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Lent Term Week 4 (lectures 27 and 28): In this question you will look at diagonalisablity and orthogonal diagonalisability.

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", "rulesets": {}, "variables": {"P": {"group": "Variables for parts d and e", "templateType": "anything", "description": "

This is the matrix with columns equal to our g1 , g2 , and g3.

", "name": "P", "definition": "matrix([aa , bb , 1] , [1 , 0 , -aa] , [0 , 1 , -bb])"}, "g1": {"group": "Variables for parts d and e", "templateType": "anything", "description": "

This is our first eigenvector.

", "name": "g1", "definition": "vector(aa , 1 , 0)"}, "w2": {"group": "Variables for parts d and e", "templateType": "anything", "description": "

This will be found by the student when applying the Gram-schmidt process to obtain u1 , u2 , u3 from g1 , g2 , g3.

", "name": "w2", "definition": "g2 - dot(g2 , u1)*u1"}, "g3_2_norm": {"group": "Variables for parts d and e", "templateType": "anything", "description": "

This is the absolute value of g3_2.

", "name": "g3_2_norm", "definition": "(dot(g3 , g3))^(1/2)"}, "hh": {"group": "Variables for parts d and e", "templateType": "anything", "description": "

This is the eigenvalue for our third vector.

", "name": "hh", "definition": "(1 + bb^2 + aa^2)*random(4,5,6)"}, "w2_norm": {"group": "Variables for parts d and e", "templateType": "anything", "description": "

This is the absolute value of w2.

", "name": "w2_norm", "definition": "(dot(w2 , w2))^(1/2)"}, "gg": {"group": "Variables for parts d and e", "templateType": "anything", "description": "

This is the eigenvalue for our first two eigenvectors.

", "name": "gg", "definition": "(1 + bb^2 + aa^2)*random(1,2,3)"}, "g": {"group": "Variables for parts a to c", "templateType": "anything", "description": "

This is the eigenvalue of our second eigenvector.

", "name": "g", "definition": "(a*j*d + c*b*k) * (f + random(1..2)) "}, "a": {"group": "Variables for parts a to c", "templateType": "anything", "description": "

This is one of the entries for our first eigenvector.

", "name": "a", "definition": "random(-3..3 except(0))"}, "u2": {"group": "Variables for parts d and e", "templateType": "anything", "description": "

u1 , u2 , and u3 are what we get when we apply the Gram-Schmidt process to the g1 , g2 , and g3.

", "name": "u2", "definition": "(1/(((aa^2)*(bb^2) + 2*(aa^2) + bb^2 + aa^4 + 1 )^(1/2))) * vector(bb , -aa*bb , 1 + aa^2)"}, "f1": {"group": "Variables for parts a to c", "templateType": "anything", "description": "

This is our first eigenvector.

", "name": "f1", "definition": "vector(a , b , 0)"}, "A_T": {"group": "Variables for parts a to c", "templateType": "anything", "description": "

This is the matrix for the transformation T which is defined by T(f1) = f * f1 , T(f2) = g * f2 , T(f3) = h * f3. Note that f,g, and h have been chosen so that they are multiples of the determinant of P_C, and hence A_T = P_C * DD * P_C_inv has integer entries.

", "name": "A_T", "definition": "P_C * DD * P_C_inv"}, "f": {"group": "Variables for parts a to c", "templateType": "anything", "description": "

This is the eigenvalue of our first eigenvector.

", "name": "f", "definition": "0"}, "g3": {"group": "Variables for parts d and e", "templateType": "anything", "description": "

This is our third eigenvector.

", "name": "g3", "definition": "vector(1 , -aa , -bb)"}, "PP_D": {"group": "Variables for parts d and e", "templateType": "anything", "description": "

This is the orthogonal matrix with u1 , u2 , and u3 as the columns.

", "name": "PP_D", "definition": "matrix([aa/((1 + aa^2)^(1/2)) , bb/(((aa^2)*(bb^2) + 2*(aa^2) + bb^2 + aa^4 + 1 )^(1/2)) , 1/((1+ aa^2 +bb^2)^(1/2))] , [1/((1 + aa^2)^(1/2)) , -bb*aa/(((aa^2)*(bb^2) + 2*(aa^2) + bb^2 + aa^4 + 1 )^(1/2)) , -aa/((1+ aa^2 +bb^2)^(1/2))] , [0 , (1+aa^2)/(((aa^2)*(bb^2) + 2*(aa^2) + bb^2 + aa^4 + 1 )^(1/2)) , -bb/((1+ aa^2 +bb^2)^(1/2))])"}, "h": {"group": "Variables for parts a to c", "templateType": "anything", "description": "

This is the eigenvalue of our third eigenvector. Note that |f|<|g|<|h| by definition

", "name": "h", "definition": "(a*j*d + c*b*k) * (f + random(3..4))"}, "AA_T": {"group": "Variables for parts d and e", "templateType": "anything", "description": "

This is our matric A_T for parts d and e. It satisfies A_T g1 = gg * g1 and A_T g2 = hh * g2.

", "name": "AA_T", "definition": "P * DDD * P_inv"}, "P_C": {"group": "Variables for parts a to c", "templateType": "anything", "description": "

This is the transition matrix from basis C to the standard basis B.

", "name": "P_C", "definition": "matrix([a , c , 0] , [b , 0 , j] , [0 , d , k])"}, "N_hh": {"group": "Variables for parts d and e", "templateType": "anything", "description": "

This is the RRE form of A_T - hh * I

", "name": "N_hh", "definition": "matrix([1 , 0 , 1/bb] , [0 , 1 , -aa/bb] , [0 , 0 , 0])"}, "j": {"group": "Variables for parts a to c", "templateType": "anything", "description": "

This is one of the entries for our third eigenvector.

", "name": "j", "definition": "random(-3..3 except(0))"}, "DD": {"group": "Variables for parts a to c", "templateType": "anything", "description": "

This is the diagonal matrix consisting of the eigenvalues in ascending order.

", "name": "DD", "definition": "matrix([f , 0 , 0] , [0 , g , 0] , [0 , 0 , h])"}, "P_C_inv": {"group": "Variables for parts a to c", "templateType": "anything", "description": "

This is the inverse of P_C

", "name": "P_C_inv", "definition": "(1/(a*j*d + c*b*k)) * matrix([j*d , c*k , -c*j] , [b*k , -a*k , a*j] , [-b*d , a*d , b*c])"}, "aa": {"group": "Variables for parts d and e", "templateType": "anything", "description": "

This will be used in the definition of our eigenvectors.

", "name": "aa", "definition": "random(-3..3 except(0))"}, "c": {"group": "Variables for parts a to c", "templateType": "anything", "description": "

This is one of the entries for our second eigenvector.

", "name": "c", "definition": "random(-3..3 except(0))"}, "N_gg": {"group": "Variables for parts d and e", "templateType": "anything", "description": "

This is the RRE form of A_T - gg * I

", "name": "N_gg", "definition": "matrix([1 , -aa , -bb] , [0 , 0 , 0] , [0 , 0 , 0])"}, "d": {"group": "Variables for parts a to c", "templateType": "anything", "description": "

This is one of the entries for our second eigenvector.

", "name": "d", "definition": "random(-3..3 except(0))"}, "DDD": {"group": "Variables for parts d and e", "templateType": "anything", "description": "

This is the diagonal matrix with our eigenvalues for the diagonal entries.

", "name": "DDD", "definition": "matrix([gg , 0 , 0] , [0 , gg , 0] , [0 , 0 , hh])"}, "g2": {"group": "Variables for parts d and e", "templateType": "anything", "description": "

This is our second eigenvector.

", "name": "g2", "definition": "vector(bb , 0 , 1)"}, "u1": {"group": "Variables for parts d and e", "templateType": "anything", "description": "

u1 , u2 , and u3 are what we get when we apply the Gram-Schmidt process to the g1 , g2 , and g3.

", "name": "u1", "definition": "(1/((1+ aa^2 )^(1/2))) * vector(aa , 1 , 0)"}, "b": {"group": "Variables for parts a to c", "templateType": "anything", "description": "

This is one of the entries for our first eigenvector.

", "name": "b", "definition": "random(-3..3 except(0))"}, "g1_norm": {"group": "Variables for parts d and e", "templateType": "anything", "description": "

This is the absolute value of g1.

", "name": "g1_norm", "definition": "(dot(g1 , g1))^(1/2)"}, "u3": {"group": "Variables for parts d and e", "templateType": "anything", "description": "

u1 , u2 , and u3 are what we get when we apply the Gram-Schmidt process to the g1 , g2 , and g3.

", "name": "u3", "definition": "(1/((1+ aa^2 +bb^2)^(1/2))) * vector(1 , -aa , -bb)"}, "k": {"group": "Variables for parts a to c", "templateType": "anything", "description": "

This is one of the entries for ourthird eigenvector.

", "name": "k", "definition": "random(-3..3 except(0) except(-a*d*j/(b*c)))"}, "f3": {"group": "Variables for parts a to c", "templateType": "anything", "description": "

This is our third eigenvector. By our definition of k, this vector is linearly independent to f1 and f2. f1 , f2 , and f3 form a basis for R^3, which we call C.

", "name": "f3", "definition": "vector(0 , j , k)"}, "f2": {"group": "Variables for parts a to c", "templateType": "anything", "description": "

This is our second eigenvector. Note that it is linearly independent from f1.

", "name": "f2", "definition": "vector(c , 0 , d)"}, "bb": {"group": "Variables for parts d and e", "templateType": "anything", "description": "

This will be used in the definition of our eigenvectors.

", "name": "bb", "definition": "random(-3..3 except(0))"}, "P_inv": {"group": "Variables for parts d and e", "templateType": "anything", "description": "

This is the inverse of P.

", "name": "P_inv", "definition": "(1/(1 + bb^2 + aa^2)) * matrix([aa , 1 + bb^2 , -aa*bb] , [bb , -aa*bb , aa^2 + 1] , [1 , -aa , -bb])"}, "g3_2": {"group": "Variables for parts d and e", "templateType": "anything", "description": "

This is a rescaling of our eigenvector g3 so its matches what the student will get when they find it by using the null space of A_T - hh * I.

", "name": "g3_2", "definition": "vector(-1/bb , aa/bb , 1)"}}, "name": "MA100 LT Week 4", "functions": {}, "extensions": [], "parts": [{"showFeedbackIcon": true, "scripts": {}, "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "stepsPenalty": 0, "gaps": [{"showFeedbackIcon": true, "scripts": {}, "mustBeReducedPC": 0, "variableReplacementStrategy": "originalfirst", "notationStyles": ["plain", "en", "si-en"], "allowFractions": false, "minValue": "f", "mustBeReduced": false, "correctAnswerStyle": "plain", "showCorrectAnswer": true, "customMarkingAlgorithm": "", "type": "numberentry", "extendBaseMarkingAlgorithm": true, "variableReplacements": [], "marks": 1, "maxValue": "f", "unitTests": [], "correctAnswerFraction": false}, {"showFeedbackIcon": true, "scripts": {}, "mustBeReducedPC": 0, "variableReplacementStrategy": "originalfirst", "notationStyles": ["plain", "en", "si-en"], "allowFractions": false, "minValue": "g", "mustBeReduced": false, "correctAnswerStyle": "plain", "showCorrectAnswer": true, "customMarkingAlgorithm": "", "type": "numberentry", "extendBaseMarkingAlgorithm": true, "variableReplacements": [], "marks": 1, "maxValue": "g", "unitTests": [], "correctAnswerFraction": false}, {"showFeedbackIcon": true, "scripts": {}, "mustBeReducedPC": 0, "variableReplacementStrategy": "originalfirst", "notationStyles": ["plain", "en", "si-en"], "allowFractions": false, "minValue": "h", "mustBeReduced": false, "correctAnswerStyle": "plain", "showCorrectAnswer": true, "customMarkingAlgorithm": "", "type": "numberentry", "extendBaseMarkingAlgorithm": true, "variableReplacements": [], "marks": 1, "maxValue": "h", "unitTests": [], "correctAnswerFraction": false}], "showCorrectAnswer": true, "unitTests": [], "type": "gapfill", "extendBaseMarkingAlgorithm": true, "variableReplacements": [], "marks": 0, "sortAnswers": false, "prompt": "

Consider the linear transformation $T : \\mathbb{R}^3 \\longrightarrow \\mathbb{R}^3$ defined by $T( \\boldsymbol{v} ) = A_T \\boldsymbol{v}$, for all $\\boldsymbol{v} \\in \\mathbb{R}^3$, where $A_T = \\var{{A_T}}$.

In parts a to c, we will diagonalise the matrix $A_T$. That is, we will find a diagonal matrix $D$ and an invertible matrix $P$ such that $P^{-1} A_T P = D$.

First we will need to find the eigenvalues of $A_T$. Find the eigenvalues $k$, $l$, and $m$ of $A_T$, with $|k| < |l| < |m|$. (Press the \"Show steps\" button if you require a hint).

$k =$ [[0]] .
$l =$ [[1]] .
$m =$ [[2]] .

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Suppose $A$ is an $n \\times n$ matrix. In order to find the eigenvalues of $A$ we must solve the equation $\\det (A - \\lambda I_n) = 0$ for $\\lambda$, where $I_n$ is the $n \\times n$ identity matrix.

That is, we first find the determinant of the matrix $A - \\lambda I_n$. This will be a polynomial of $\\lambda$ with degree $n$. We then determine when this polynomial is equal to zero; i.e. we find its roots. These roots are the eigenvalues.

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Now we will find corresponding eigenvectors to the eigenvalues. Let $\\boldsymbol{f_1}$ be an eigenvector for $k$, $\\boldsymbol{f_2}$ be an eigenvector for $l$, and $\\boldsymbol{f_3}$ be an eigenvector for $m$.

Suppose we are also told that $\\boldsymbol{f_1}$, $\\boldsymbol{f_2}$, $\\boldsymbol{f_3}$ are of the form $\\boldsymbol{f_1} = \\begin{pmatrix} \\var{{a}} \\\\ * \\\\* \\end{pmatrix}$ , $\\boldsymbol{f_2} = \\begin{pmatrix} \\var{{c}} \\\\ * \\\\* \\end{pmatrix}$ , $\\boldsymbol{f_3} = \\begin{pmatrix} * \\\\ * \\\\ \\var{{k}} \\end{pmatrix}$.

Find $\\boldsymbol{f_1}$, $\\boldsymbol{f_2}$, $\\boldsymbol{f_3}$. (If you require a hint, press the \"Show steps\" button).

$\\boldsymbol{f_1} =$ [[0]] .
$\\boldsymbol{f_2} =$ [[1]] .
$\\boldsymbol{f_3} =$ [[2]] .

Suppose we have a matrix $A$ with eigenvalue $\\lambda$, and we want to find an eigenvector, $\\boldsymbol{v}$, of $A$ corresponding to the eigenvalue $\\lambda$.

For this to be the case, $\\boldsymbol{v}$ must satisfy $A \\boldsymbol{v} = \\lambda \\boldsymbol{v}$. This is equivaluent to satisfying $(A - \\lambda I) \\boldsymbol{v} = \\boldsymbol{0}$. That is, $\\boldsymbol{v}$ must be in the null space of $A - \\lambda I$: $\\boldsymbol{v} \\in N(A-\\lambda I)$.

Now, we recall that the null space of $A - \\lambda I$ is the same as the null space of the reduced row echelon form of $A - \\lambda I$; i.e. $N(A-\\lambda I) = N \\Big(\\textbf{RRE} (A-\\lambda I) \\Big)$. It is easier to deal with the reduced row echelon form here, so the first step is to find $\\textbf{RRE} (A-\\lambda I)$.

Once we have found $\\textbf{RRE} (A-\\lambda I)$, we can find $N \\Big(\\textbf{RRE} (A-\\lambda I) \\Big)$ in the same way that we would find the set of solutions to $\\Big(\\textbf{RRE} (A-\\lambda I) \\Big) \\boldsymbol{x} = \\boldsymbol{0}$.

$N(A-\\lambda I)$ (which is equal to $N \\Big(\\textbf{RRE} (A-\\lambda I) \\Big)$) is called the eigenspace of the eigenvalue $\\lambda$, and it contains all eigenvectors of the eigenvalue $\\lambda$.

For the question in above, we just need to find the eigenspaces for the eigenvalues $k$, $l$, and $m$, and then pick an eigenvector $\\boldsymbol{f_1}$, $\\boldsymbol{f_2}$, and $\\boldsymbol{f_3}$ from each eigenspace, respectively, so that they satisfy the conditions given in the question: $\\boldsymbol{f_1}$, $\\boldsymbol{f_2}$, $\\boldsymbol{f_3}$ are of the form $\\boldsymbol{f_1} = \\begin{pmatrix} \\var{{a}} \\\\ * \\\\* \\end{pmatrix}$ , $\\boldsymbol{f_2} = \\begin{pmatrix} \\var{{c}} \\\\ * \\\\* \\end{pmatrix}$ , $\\boldsymbol{f_3} = \\begin{pmatrix}* \\\\ * \\\\ \\var{{k}} \\end{pmatrix}$.

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Now that we have found the eigenvalues of $A_T$ and corresponding eigenvectors, we can find a diagonal matrix $D$ and an invertible matrix $P$ such that $P^{-1} A_T P = D$. As discussed in the lecture notes, this is possible because there is a basis, $C$, of $\\mathbb{R}^3$ which consists of eigenvectors of $A_T$; namely, $C = \\{ \\boldsymbol{f_1} , \\boldsymbol{f_2} , \\boldsymbol{f_3} \\}$. (We know $\\boldsymbol{f_1} , \\boldsymbol{f_2} , \\boldsymbol{f_3}$ are linearly independent because they come from different eigenvalues; due to this, and the fact that they are three vectors, they must also span $\\mathbb{R^3}$, and therefore form a basis for $\\mathbb{R}^3$.) See section 27.1 of the lecture notes for more details.

Given what we have found in parts a and b, we can find $P$ and $D$ immediately, without even doing any further calculations. What are $P$ and $D$? (If you require a hint, please press the \"Show steps\" button below).

$P= $ [[0]] .
$D =$ [[1]] .

Suppose we have an $n \\times n$ matrix $A$ with $n$ linearly independent eigenvectors $\\boldsymbol{v_1} , \\boldsymbol{v_2} , \\ldots , \\boldsymbol{v_n}$ and corresponding eigenvalues $\\lambda_1 , \\lambda_2 , \\ldots , \\lambda_n$ (i.e. these eigenvectors form a basis of $\\mathbb{R}^n$).

Let $P = (\\boldsymbol{v_1} \\; \\boldsymbol{v_2} \\; \\ldots \\; \\boldsymbol{v_n})$ and let $D$ be the $n \\times n$ diagonal matrix with $\\lambda_i$ in the $(i,i)$ entry. Then, $P^{-1} A P = D$.

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Now consider a new linear transformation $T : \\mathbb{R}^3 \\longrightarrow \\mathbb{R}^3$ defined by $T( \\boldsymbol{v} ) = A_T \\boldsymbol{v}$, for all $\\boldsymbol{v} \\in \\mathbb{R}^3$, where $A_T = \\var{{AA_T}}$. We are also told that the eigenvalues of $A_T$ are $\\var{{gg}}$ and $\\var{{hh}}$.

In parts d to f, we will orthogonally diagonalise the matrix $A_T$. That is, we will find a diagonal matrix $D$ and an orthogonal matrix $P$ such that $P^{\\text{T}} A_T P = D$. (Recall that an invertible square matrix $M$ is orthogonal if and only if $M^{-1} = M^{\\text{T}}$).

To do this we will need to find an orthonormal basis of $\\mathbb{R}^3$ consisting of eigenvectors of $A_T$. First we will find a regular basis of eigenvectors, and then we will orthonormalise it by applying the Gram-Schmidt process. We have already been told what the eigenvalues are, so this will make it easier to find the eigenspaces/eigenvectors.

i) The first eigenvalue is $\\var{{gg}}$. We know that the eigenspace of $\\var{{gg}}$ is $N(A_T - (\\var{gg}) I_3 )$, which is the same as $N \\Big(\\textbf{RRE} \\big( A_T - (\\var{gg}) I_3 \\big) \\Big)$. It is much easier to deal with the latter than the former.

What is $\\textbf{RRE} \\big( A_T - (\\var{gg}) I_3 \\big)$?
$\\textbf{RRE} \\big( A_T - (\\var{gg}) I_3 \\big) =$ [[0]] .

$N \\Big(\\textbf{RRE} \\big( A_T - (\\var{gg}) I_3 \\big) \\Big)$ is the set of all solutions to $\\textbf{RRE} \\big( A_T - (\\var{gg}) I_3 \\big) \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} = \\boldsymbol{0}$. In an attempt to find the general solution of this system, we see that $\\textbf{RRE} \\big( A_T - (\\var{gg}) I_3 \\big)$ has the free variables $y$ and $z$. Letting $y=s$ and $z=t$, we see that the general solution is
$\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} =$ [[1]]$s +$ [[2]]$t$.

So, $\\Bigg\\{$ [[1]] , [[2]] $\\Bigg\\}$ is a basis for $N \\Big(\\textbf{RRE} \\big( A_T - (\\var{gg}) I_3 \\big) \\Big)$, and hence for $N(A_T - (\\var{gg}) I_3 )$ as well.

Let $\\boldsymbol{g_1} =$ [[1]] and $\\boldsymbol{g_2} =$ [[2]] be our first two eigenvectors.

ii) Now let us look at the second eigenvalue, $\\var{{hh}}$. In a similar fashion as in part a, we first find $\\textbf{RRE} \\big( A_T - (\\var{hh}) I_3 \\big)$.

$\\textbf{RRE} \\big( A_T - (\\var{hh}) I_3 \\big) =$ [[3]] .

In an attempt to find the general solution of the system $\\textbf{RRE} \\big( A_T - (\\var{hh}) I_3 \\big) \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} = \\boldsymbol{0}$, we see that $\\textbf{RRE} \\big( A_T - (\\var{hh}) I_3 \\big)$ has the free variable $z$. Letting $z=t$, we see that the general solution is
$\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} =$ [[4]]$t$.

So, $\\Bigg\\{$ [[4]] $\\Bigg\\}$ is a basis for $N \\Big(\\textbf{RRE} \\big( A_T - (\\var{hh}) I_3 \\big) \\Big)$, and hence for $N(A_T - (\\var{hh}) I_3 )$ as well.

Let $\\boldsymbol{g_3} =$ [[4]] be our third eigenvector.

iii) Now that we have a basis of eigenvectors, $\\{ \\boldsymbol{g_1} , \\boldsymbol{g_2} , \\boldsymbol{g_3} \\}$, we will attempt to apply the Gram-Schmidt process to obtain an orthonormal basis.

The first step is to normalise $\\boldsymbol{g_1}$.
$\\boldsymbol{\\| g_1 \\|} =$ [[5]] . (If you wish to write something like $\\sqrt{2}$, then please write 2^(1/2) ).
Hence $\\boldsymbol{u_1} := \\frac{1}{\\boldsymbol{\\| g_1 \\|}} \\boldsymbol{g_1} = \\Big(1\\Big/$ [[5]]$\\Big)$ [[1]].

iv) Now we wish to find $\\boldsymbol{u_2}$.

$\\boldsymbol{w_2} := \\boldsymbol{g_2} - \\langle \\boldsymbol{g_2} , \\boldsymbol{u_1} \\rangle \\boldsymbol{u_1} =$ [[6]] .

$ \\|\\boldsymbol{w_2} \\| =$ [[7]] .

Hence, $\\boldsymbol{u_2} := \\frac{1}{\\boldsymbol{\\| w_2 \\|}} \\boldsymbol{w_2} = \\Big(1\\Big/$ [[7]] $\\Big)$ [[6]] .

v) Finally, we need to find $\\boldsymbol{u_3}$.

We note that $\\boldsymbol{g_3}$ is orthogonal to $\\boldsymbol{u_1}$. and $\\boldsymbol{u_2}$, and hence we only need to normalise it.

$ \\|\\boldsymbol{g_3} \\| =$ [[8]] .

Hence, $\\boldsymbol{u_3} := \\frac{1}{\\boldsymbol{\\| g_3 \\|}} \\boldsymbol{g_3} =\\Big(1\\Big/$ [[8]] $\\Big)$ [[4]] .

The set $ \\{ \\boldsymbol{u_1} , \\boldsymbol{u_2} , \\boldsymbol{u_3} \\}$ is an orthonormal basis for $\\mathbb{R}^3$ consisting of eigenvectors of $A_T$.

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Now that we have found and orthonormal basis for $\\mathbb{R}^3$ consisting of eigenvectors of $A_T$, we can find a diagonal matrix $D$ and an orthogonal matrix $P$ such that $P^{\\text{T}} A_T P = D$.

Given what we have found in parts a and b, we can find $P$ and $D$ immediately, without even doing any further calculations. Suppose we are told that $P = ( \\boldsymbol{u_1} \\; , \\; \\boldsymbol{u_2} \\; , \\; \\boldsymbol{u_3})$. What will $D$ be then? (If you require a hint, please press the \"Show steps\" button below).

$D =$ [[0]] .

Suppose we have an $n \\times n$ matrix $A$ with $n$ orthogonal eigenvectors $\\boldsymbol{v_1} , \\boldsymbol{v_2} , \\ldots , \\boldsymbol{v_n}$ and corresponding eigenvalues $\\lambda_1 , \\lambda_2 , \\ldots , \\lambda_n$.

Let $P = (\\boldsymbol{v_1} \\; \\boldsymbol{v_2} \\; \\ldots \\; \\boldsymbol{v_n})$ and let $D$ be the $n \\times n$ diagonal matrix with $\\lambda_i$ in the $(i,i)$ entry. Then, $P^{\\text{T}} A P = D$.

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This is the question for Lent Term week 4 of the MA100 course at the LSE. It looks at material from chapters 27 and 28.

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