R}) = 1$. Since $\text{Rank} = \text{dim}(\text{codomain})$, $T$ is onto.
    • Conclusion: $T$ is onto but not one-to-one.

Final Answer: The correct options are:

  • $\text{Rank}(T) = 1$
  • $\text{Nullity}(T) = 2$
  • $T$ is onto but not one to one.

❓ Question 8

  1. Choose the set of correct statements. (MSQ)
  • If 4 eggs and 2 (in hundreds) grams of sugar is used, and no wheat is used, then 9 breads are produced.
  • If 4 eggs and 2 (in hundreds) grams of sugar is used, and no wheat is used, then no bread is produced.
  • If only 3 (in hundreds) grams of wheat is used, then 6 breads are produced.
  • If 3 (in hundreds) grams of wheat and 1 (in hundred) grams of sugar is used, and no egg is used, then no bread is produced.
  • If 3 (in hundreds) grams of wheat and 1 (in hundred) grams of sugar is used, and no egg is used, then 6 breads are produced.

✅ Detailed Solution

We test each statement using our formula $T(x, y, z) = 2y$ (Breads = $2 \times$ wheat).

  • Statement 1: Input is $x=4, z=2, y=0$.
    • $T(4, 0, 2) = 2(0) = 0$ breads.
    • The statement says 9 breads. False.
  • Statement 2: Input is $x=4, z=2, y=0$.
    • $T(4, 0, 2) = 2(0) = 0$ breads.
    • The statement says 0 breads. True.
  • Statement 3: Input is $y=3, x=0, z=0$.
    • $T(0, 3, 0) = 2(3) = 6$ breads.
    • The statement says 6 breads. True.
  • Statement 4: Input is $y=3, z=1, x=0$.
    • $T(0, 3, 1) = 2(3) = 6$ breads.
    • The statement says 0 breads. False.
  • Statement 5: Input is $y=3, z=1, x=0$.
    • $T(0, 3, 1) = 2(3) = 6$ breads.
    • The statement says 6 breads. True.

Final Answer: The correct statements are the 2nd, 3rd, and 5th ones.

❓ Question 9

  1. How many breads are produced by the machine from 6 eggs, 10 (in hundreds) grams of wheat, and 5 (in hundreds) grams of sugar? (NAT)

✅ Detailed Solution

We need to calculate $T(x, y, z)$ for the input $(6, 10, 5)$.

  • $x = 6$ (eggs)
  • $y = 10$ (wheat)
  • $z = 5$ (sugar)

Using the formula $T(x, y, z) = 2y$: $T(6, 10, 5) = 2(10) = 20$

Final Answer: 20

❓ Question 10

  1. Let $T$ be a linear transformation from $\mathbb{R}^8$ to $\mathbb{R}^7$. Suppose a basis for the null space of $T$ has 6 vectors. How many linearly independent vectors are needed to form a basis for the range of $T$?

💡 Concepts Involved

  • Rank-Nullity Theorem: For any linear transformation $T: V \to W$, the following relationship holds: $\text{rank}(T) + \text{nullity}(T) = \text{dim}(V)$
  • Nullity: The dimension of the null space (kernel). The problem states a basis for the null space has 6 vectors, so $\text{nullity}(T) = 6$.
  • Rank: The dimension of the range (image). This is what the question is asking for, as it’s the number of linearly independent vectors needed for a basis of the range.
  • Domain: The domain is $V = \mathbb{R}^8$, so $\text{dim}(V) = 8$.

✅ Detailed Solution

  1. Identify the given values:
    • $\text{dim}(\text{domain}) = \text{dim}(\mathbb{R}^8) = 8$
    • $\text{nullity}(T) = 6$
  2. Apply the Rank-Nullity Theorem: $\text{rank}(T) + \text{nullity}(T) = \text{dim}(\text{domain})$ $\text{rank}(T) + 6 = 8$
  3. Solve for the rank: $\text{rank}(T) = 8 - 6 = 2$

The number of linearly independent vectors needed for a basis of the range is the rank.

Final Answer: 2

❓ Question 11

  1. Let $M$ be the set of all skew-symmetric matrices of order 4. Then $M$ forms a vector space under matrix addition and scalar multiplication. Let $T$ be a linear transformation from $M$ to $\mathbb{R}$ defined by $T(A) = 5 \operatorname{tr}(A)$, where $\operatorname{tr}(A) = \text{trace of } A$. What is the nullity of $T$?

💡 Concepts Involved

  • Skew-Symmetric Matrix: A matrix $A$ is skew-symmetric if $A^T = -A$.
  • Trace ($\operatorname{tr}$): The trace of a matrix is the sum of its diagonal elements.
  • Property of Skew-Symmetric Matrices: For any skew-symmetric matrix, all of its diagonal elements must be 0. (Because $a_{ii} = -a_{ii} \implies 2a_{ii} = 0 \implies a_{ii} = 0$).
  • Nullity: The dimension of the kernel (null space) of $T$. The kernel is the set of all inputs $A$ such that $T(A) = 0$.
  • Dimension of $M$: The dimension of the vector space of $n \times n$ skew-symmetric matrices is $\frac{n(n-1)}{2}$.

✅ Detailed Solution

  1. Analyze the Transformation $T(A)$:

    • The domain is $M$, the set of all $4 \times 4$ skew-symmetric matrices.
    • Let $A$ be any matrix in $M$.
    • By definition of a skew-symmetric matrix, all its diagonal elements are 0.
    • Therefore, the trace of $A$ is $\operatorname{tr}(A) = a_{11} + a_{22} + a_{33} + a_{44} = 0 + 0 + 0 + 0 = 0$.
    • The transformation $T$ is $T(A) = 5 \operatorname{tr}(A) = 5(0) = 0$.
    • This means that $T$ is the zero transformation: it maps every input matrix $A$ in $M$ to the output 0.

  2. Find the Nullity:

    • The nullity is the dimension of the null space (kernel).
    • The kernel is the set of all $A \in M$ such that $T(A) = 0$.
    • Since $T(A) = 0$ for all $A \in M$, the kernel is the entire domain $M$.
    • $\text{Kernel}(T) = M$.
    • Therefore, $\text{nullity}(T) = \text{dim}(\text{Kernel}(T)) = \text{dim}(M)$.

  3. Calculate $\text{dim}(M)$:

    • $M$ is the space of $4 \times 4$ skew-symmetric matrices, so $n = 4$.
    • $\text{dim}(M) = \frac{n(n-1)}{2} = \frac{4(4-1)}{2} = \frac{4 \times 3}{2} = \frac{12}{2} = 6$.

  4. Conclusion:

    • $\text{nullity}(T) = \text{dim}(M) = 6$.

Final Answer: 6

❓ Question 12

  1. Let $T: \mathbb{R}^3 \to \mathbb{R}^2$ be defined by $T(x, y, z) = (2x + 7y, 4z)$. Choose the correct options about $T$.
  • The matrix of $T$ is $\begin{bmatrix} 2 & 0 \ 7 & 0 \ 0 & 4 \end{bmatrix}$.
  • A basis for the range of $T$ is ${(2, 2)}$.
  • A basis for the null space of $T$ is ${(1, -2/7, 0)}$.
  • A basis for the range of $T$ is ${(2, 7), (0, 4)}$.
  • A basis for the null space of $T$ is ${(1, -2/7, 0), (0, 0, 4)}$.
  • The matrix of $T$ is $\begin{bmatrix} 2 & 7 & 0 \ 0 & 0 & 4 \end{bmatrix}$.

💡 Concepts Involved

  • Standard Matrix: For $T: \mathbb{R}^n \to \mathbb{R}^m$, the standard matrix $A$ is an $m \times n$ matrix whose columns are the transformations of the standard basis vectors $e_1, e_2, \dots, e_n$.
  • Range (Image Space): The span of the columns of the matrix $A$. Its dimension is the rank.
  • Null Space (Kernel): The set of all input vectors $v$ such that $T(v) = 0$ (or $Av = 0$). Its dimension is the nullity.

✅ Detailed Solution

  1. Find the Standard Matrix:

    • The domain is $\mathbb{R}^3$, codomain is $\mathbb{R}^2$. The matrix will be $2 \times 3$.
    • $e_1 = (1, 0, 0) \implies T(1, 0, 0) = (2(1) + 7(0), 4(0)) = (2, 0)$
    • $e_2 = (0, 1, 0) \implies T(0, 1, 0) = (2(0) + 7(1), 4(0)) = (7, 0)$
    • $e_3 = (0, 0, 1) \implies T(0, 0, 1) = (2(0) + 7(0), 4(1)) = (0, 4)$
    • The matrix $A$ is formed by these column vectors: $A = \begin{bmatrix} 2 & 7 & 0 \ 0 & 0 & 4 \end{bmatrix}$
    • This means “The matrix of $T$ is $\begin{bmatrix} 2 & 7 & 0 \ 0 & 0 & 4 \end{bmatrix}$” is True.
    • And “The matrix of $T$ is $\begin{bmatrix} 2 & 0 \ 7 & 0 \ 0 & 4 \end{bmatrix}$” is False.

  2. Find the Range (Image Space):

    • The range is the span of the columns: $\text{span}{(2, 0), (7, 0), (0, 4)}$.
    • Notice $(7, 0)$ is just $3.5 \times (2, 0)$, so it’s redundant.
    • $\text{Range} = \text{span}{(2, 0), (0, 4)}$.
    • These two vectors are linearly independent and span all of $\mathbb{R}^2$. So, $\text{Rank}(T) = 2$.
    • Any set of 2 linearly independent vectors in $\mathbb{R}^2$ is a basis for the range.
    • Option: “A basis for the range of $T$ is ${(2, 2)}$.” False. This set only has one vector; the dimension is 2.
    • Option: “A basis for the range of $T$ is ${(2, 7), (0, 4)}$.” This set contains two vectors. Are they linearly independent? Yes (one is not a multiple of the other). Since the range is $\mathbb{R}^2$ (dim 2), any 2 independent vectors form a basis. True.

  3. Find the Null Space (Kernel):

    • We need to solve $T(x, y, z) = (0, 0)$.
    • $(2x + 7y, 4z) = (0, 0)$
    • This gives two equations:
      1. $4z = 0 \implies z = 0$
      2. $2x + 7y = 0 \implies 2x = -7y \implies x = -7/2 y$
    • A vector in the null space has the form $(x, y, z) = (-7/2 y, y, 0)$.
    • We can factor out $y$: $y(-7/2, 1, 0)$.
    • A basis for the null space is ${(-7/2, 1, 0)}$. The nullity is 1.
    • Option: “A basis for the null space of $T$ is ${(1, -2/7, 0)}$.”
      • Let’s check if this vector is a multiple of our basis vector.
      • Is $c(-7/2, 1, 0) = (1, -2/7, 0)$?
      • From the second component: $c \cdot 1 = -2/7 \implies c = -2/7$.
      • Check first component: $c \cdot (-7/2) = (-2/7) \cdot (-7/2) = 1$. (It matches).
      • Since this is a non-zero scalar multiple of our basis vector, it is also a valid basis. True.
    • Option: “A basis for the null space of $T$ is ${(1, -2/7, 0), (0, 0, 4)}$.” False. The nullity is 1, so the basis can only have one vector. Also, $T(0, 0, 4) = (0, 16) \neq (0, 0)$.

Final Answer: The correct options are:

  • A basis for the null space of $T$ is ${(1, -2/7, 0)}$.
  • A basis for the range of $T$ is ${(2, 7), (0, 4)}$.
  • The matrix of $T$ is $\begin{bmatrix} 2 & 7 & 0 \ 0 & 0 & 4 \end{bmatrix}$.
Graded Assignment 5Graded Assignment 7