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Lent Term Week 2 (lectures 23 and 24): In this question, you will look at linear transformations, the kernal and range of a linear transformation, and changing bases.

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", "variables": {"k3": {"description": "

This is one of the basis vectors of the kernal of T in part b.

", "definition": "vector(-d/a , 0 , 0 , 1)", "group": "Variables for part b", "templateType": "anything", "name": "k3"}, "A1": {"description": "

This is the matrix for the transformation T in part b.

", "definition": "matrix([-a*g , -b*g , -c*g , -d*g] , [a*f , b*f , c*f , d*f])", "group": "Variables for part b", "templateType": "anything", "name": "A1"}, "l": {"description": "

This is one of the entries of the first vector in the basis C.

", "definition": "random(-5..5 except(0))", "group": "Variables for parts c and d", "templateType": "anything", "name": "l"}, "A_BB": {"description": "

This is the matrix A with respect to basis B for the domain and basis B for the codomain.

", "definition": "matrix([h , j] , [ii , k])", "group": "Variables for parts c and d", "templateType": "anything", "name": "A_BB"}, "k2": {"description": "

This is one of the basis vectors of the kernal of T in part b.

", "definition": "vector(-c/a , 0 , 1 , 0)", "group": "Variables for part b", "templateType": "anything", "name": "k2"}, "g1": {"description": "

This is the first vector of the basis D.

", "definition": "vector(q , r)", "group": "Variables for parts c and d", "templateType": "anything", "name": "g1"}, "t": {"description": "

This is one of the entries of thesecondvector in the basis D. It is chosen such that g1 and g2 really do form a basis (i.e. are linearly independent).

", "definition": "random(-5..5 except(0) except(s*r/q))", "group": "Variables for parts c and d", "templateType": "anything", "name": "t"}, "A_CD": {"description": "

This is the matrix A with respect to basis C for the domain and basis D for the codomain.

", "definition": "M_BD * matrix([h*l + j*m , h*n + j*p] , [ii*l + k*m , ii*n + k*p])", "group": "Variables for parts c and d", "templateType": "anything", "name": "A_CD"}, "k": {"description": "

This will be used in the definition of the transformation T in part c.

", "definition": "random(-5..5 except(0))", "group": "Variables for parts c and d", "templateType": "anything", "name": "k"}, "m": {"description": "

This is one of the entries of the first vector in the basis C.

", "definition": "random(-5..5 except(0))", "group": "Variables for parts c and d", "templateType": "anything", "name": "m"}, "s": {"description": "

This is one of the entries of the second vector in the basis D.

", "definition": "random(-5..5 except(0))", "group": "Variables for parts c and d", "templateType": "anything", "name": "s"}, "T_f2_D": {"description": "

This is the vector T(f2) with respect to the basis D.

", "definition": "M_BD * T_f2", "group": "Variables for parts c and d", "templateType": "anything", "name": "T_f2_D"}, "ii": {"description": "

This will be used in the definition of the transformation T in part c.

", "definition": "random(-5..5 except(0))", "group": "Variables for parts c and d", "templateType": "anything", "name": "ii"}, "f": {"description": "

This will appear in the cartesian equation of the range of T in part b.

", "definition": "random(-5..5 except(0))", "group": "Variables for part b", "templateType": "anything", "name": "f"}, "h": {"description": "

This will be used in the definition of the transformation T in part c.

", "definition": "random(-5..5 except(0))", "group": "Variables for parts c and d", "templateType": "anything", "name": "h"}, "f2": {"description": "

This is the second vector of the basis C.

", "definition": "vector(n , p)", "group": "Variables for parts c and d", "templateType": "anything", "name": "f2"}, "n": {"description": "

This is one of the entries of the second vector in the basis C.

", "definition": "random(-5..5 except(0))", "group": "Variables for parts c and d", "templateType": "anything", "name": "n"}, "j": {"description": "

This will be used in the definition of the transformation T in part c.

", "definition": "random(-5..5 except(0))", "group": "Variables for parts c and d", "templateType": "anything", "name": "j"}, "k1": {"description": "

This is one of the basis vectors of the kernal of T in part b.

", "definition": "vector(-b/a , 1 , 0 , 0)", "group": "Variables for part b", "templateType": "anything", "name": "k1"}, "c": {"description": "

This will appear in the cartesian equation of the kernal of T in part b.

", "definition": "random(-5..5 except(0))", "group": "Variables for part b", "templateType": "anything", "name": "c"}, "d": {"description": "

This will appear in the cartesian equation of the kernal of T in part b.

", "definition": "random(-5..5 except(0))", "group": "Variables for part b", "templateType": "anything", "name": "d"}, "a": {"description": "

This will appear in the cartesian equation of the kernal of T in part b.

", "definition": "random(-5..5 except(0))", "group": "Variables for part b", "templateType": "anything", "name": "a"}, "T_f1_D": {"description": "

This is the vector T(f1) with respect to the basis D.

", "definition": "M_BD * T_f1", "group": "Variables for parts c and d", "templateType": "anything", "name": "T_f1_D"}, "f1": {"description": "

This is the first vector of the basis C.

", "definition": "vector(l , m)", "group": "Variables for parts c and d", "templateType": "anything", "name": "f1"}, "T_f2": {"description": "

This is the vector T(f1).

", "definition": "vector(h*n + j*p , ii*n + k*p)", "group": "Variables for parts c and d", "templateType": "anything", "name": "T_f2"}, "M_BD": {"description": "

This is thetransition matrix from basis B to D.

", "definition": "(1/(q*t - r*s))*matrix([t , -s] , [-r , q])", "group": "Variables for parts c and d", "templateType": "anything", "name": "M_BD"}, "q": {"description": "

This is one of the entries of the first vector in the basis D.

", "definition": "random(-5..5 except(0))", "group": "Variables for parts c and d", "templateType": "anything", "name": "q"}, "r": {"description": "

This is one of the entries of the first vector in the basis D.

", "definition": "random(-5..5 except(0))", "group": "Variables for parts c and d", "templateType": "anything", "name": "r"}, "g2": {"description": "

This is the second vector of the basis D.

", "definition": "vector(s , t)", "group": "Variables for parts c and d", "templateType": "anything", "name": "g2"}, "b": {"description": "

This will appear in the cartesian equation of the kernal of T in part b.

", "definition": "random(-5..5 except(0))", "group": "Variables for part b", "templateType": "anything", "name": "b"}, "p": {"description": "

This is one of the entries of the second vector in the basis C. It is chosen such that f1 and f2 really do form a basis (i.e. are linearly independent).

", "definition": "random(-5..5 except(0) except(n*m/l))", "group": "Variables for parts c and d", "templateType": "anything", "name": "p"}, "g": {"description": "

This will appear in the cartesian equation of the range of T in part b.

", "definition": "random(-5..5 except(0))", "group": "Variables for part b", "templateType": "anything", "name": "g"}, "r1": {"description": "

This is one of the basis vectors of the range of T in part b.

", "definition": "vector(-g/f , 1 )", "group": "Variables for part b", "templateType": "anything", "name": "r1"}, "T_f1": {"description": "

This is the vector T(f1).

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For parts ii and iii, it may be helpful to recall the following:

Suppose $W$ is a subspace of $\\mathbb{R}^n$, and $W$ has $m$ non-redundant equations in its cartesian description (where $m \\leq n$). Then the dimension of $W$ is $n-m$. That is, each equation is one restriction and it reduces the dimension by $1$.

By \"non-redundant\" we mean that there is no equation which can be written in terms of the other $m-1$ equations. If there were such an equation, say the last one, then it would not be an additional restriction and would not reduce the dimension of $W$ any further.

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Suppose $T$ is a linear transformation from a finite-dimensional vector space $V$ to a vector space $W$. We recall that the rank-nullity theorem tells us that $\\text{rank} (T) + \\text{nullity} (T) = \\text{dim} (V)$. That is, $\\text{dim} (R(T)) + \\text{dim}(\\text{Ker} (T)) = \\text{dim} (V)$.

In the following, we will be given information about the kernal and range of a possible linear transformation and we must decide, based on the rank-nullity theorem, whether such a linear transformation can really exist. (Press the \"Show steps\" button if you require a hint).

i) Suppose we are told that $T:\\mathbb{R}^3 \\longrightarrow \\mathbb{R}^3$ has a kernal with basis $\\Bigg\\{ \\begin{pmatrix} 1 \\\\ 1 \\\\ 7 \\end{pmatrix} , \\begin{pmatrix} 3 \\\\ 4 \\\\ 3 \\end{pmatrix} \\Bigg\\}$, and its range has basis $\\Bigg\\{ \\begin{pmatrix} 0 \\\\ 5 \\\\ 1 \\end{pmatrix} \\Bigg\\}$.

In this case, we have $\\text{dim} (R(T)) =$ [[0]] , $\\text{dim}(\\text{Ker} (T)) =$ [[1]] , and $\\text{dim} (V) =$ [[2]] . Hence, we can see that the rank-nullity theorem [[3]] hold here, and so the linear transformation $T$ [[4]] really exist.

ii) Suppose we are told that $T:\\mathbb{R}^3 \\longrightarrow \\mathbb{R}^3$ has $\\text{ker} (T) = \\Bigg\\{ \\begin{pmatrix} x_1 \\\\ x_2 \\\\ x_3 \\end{pmatrix} \\in \\mathbb{R}^3 \\Big| x_1 + x_2 + 7x_3 = 0 \\Bigg\\}$ and $\\text{R} (T) = \\Bigg\\{ \\begin{pmatrix} x_1 \\\\ x_2 \\\\ x_3 \\end{pmatrix} \\in \\mathbb{R}^3 \\Big| 3x_1 + 4x_2 + 3x_3 = 0 \\; \\text{ and } \\; 5x_2 + x_3 = 0 \\Bigg\\}$.

In this case, we have $\\text{dim} (R(T)) =$ [[5]] , $\\text{dim}(\\text{Ker} (T)) =$ [[6]] , and $\\text{dim} (V) =$ [[7]] . Hence, we can see that the rank-nullity theorem [[8]] hold here, and so the linear transformation $T$ [[9]] really exist.

iii) Suppose we are told that $T:\\mathbb{R}^4 \\longrightarrow \\mathbb{R}^2$ has $\\text{ker} (T) = \\Bigg\\{ \\begin{pmatrix} x_1 \\\\ x_2 \\\\ x_3 \\\\ x_4 \\end{pmatrix} \\in \\mathbb{R}^4 \\Big| 2 x_1 + 4 x_2 + x_4 = 0 \\; \\text{ and } \\; x_1 + 4x_3 = 0 \\Bigg\\}$ and $\\text{R} (T) = \\Bigg\\{ \\begin{pmatrix} x_1 \\\\ x_2 \\end{pmatrix} \\in \\mathbb{R}^2 \\Big| 5x_1 + 2x_2 = 0 \\Bigg\\}$.

In this case, we have $\\text{dim} (R(T)) =$ [[10]] , $\\text{dim}(\\text{Ker} (T)) =$ [[11]] , and $\\text{dim} (V) =$ [[12]] . Hence, we can see that the rank-nullity theorem [[13]] hold here, and so the linear transformation $T$ [[14]] really exist.

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Let $\\boldsymbol{c_i}$ respresent the $i^{\\text{th}}$ column of $A$. That is, $A = ( \\boldsymbol{c_1} \\; \\boldsymbol{c_2} \\; \\boldsymbol{c_3} \\; \\boldsymbol{c_4} )$. Note that we are told in the question that $\\boldsymbol{c_1} = \\begin{pmatrix} \\simplify{-{a}*{g}} \\\\ \\simplify{{a}*{f}} \\end{pmatrix}$.

From part ii we know that the range of $T$ must be equal to $\\text{Lin} (\\boldsymbol{r_1})$.
That is, $A \\boldsymbol{x} \\in \\text{Lin} (\\boldsymbol{r_1})$ for all $\\boldsymbol{x} \\in \\mathbb{R}^4$.
This is equivalent to saying that $x_1 \\boldsymbol{c_1} + x_2 \\boldsymbol{c_2} + x_3 \\boldsymbol{c_3} + x_4 \\boldsymbol{c_4} \\in \\text{Lin} (\\boldsymbol{r_1})$ for all $x_1 , x_2 , x_3 , x_4 \\in \\mathbb{R}$.
Hence, we can see that all $\\boldsymbol{c_1} , \\boldsymbol{c_2} , \\boldsymbol{c_3} , \\boldsymbol{c_4}$ must all be multiples of $\\boldsymbol{r_1}$. Notice that the given value of $\\boldsymbol{c_1}$ is such a multiple.
This helps, but we must still use the rest of the information we are given to determine $A$ explicitely.

From part i we know that the kernal of $A$ is $\\text{Lin}\\{ \\boldsymbol{k_1} ,\\boldsymbol{k_2} ,\\boldsymbol{k_3} \\}$.
You should prove to yourself that this is equivalent to saying that $A \\boldsymbol{k_1} = \\boldsymbol{0}$ , $A \\boldsymbol{k_2} = \\boldsymbol{0}$ , and $A \\boldsymbol{k_3} = \\boldsymbol{0}$.
That is, it is equivalent to saying that $( \\boldsymbol{c_1} \\; \\boldsymbol{c_2} \\; \\boldsymbol{c_3} \\; \\boldsymbol{c_4} ) \\boldsymbol{k_1} = \\boldsymbol{0}$ , $( \\boldsymbol{c_1} \\; \\boldsymbol{c_2} \\; \\boldsymbol{c_3} \\; \\boldsymbol{c_4} ) \\boldsymbol{k_2} = \\boldsymbol{0}$ , and $( \\boldsymbol{c_1} \\; \\boldsymbol{c_2} \\; \\boldsymbol{c_3} \\; \\boldsymbol{c_4} ) \\boldsymbol{k_3} = \\boldsymbol{0}$.
By using the explicit values of $\\boldsymbol{k_1} ,\\boldsymbol{k_2} ,\\boldsymbol{k_3}$ that we found in part i, and expanding the equations above, we can obtain equations which relate $\\boldsymbol{c_1} , \\boldsymbol{c_2} , \\boldsymbol{c_3} , \\boldsymbol{c_4}$ to each other. Since we already know $\\boldsymbol{c_1}$, we can obtain $\\boldsymbol{c_2} , \\boldsymbol{c_3} , \\boldsymbol{c_4}$ from these equations.

Suppose we are told that a linear transformation $T: \\mathbb{R}^4 \\longrightarrow \\mathbb{R}^2$ has

$\\text{ker} (T) = \\Bigg\\{ \\begin{pmatrix} x_1 \\\\ x_2 \\\\ x_3 \\\\ x_4 \\end{pmatrix} \\in \\mathbb{R}^4 \\Big| \\simplify[!noLeadingMinus]{{a}*x_1 + {b}*x_2 + {c}*x_3 + {d}*x_4} = 0 \\Bigg\\}$

and

$\\text{R} (T) = \\Bigg\\{ \\begin{pmatrix} x_1 \\\\ x_2 \\end{pmatrix} \\in \\mathbb{R}^2 \\Big| \\simplify[!noLeadingMinus]{{f}*x_1 + {g}*x_2} = 0 \\Bigg\\}$.

i) Consider the cartesian equation of $\\text{ker} (T)$. By letting $x_2 = r$ , $x_3 = s$ , and $x_4 = t$, we can see that the parametric description of $\\text{ker} (T)$ is given by

$\\begin{pmatrix} x_1 \\\\ x_2 \\\\ x_3 \\\\ x_4 \\end{pmatrix} =$ [[0]] $r +$ [[1]] $s +$ [[2]] $t$.

Hence, $B_1 = \\{ \\boldsymbol{k_1} , \\boldsymbol{k_2} , \\boldsymbol{k_3} \\} = \\Bigg\\{$ [[0]] , [[1]] , [[2]] $\\Bigg\\}$ is a basis for $\\text{ker} (T)$.

ii) Now consider the cartesian equation of $\\text{R}(T)$. By letting $x_2 = u$, we can see that the parametric description of $\\text{R}(T)$ is given by

$\\begin{pmatrix} x_1 \\\\ x_2 \\end{pmatrix} =$ [[3]] $u$.

Hence, $B_2 = \\{\\boldsymbol{r_1} \\} = \\Bigg\\{$ [[3]]$\\Bigg\\}$ is a basis for $\\text{R} (T)$.

ii) Suppose further that there is a $2 \\times 4$ real matrix $A$ such that $T (\\boldsymbol{x}) = A \\boldsymbol{x}$ for all $\\boldsymbol{x} \\in \\mathbb{R}^4$, and such that the first column of $A$ is $\\begin{pmatrix} \\simplify{-{a}*{g}} \\\\ \\simplify{{a}*{f}} \\end{pmatrix}$. Use what we have learnt from parts i and ii to determine $A$ explicitely. (If you require a hint then please press the \"Show steps\" button below).

$A=$ [[4]] .

Recall that $A_T$ is equal to the matrix $(T ( \\boldsymbol{e_1}) \\; T(\\boldsymbol{e_2}))$, where $T(\\boldsymbol{e_1})$ and $T(\\boldsymbol{e_2})$ are in vector form.

Consider the linear transformation $T: \\mathbb{R}^2 \\longrightarrow \\mathbb{R}^2$ defined by its effect on the standard basis $B = \\{ \\boldsymbol{e_1} , \\boldsymbol{e_2} \\} = \\bigg\\{ \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} , \\begin{pmatrix} 0 \\\\ 1 \\end{pmatrix} \\bigg\\}$:

$T(\\boldsymbol{e_1}) = (\\var{h}) \\boldsymbol{e_1} + (\\var{ii}) \\boldsymbol{e_2}$
$T(\\boldsymbol{e_2}) = (\\var{j}) \\boldsymbol{e_1} + (\\var{k}) \\boldsymbol{e_2}$

Find the matrix $A_T$ representing the transformation $T$. that is, find the matrix $A_T^{B\\rightarrow B}$. (Press the \"Show steps\" button if you require a hint).

$A_T =$ [[0]] .

We recall that the matrix $A_T^{C \\rightarrow D}$ is given by $A_T^{C \\rightarrow D} = \\Big( \\big( T(\\boldsymbol{f_1}) \\big)_D \\; , \\; \\big( T(\\boldsymbol{f_2}) \\big)_D \\Big)$, where $\\big( T(\\boldsymbol{f_1}) \\big)_D$ and $\\big( T(\\boldsymbol{f_2}) \\big)_D$ are the vectors $T(\\boldsymbol{f_1})$ and $T(\\boldsymbol{f_1})$ represented in terms of the basis $D$, respectively.

We should first find $T(\\boldsymbol{f_1})$ and $T(\\boldsymbol{f_1})$ with respect to the standard basis, and then with respect to the basis $D$. From what we are told in part c, we have that

$T(\\boldsymbol{f_1}) = T((\\var{{l}}) \\boldsymbol{e_1} + (\\var{{m}}) \\boldsymbol{e_2}) =$

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$T(\\boldsymbol{f_2}) =T((\\var{{n}}) \\boldsymbol{e_1} + (\\var{{p}}) \\boldsymbol{e_2}) =$

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Now we wish to express $T(\\boldsymbol{f_1})$ and $T(\\boldsymbol{f_1})$ with respect to the basis $D$. That is, we wish to find $\\lambda_1 , \\lambda_2$ and $\\mu_1 , \\mu_2$ such that $T(\\boldsymbol{f_1}) = \\lambda_1 \\boldsymbol{g_1} + \\lambda_2 \\boldsymbol{g_2}$ and $T(\\boldsymbol{f_2}) =\\mu_1 \\boldsymbol{g_1} +\\mu_2 \\boldsymbol{g_2}$. Then, we would have that $\\big( T(\\boldsymbol{f_1}) \\big)_D = \\begin{pmatrix} \\lambda_1 \\\\ \\lambda_2 \\end{pmatrix}_D$ and $\\big( T(\\boldsymbol{f_2}) \\big)_D = \\begin{pmatrix} \\mu_1 \\\\ \\mu_2 \\end{pmatrix}_D$.

Notice that we can express the equation $T(\\boldsymbol{f_1}) =\\lambda_1\\boldsymbol{g_1} +\\lambda_2\\boldsymbol{g_2}$ as $T(\\boldsymbol{f_1}) = M \\begin{pmatrix} \\lambda_1 \\\\ \\lambda_2 \\end{pmatrix}$, and the equation $T(\\boldsymbol{f_2}) =\\mu_1\\boldsymbol{g_1} +\\mu_2\\boldsymbol{g_2}$ as $T(\\boldsymbol{f_2}) = M \\begin{pmatrix} \\mu_1 \\\\ \\mu_2 \\end{pmatrix}$, where $M = ( \\boldsymbol{g_1} , \\boldsymbol{g_2})$.

If we calculate $M^{-1}$ then it will be easy for us to determine $\\begin{pmatrix} \\lambda_1 \\\\ \\lambda_2 \\end{pmatrix}$ and $\\begin{pmatrix} \\mu_1 \\\\ \\mu_2 \\end{pmatrix}$.

$M^{-1} =$

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Hence, we have that $\\big( T(\\boldsymbol{f_1}) \\big)_D = \\begin{pmatrix} \\lambda_1 \\\\ \\lambda_2 \\end{pmatrix}_D = M^{-1} T(\\boldsymbol{f_1}) =$

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and $\\big( T(\\boldsymbol{f_2}) \\big)_D = \\begin{pmatrix} \\mu_1 \\\\ \\mu_2 \\end{pmatrix}_D = M^{-1}T(\\boldsymbol{f_2}) =$

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Hence, we have that $A_T^{C \\rightarrow D} = \\Big( \\big( T(\\boldsymbol{f_1}) \\big)_D \\; , \\; \\big(T(\\boldsymbol{f_2}) \\big)_D \\Big) =$

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Now consider the bases $C = \\{ \\boldsymbol{f_1} , \\boldsymbol{f_2} \\} = \\bigg\\{ \\begin{pmatrix} \\var{{l}} \\\\ \\var{{m}} \\end{pmatrix} , \\begin{pmatrix} \\var{{n}} \\\\ \\var{{p}} \\end{pmatrix} \\bigg\\}$ and $D = \\{ \\boldsymbol{g_1} , \\boldsymbol{g_2} \\} = \\bigg\\{ \\begin{pmatrix} \\var{{q}} \\\\ \\var{{r}} \\end{pmatrix} , \\begin{pmatrix} \\var{{s}} \\\\ \\var{{t}} \\end{pmatrix} \\bigg\\}$.

Find the matrix $A_T^{C \\rightarrow D}$ representing the same transformation $T$, but with respect to the non-standard basis $C$ for the domain of $T$ and the non-standard basis $D$ for the codomain of $T$. (Press the \"Show steps\" button for a breakdown of this question into steps).

$A_T^{C \\rightarrow D} =$ [[0]] .

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This is the question for Lent Term week 2 of the MA100 course at the LSE. It looks at material from chapters 23 and 24.

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