What is the Maximum Value of $x^T Ax$?

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Let $A$ be the matrix $begin{bmatrix}3 &1 \ 1&2end{bmatrix}$. What is the maximum value of $x^T Ax$ where the maximum is taken over all $x$ that are the unit eigenvectors of $A$?

1. $5$


2. $frac{(5 + sqrt{5})}{2}$


3. $3$


4. $frac{(5 - sqrt{5})}{2}$

---

Eigenvalues of $A$ are $frac{(5 pm sqrt{5})}{2}$

My question is: what is $x^T Ax$ ? Can you explain little bit please?

linear-algebra matrices eigenvalues-eigenvectors

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edited Dec 28 '15 at 12:56

H. R.

9,31093261

asked Nov 30 '15 at 12:12

Mithlesh Upadhyay

2,88882661

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $x=(x_1,x_2)^T$ then $x^TAx=(x_1,x_2)A(x_1,x_2)^T$
  – Kushal Bhuyan
  Nov 30 '15 at 12:14
  

  
  
  
  

add a comment | 

up vote
3
down vote

favorite

2

Let $A$ be the matrix $begin{bmatrix}3 &1 \ 1&2end{bmatrix}$. What is the maximum value of $x^T Ax$ where the maximum is taken over all $x$ that are the unit eigenvectors of $A$?

1. $5$


2. $frac{(5 + sqrt{5})}{2}$


3. $3$


4. $frac{(5 - sqrt{5})}{2}$

---

Eigenvalues of $A$ are $frac{(5 pm sqrt{5})}{2}$

My question is: what is $x^T Ax$ ? Can you explain little bit please?

linear-algebra matrices eigenvalues-eigenvectors

share|cite|improve this question

edited Dec 28 '15 at 12:56

H. R.

9,31093261

asked Nov 30 '15 at 12:12

Mithlesh Upadhyay

2,88882661

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $x=(x_1,x_2)^T$ then $x^TAx=(x_1,x_2)A(x_1,x_2)^T$
  – Kushal Bhuyan
  Nov 30 '15 at 12:14
  

  
  
  
  

add a comment | 

up vote
3
down vote

favorite

2

up vote
3
down vote

favorite

2

2

Let $A$ be the matrix $begin{bmatrix}3 &1 \ 1&2end{bmatrix}$. What is the maximum value of $x^T Ax$ where the maximum is taken over all $x$ that are the unit eigenvectors of $A$?

1. $5$


2. $frac{(5 + sqrt{5})}{2}$


3. $3$


4. $frac{(5 - sqrt{5})}{2}$

---

Eigenvalues of $A$ are $frac{(5 pm sqrt{5})}{2}$

My question is: what is $x^T Ax$ ? Can you explain little bit please?

linear-algebra matrices eigenvalues-eigenvectors

share|cite|improve this question

edited Dec 28 '15 at 12:56

H. R.

9,31093261

asked Nov 30 '15 at 12:12

Mithlesh Upadhyay

2,88882661

Let $A$ be the matrix $begin{bmatrix}3 &1 \ 1&2end{bmatrix}$. What is the maximum value of $x^T Ax$ where the maximum is taken over all $x$ that are the unit eigenvectors of $A$?

1. $5$


2. $frac{(5 + sqrt{5})}{2}$


3. $3$


4. $frac{(5 - sqrt{5})}{2}$

---

Eigenvalues of $A$ are $frac{(5 pm sqrt{5})}{2}$

My question is: what is $x^T Ax$ ? Can you explain little bit please?

linear-algebra matrices eigenvalues-eigenvectors

linear-algebra matrices eigenvalues-eigenvectors

share|cite|improve this question

edited Dec 28 '15 at 12:56

H. R.

9,31093261

asked Nov 30 '15 at 12:12

Mithlesh Upadhyay

2,88882661

share|cite|improve this question

edited Dec 28 '15 at 12:56

H. R.

9,31093261

asked Nov 30 '15 at 12:12

Mithlesh Upadhyay

2,88882661

share|cite|improve this question

share|cite|improve this question

edited Dec 28 '15 at 12:56

H. R.

9,31093261

edited Dec 28 '15 at 12:56

H. R.

9,31093261

edited Dec 28 '15 at 12:56

H. R.

9,31093261

9,31093261

asked Nov 30 '15 at 12:12

Mithlesh Upadhyay

2,88882661

asked Nov 30 '15 at 12:12

Mithlesh Upadhyay

2,88882661

asked Nov 30 '15 at 12:12

Mithlesh Upadhyay

2,88882661

2,88882661

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $x=(x_1,x_2)^T$ then $x^TAx=(x_1,x_2)A(x_1,x_2)^T$
  – Kushal Bhuyan
  Nov 30 '15 at 12:14
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $x=(x_1,x_2)^T$ then $x^TAx=(x_1,x_2)A(x_1,x_2)^T$
  – Kushal Bhuyan
  Nov 30 '15 at 12:14
  

  
  
  
  

1

1

$x=(x_1,x_2)^T$ then $x^TAx=(x_1,x_2)A(x_1,x_2)^T$
– Kushal Bhuyan
Nov 30 '15 at 12:14

$x=(x_1,x_2)^T$ then $x^TAx=(x_1,x_2)A(x_1,x_2)^T$
– Kushal Bhuyan
Nov 30 '15 at 12:14

add a comment | 

5 Answers
5

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up vote
3
down vote



accepted

If $xneq 0$ is an eigenvector of $A$ with unit length and associated eigenvalue $lambda$, then $x^TAx=x^Tlambda x=lambda x^Tx = lambda$. So the maximum value of $x^T Ax$ where the maximum is taken over all $x$ that are the unit eigenvectors of $A$ is simply the value of the largest eigenvalue of $A$.

As pointed out by @Ant, the quantity $x^TAx$ arise in the Rayleigh quotient associated to $A$ (note that $A$ is symmetric). In fact, one can show that the eigenvectors (resp. eigenvalues) of a symmetric matrix $M$ are the critical points (resp. values) of the function
$$R(x)=frac{x^TMx}{x^Tx}$$
which is called the Rayleigh quotient associated to $M$.

share|cite|improve this answer

edited Nov 30 '15 at 14:51

answered Nov 30 '15 at 12:41

Surb

37k94375

add a comment | 

up vote
3
down vote

This is an immediate application of this result from Raleigh.. https://en.m.wikipedia.org/wiki/Rayleigh_quotient

share|cite|improve this answer

answered Nov 30 '15 at 12:33

Ant

17.3k22873

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for nice article/answer.
  – Mithlesh Upadhyay
  Nov 30 '15 at 12:49
  
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

OK, Let us say that the eigenvalue of $A$ is $lambda$ and then call the corresponding eigenvector $x_{lambda}$ and hence

$$Ax_{lambda}=lambda x_{lambda}$$

and observe that

$$x_lambda ^TA{x_lambda } = lambda x_lambda ^T{x_lambda } = lambda (1) = lambda $$

so you should find the maximum eigenvalue of $A$. So the maximum value of the quadratic form will be $frac{5 + sqrt{5}}{2}$.

share|cite|improve this answer

edited Nov 30 '15 at 13:50

answered Nov 30 '15 at 12:41

H. R.

9,31093261

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thanks for beautiful explanation.
  – Mithlesh Upadhyay
  Nov 30 '15 at 12:46
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @MithleshUpadhyay: You are welcome! :)
  – H. R.
  Nov 30 '15 at 13:50
  

  
  
  
  

add a comment | 

up vote
1
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$$x^TAx=begin{bmatrix} x_1 & x_2 end{bmatrix}begin{bmatrix} 3 & 1 \ 1 & 2 end{bmatrix}begin{bmatrix} x_1 \ x_2 end{bmatrix}= 3x_1^2 + 2x_1x_2 + 2x_2^2$$

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answered Nov 30 '15 at 12:20

Nigel Overmars

2,59911225

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I need to find maximum value, from this function, rt?
  – Mithlesh Upadhyay
  Nov 30 '15 at 12:24
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Using the (unit) eigenvectors, yes.
  – Nigel Overmars
  Nov 30 '15 at 12:25
  

  
  
  
  

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0
down vote

Here $A=begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}$

For finding the eigen value:

Chracteristic equation:$|A-lambda I|=0$ where $I$ is called identity matrix.

                                     $Rightarrow|A-lambda I|=begin{vmatrix} 3 &1 \ 1 & 2 end{vmatrix}-lambdabegin{vmatrix}  1&0 \ 0 & 1 end{vmatrix}=0$

                                     $Rightarrow begin{vmatrix} 3 &1 \ 1 & 2 end{vmatrix}-begin{vmatrix}  lambda&0 \ 0 & lambda end{vmatrix}=0$

                                     $Rightarrow begin{vmatrix} 3-lambda &1 \ 1 & 2-lambda end{vmatrix}=0$

                                     $Rightarrow(3-lambda)(2-lambda)-1=0$

                                      $Rightarrow 6-3lambda-2lambda+lambda^{2}-1=0$

                                       $Rightarrow lambda^{2}-5lambda+5=0$

Now$,lambda =frac{-(-5)pm sqrt{25-20}}{2}$

 $lambda =frac{5pm sqrt{5}}{2}$

 So$,lambda_{1} =frac{5+sqrt{5}}{2}$ and $lambda_{2} =frac{5-sqrt{5}}{2}$

Let Eigen vector $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}_{2times 1}$

Now find the Eigen Vectors:

           $AX=lambda X$

       $Rightarrow AX-lambda X=begin{bmatrix} 0 end{bmatrix}$

       $Rightarrow(A-lambda I )X=begin{bmatrix} 0 end{bmatrix}$

  $Rightarrow begin{bmatrix} 3-lambda &1 \ 1 & 2-lambda end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$------------$>(1)$

           Put $lambda=frac{5+sqrt{5}}{2}$

  $Rightarrow begin{bmatrix} 3-(frac{5+sqrt{5}}{2}) &1 \ 1 & 2-(frac{5+sqrt{5}}{2}) end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

  $Rightarrow begin{bmatrix}frac{6-5-sqrt{5}}{2} &1 \ 1 &frac{4-5-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

  $Rightarrow begin{bmatrix}frac{1-sqrt{5}}{2} &1 \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

Perform some row operation

$R_{1}rightarrowfrac{1+sqrt{5}}{2}R_{1}$

$Rightarrow begin{bmatrix}(frac{1-sqrt{5}}{2}).(frac{1+sqrt{5}}{2})&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

$Rightarrow begin{bmatrix}frac{1-5}{4}&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

$Rightarrow begin{bmatrix}-1&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

Again do some operation $R_{2}rightarrow R_{2}+R_{1}$

$Rightarrow begin{bmatrix}-1&frac{1+sqrt{5}}{2} \ 0 &0 end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

Here $Rank (A)=1$ and $number$  $of$  $unknowns$ $ =2$

clearly see $r(A)<UK(1<2),$So $UK-r(A)=2-1=1$ value assign to the one unknowns.[This is case of infinite solutions]

So$,x_{2}=k$ and $-x_{1}+(frac{1+sqrt{5}}{2})x_{2}=0$

                                  $ -x_{1}=-(frac{1+sqrt{5}}{2})k$

                                   $ x_{1}=(frac{1+sqrt{5}}{2})k$

Therefor $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix}(frac{1+sqrt{5}}{2})k \k end{bmatrix}$

               $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=k begin{bmatrix}(frac{1+sqrt{5}}{2}) \1 end{bmatrix}$

Now, find the unit vectors of $A:$

$hat{u}=frac{vec{X}}{|vec{X}|}$

$|vec{X}|=sqrt{(frac{1+sqrt{5}}{2})^{2}+1^{2}}=1.9$

$hat{u}=frac{begin{bmatrix}(frac{1+sqrt{5}}{2}) \1 end{bmatrix}}{1.9}$

$hat{u}=begin{bmatrix} 0.85\ 0.52 end{bmatrix}$

here in given question  $hat{u}=x=begin{bmatrix} 0.85 \0.52 end{bmatrix}$ and $x^{T}=begin{bmatrix} 0.85 &0.52 end{bmatrix}$

Now $x^{T}Ax=begin{bmatrix} 0.85 &0.52 end{bmatrix}_{1times 2}times begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}times begin{bmatrix} 0.85 \0.52 end{bmatrix}_{2times 1}$

$x^{T}Ax=begin{bmatrix} 3.6 end{bmatrix}_{1times 1}$

and similarly do for $lambda=frac{5-sqrt{5}}{2}$

 $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=k begin{bmatrix}(frac{1-sqrt{5}}{2}) \1 end{bmatrix}$

Now, find the unit vectors of $A:$

$hat{u}=frac{vec{X}}{|vec{X}|}$

$|vec{X}|=sqrt{(frac{1-sqrt{5}}{2})^{2}+1^{2}}=1.2$

$hat{u}=frac{begin{bmatrix}(frac{1-sqrt{5}}{2}) \1 end{bmatrix}}{1.2}$

$hat{u}=begin{bmatrix} -0.52\ 0.83 end{bmatrix}$

here in given question  $hat{u}=x=begin{bmatrix} -0.52\0.83 end{bmatrix}$ and $x^{T}=begin{bmatrix} -0.52 &0.83 end{bmatrix}$

Now $x^{T}Ax=begin{bmatrix} -0.52 &0.83 end{bmatrix}_{1times 2}times begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}times begin{bmatrix} -0.52 \0.83 end{bmatrix}_{2times 1}$

$x^{T}Ax=begin{bmatrix} 1.33 end{bmatrix}_{1times 1}$

So,I got maximum value $x^{T}Ax=begin{bmatrix} 3.6 end{bmatrix}_{1times 1}=begin{bmatrix} frac{5+sqrt{5}}{2} end{bmatrix}$

and minimum value $x^{T}Ax=begin{bmatrix} 1.33 end{bmatrix}_{1times 1}=begin{bmatrix} frac{5-sqrt{5}}{2} end{bmatrix}$

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answered yesterday

Lakshman Patel

11

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

If $xneq 0$ is an eigenvector of $A$ with unit length and associated eigenvalue $lambda$, then $x^TAx=x^Tlambda x=lambda x^Tx = lambda$. So the maximum value of $x^T Ax$ where the maximum is taken over all $x$ that are the unit eigenvectors of $A$ is simply the value of the largest eigenvalue of $A$.

As pointed out by @Ant, the quantity $x^TAx$ arise in the Rayleigh quotient associated to $A$ (note that $A$ is symmetric). In fact, one can show that the eigenvectors (resp. eigenvalues) of a symmetric matrix $M$ are the critical points (resp. values) of the function
$$R(x)=frac{x^TMx}{x^Tx}$$
which is called the Rayleigh quotient associated to $M$.

share|cite|improve this answer

edited Nov 30 '15 at 14:51

answered Nov 30 '15 at 12:41

Surb

37k94375

add a comment | 

up vote
3
down vote



accepted

If $xneq 0$ is an eigenvector of $A$ with unit length and associated eigenvalue $lambda$, then $x^TAx=x^Tlambda x=lambda x^Tx = lambda$. So the maximum value of $x^T Ax$ where the maximum is taken over all $x$ that are the unit eigenvectors of $A$ is simply the value of the largest eigenvalue of $A$.

As pointed out by @Ant, the quantity $x^TAx$ arise in the Rayleigh quotient associated to $A$ (note that $A$ is symmetric). In fact, one can show that the eigenvectors (resp. eigenvalues) of a symmetric matrix $M$ are the critical points (resp. values) of the function
$$R(x)=frac{x^TMx}{x^Tx}$$
which is called the Rayleigh quotient associated to $M$.

share|cite|improve this answer

edited Nov 30 '15 at 14:51

answered Nov 30 '15 at 12:41

Surb

37k94375

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

If $xneq 0$ is an eigenvector of $A$ with unit length and associated eigenvalue $lambda$, then $x^TAx=x^Tlambda x=lambda x^Tx = lambda$. So the maximum value of $x^T Ax$ where the maximum is taken over all $x$ that are the unit eigenvectors of $A$ is simply the value of the largest eigenvalue of $A$.

As pointed out by @Ant, the quantity $x^TAx$ arise in the Rayleigh quotient associated to $A$ (note that $A$ is symmetric). In fact, one can show that the eigenvectors (resp. eigenvalues) of a symmetric matrix $M$ are the critical points (resp. values) of the function
$$R(x)=frac{x^TMx}{x^Tx}$$
which is called the Rayleigh quotient associated to $M$.

share|cite|improve this answer

edited Nov 30 '15 at 14:51

answered Nov 30 '15 at 12:41

Surb

37k94375

If $xneq 0$ is an eigenvector of $A$ with unit length and associated eigenvalue $lambda$, then $x^TAx=x^Tlambda x=lambda x^Tx = lambda$. So the maximum value of $x^T Ax$ where the maximum is taken over all $x$ that are the unit eigenvectors of $A$ is simply the value of the largest eigenvalue of $A$.

As pointed out by @Ant, the quantity $x^TAx$ arise in the Rayleigh quotient associated to $A$ (note that $A$ is symmetric). In fact, one can show that the eigenvectors (resp. eigenvalues) of a symmetric matrix $M$ are the critical points (resp. values) of the function
$$R(x)=frac{x^TMx}{x^Tx}$$
which is called the Rayleigh quotient associated to $M$.

share|cite|improve this answer

edited Nov 30 '15 at 14:51

answered Nov 30 '15 at 12:41

Surb

37k94375

share|cite|improve this answer

share|cite|improve this answer

edited Nov 30 '15 at 14:51

edited Nov 30 '15 at 14:51

edited Nov 30 '15 at 14:51

answered Nov 30 '15 at 12:41

Surb

37k94375

answered Nov 30 '15 at 12:41

Surb

37k94375

answered Nov 30 '15 at 12:41

Surb

37k94375

37k94375

add a comment | 

add a comment | 

up vote
3
down vote

This is an immediate application of this result from Raleigh.. https://en.m.wikipedia.org/wiki/Rayleigh_quotient

share|cite|improve this answer

answered Nov 30 '15 at 12:33

Ant

17.3k22873

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for nice article/answer.
  – Mithlesh Upadhyay
  Nov 30 '15 at 12:49
  
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

This is an immediate application of this result from Raleigh.. https://en.m.wikipedia.org/wiki/Rayleigh_quotient

share|cite|improve this answer

answered Nov 30 '15 at 12:33

Ant

17.3k22873

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for nice article/answer.
  – Mithlesh Upadhyay
  Nov 30 '15 at 12:49
  
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

This is an immediate application of this result from Raleigh.. https://en.m.wikipedia.org/wiki/Rayleigh_quotient

share|cite|improve this answer

answered Nov 30 '15 at 12:33

Ant

17.3k22873

This is an immediate application of this result from Raleigh.. https://en.m.wikipedia.org/wiki/Rayleigh_quotient

share|cite|improve this answer

answered Nov 30 '15 at 12:33

Ant

17.3k22873

share|cite|improve this answer

share|cite|improve this answer

answered Nov 30 '15 at 12:33

Ant

17.3k22873

answered Nov 30 '15 at 12:33

Ant

17.3k22873

answered Nov 30 '15 at 12:33

Ant

17.3k22873

17.3k22873

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for nice article/answer.
  – Mithlesh Upadhyay
  Nov 30 '15 at 12:49
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for nice article/answer.
  – Mithlesh Upadhyay
  Nov 30 '15 at 12:49
  
  
  

  
  
  
  

Thanks for nice article/answer.
– Mithlesh Upadhyay
Nov 30 '15 at 12:49

Thanks for nice article/answer.
– Mithlesh Upadhyay
Nov 30 '15 at 12:49

add a comment | 

up vote
2
down vote

OK, Let us say that the eigenvalue of $A$ is $lambda$ and then call the corresponding eigenvector $x_{lambda}$ and hence

$$Ax_{lambda}=lambda x_{lambda}$$

and observe that

$$x_lambda ^TA{x_lambda } = lambda x_lambda ^T{x_lambda } = lambda (1) = lambda $$

so you should find the maximum eigenvalue of $A$. So the maximum value of the quadratic form will be $frac{5 + sqrt{5}}{2}$.

share|cite|improve this answer

edited Nov 30 '15 at 13:50

answered Nov 30 '15 at 12:41

H. R.

9,31093261

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thanks for beautiful explanation.
  – Mithlesh Upadhyay
  Nov 30 '15 at 12:46
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @MithleshUpadhyay: You are welcome! :)
  – H. R.
  Nov 30 '15 at 13:50
  

  
  
  
  

add a comment | 

up vote
2
down vote

OK, Let us say that the eigenvalue of $A$ is $lambda$ and then call the corresponding eigenvector $x_{lambda}$ and hence

$$Ax_{lambda}=lambda x_{lambda}$$

and observe that

$$x_lambda ^TA{x_lambda } = lambda x_lambda ^T{x_lambda } = lambda (1) = lambda $$

so you should find the maximum eigenvalue of $A$. So the maximum value of the quadratic form will be $frac{5 + sqrt{5}}{2}$.

share|cite|improve this answer

edited Nov 30 '15 at 13:50

answered Nov 30 '15 at 12:41

H. R.

9,31093261

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thanks for beautiful explanation.
  – Mithlesh Upadhyay
  Nov 30 '15 at 12:46
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @MithleshUpadhyay: You are welcome! :)
  – H. R.
  Nov 30 '15 at 13:50
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

OK, Let us say that the eigenvalue of $A$ is $lambda$ and then call the corresponding eigenvector $x_{lambda}$ and hence

$$Ax_{lambda}=lambda x_{lambda}$$

and observe that

$$x_lambda ^TA{x_lambda } = lambda x_lambda ^T{x_lambda } = lambda (1) = lambda $$

so you should find the maximum eigenvalue of $A$. So the maximum value of the quadratic form will be $frac{5 + sqrt{5}}{2}$.

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edited Nov 30 '15 at 13:50

answered Nov 30 '15 at 12:41

H. R.

9,31093261

OK, Let us say that the eigenvalue of $A$ is $lambda$ and then call the corresponding eigenvector $x_{lambda}$ and hence

$$Ax_{lambda}=lambda x_{lambda}$$

and observe that

$$x_lambda ^TA{x_lambda } = lambda x_lambda ^T{x_lambda } = lambda (1) = lambda $$

so you should find the maximum eigenvalue of $A$. So the maximum value of the quadratic form will be $frac{5 + sqrt{5}}{2}$.

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edited Nov 30 '15 at 13:50

answered Nov 30 '15 at 12:41

H. R.

9,31093261

share|cite|improve this answer

share|cite|improve this answer

edited Nov 30 '15 at 13:50

edited Nov 30 '15 at 13:50

edited Nov 30 '15 at 13:50

answered Nov 30 '15 at 12:41

H. R.

9,31093261

answered Nov 30 '15 at 12:41

H. R.

9,31093261

answered Nov 30 '15 at 12:41

H. R.

9,31093261

9,31093261

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thanks for beautiful explanation.
  – Mithlesh Upadhyay
  Nov 30 '15 at 12:46
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @MithleshUpadhyay: You are welcome! :)
  – H. R.
  Nov 30 '15 at 13:50
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thanks for beautiful explanation.
  – Mithlesh Upadhyay
  Nov 30 '15 at 12:46
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @MithleshUpadhyay: You are welcome! :)
  – H. R.
  Nov 30 '15 at 13:50
  

  
  
  
  

1

1

Thanks for beautiful explanation.
– Mithlesh Upadhyay
Nov 30 '15 at 12:46

Thanks for beautiful explanation.
– Mithlesh Upadhyay
Nov 30 '15 at 12:46

1

1

@MithleshUpadhyay: You are welcome! :)
– H. R.
Nov 30 '15 at 13:50

@MithleshUpadhyay: You are welcome! :)
– H. R.
Nov 30 '15 at 13:50

add a comment | 

up vote
1
down vote

$$x^TAx=begin{bmatrix} x_1 & x_2 end{bmatrix}begin{bmatrix} 3 & 1 \ 1 & 2 end{bmatrix}begin{bmatrix} x_1 \ x_2 end{bmatrix}= 3x_1^2 + 2x_1x_2 + 2x_2^2$$

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answered Nov 30 '15 at 12:20

Nigel Overmars

2,59911225

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I need to find maximum value, from this function, rt?
  – Mithlesh Upadhyay
  Nov 30 '15 at 12:24
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Using the (unit) eigenvectors, yes.
  – Nigel Overmars
  Nov 30 '15 at 12:25
  

  
  
  
  

add a comment | 

up vote
1
down vote

$$x^TAx=begin{bmatrix} x_1 & x_2 end{bmatrix}begin{bmatrix} 3 & 1 \ 1 & 2 end{bmatrix}begin{bmatrix} x_1 \ x_2 end{bmatrix}= 3x_1^2 + 2x_1x_2 + 2x_2^2$$

share|cite|improve this answer

answered Nov 30 '15 at 12:20

Nigel Overmars

2,59911225

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I need to find maximum value, from this function, rt?
  – Mithlesh Upadhyay
  Nov 30 '15 at 12:24
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Using the (unit) eigenvectors, yes.
  – Nigel Overmars
  Nov 30 '15 at 12:25
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

$$x^TAx=begin{bmatrix} x_1 & x_2 end{bmatrix}begin{bmatrix} 3 & 1 \ 1 & 2 end{bmatrix}begin{bmatrix} x_1 \ x_2 end{bmatrix}= 3x_1^2 + 2x_1x_2 + 2x_2^2$$

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answered Nov 30 '15 at 12:20

Nigel Overmars

2,59911225

$$x^TAx=begin{bmatrix} x_1 & x_2 end{bmatrix}begin{bmatrix} 3 & 1 \ 1 & 2 end{bmatrix}begin{bmatrix} x_1 \ x_2 end{bmatrix}= 3x_1^2 + 2x_1x_2 + 2x_2^2$$

share|cite|improve this answer

answered Nov 30 '15 at 12:20

Nigel Overmars

2,59911225

share|cite|improve this answer

share|cite|improve this answer

answered Nov 30 '15 at 12:20

Nigel Overmars

2,59911225

answered Nov 30 '15 at 12:20

Nigel Overmars

2,59911225

answered Nov 30 '15 at 12:20

Nigel Overmars

2,59911225

2,59911225

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I need to find maximum value, from this function, rt?
  – Mithlesh Upadhyay
  Nov 30 '15 at 12:24
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Using the (unit) eigenvectors, yes.
  – Nigel Overmars
  Nov 30 '15 at 12:25
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I need to find maximum value, from this function, rt?
  – Mithlesh Upadhyay
  Nov 30 '15 at 12:24
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Using the (unit) eigenvectors, yes.
  – Nigel Overmars
  Nov 30 '15 at 12:25
  

  
  
  
  

I need to find maximum value, from this function, rt?
– Mithlesh Upadhyay
Nov 30 '15 at 12:24

I need to find maximum value, from this function, rt?
– Mithlesh Upadhyay
Nov 30 '15 at 12:24

Using the (unit) eigenvectors, yes.
– Nigel Overmars
Nov 30 '15 at 12:25

Using the (unit) eigenvectors, yes.
– Nigel Overmars
Nov 30 '15 at 12:25

add a comment | 

up vote
0
down vote

Here $A=begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}$

For finding the eigen value:

Chracteristic equation:$|A-lambda I|=0$ where $I$ is called identity matrix.

                                     $Rightarrow|A-lambda I|=begin{vmatrix} 3 &1 \ 1 & 2 end{vmatrix}-lambdabegin{vmatrix}  1&0 \ 0 & 1 end{vmatrix}=0$

                                     $Rightarrow begin{vmatrix} 3 &1 \ 1 & 2 end{vmatrix}-begin{vmatrix}  lambda&0 \ 0 & lambda end{vmatrix}=0$

                                     $Rightarrow begin{vmatrix} 3-lambda &1 \ 1 & 2-lambda end{vmatrix}=0$

                                     $Rightarrow(3-lambda)(2-lambda)-1=0$

                                      $Rightarrow 6-3lambda-2lambda+lambda^{2}-1=0$

                                       $Rightarrow lambda^{2}-5lambda+5=0$

Now$,lambda =frac{-(-5)pm sqrt{25-20}}{2}$

 $lambda =frac{5pm sqrt{5}}{2}$

 So$,lambda_{1} =frac{5+sqrt{5}}{2}$ and $lambda_{2} =frac{5-sqrt{5}}{2}$

Let Eigen vector $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}_{2times 1}$

Now find the Eigen Vectors:

           $AX=lambda X$

       $Rightarrow AX-lambda X=begin{bmatrix} 0 end{bmatrix}$

       $Rightarrow(A-lambda I )X=begin{bmatrix} 0 end{bmatrix}$

  $Rightarrow begin{bmatrix} 3-lambda &1 \ 1 & 2-lambda end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$------------$>(1)$

           Put $lambda=frac{5+sqrt{5}}{2}$

  $Rightarrow begin{bmatrix} 3-(frac{5+sqrt{5}}{2}) &1 \ 1 & 2-(frac{5+sqrt{5}}{2}) end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

  $Rightarrow begin{bmatrix}frac{6-5-sqrt{5}}{2} &1 \ 1 &frac{4-5-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

  $Rightarrow begin{bmatrix}frac{1-sqrt{5}}{2} &1 \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

Perform some row operation

$R_{1}rightarrowfrac{1+sqrt{5}}{2}R_{1}$

$Rightarrow begin{bmatrix}(frac{1-sqrt{5}}{2}).(frac{1+sqrt{5}}{2})&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

$Rightarrow begin{bmatrix}frac{1-5}{4}&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

$Rightarrow begin{bmatrix}-1&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

Again do some operation $R_{2}rightarrow R_{2}+R_{1}$

$Rightarrow begin{bmatrix}-1&frac{1+sqrt{5}}{2} \ 0 &0 end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

Here $Rank (A)=1$ and $number$  $of$  $unknowns$ $ =2$

clearly see $r(A)<UK(1<2),$So $UK-r(A)=2-1=1$ value assign to the one unknowns.[This is case of infinite solutions]

So$,x_{2}=k$ and $-x_{1}+(frac{1+sqrt{5}}{2})x_{2}=0$

                                  $ -x_{1}=-(frac{1+sqrt{5}}{2})k$

                                   $ x_{1}=(frac{1+sqrt{5}}{2})k$

Therefor $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix}(frac{1+sqrt{5}}{2})k \k end{bmatrix}$

               $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=k begin{bmatrix}(frac{1+sqrt{5}}{2}) \1 end{bmatrix}$

Now, find the unit vectors of $A:$

$hat{u}=frac{vec{X}}{|vec{X}|}$

$|vec{X}|=sqrt{(frac{1+sqrt{5}}{2})^{2}+1^{2}}=1.9$

$hat{u}=frac{begin{bmatrix}(frac{1+sqrt{5}}{2}) \1 end{bmatrix}}{1.9}$

$hat{u}=begin{bmatrix} 0.85\ 0.52 end{bmatrix}$

here in given question  $hat{u}=x=begin{bmatrix} 0.85 \0.52 end{bmatrix}$ and $x^{T}=begin{bmatrix} 0.85 &0.52 end{bmatrix}$

Now $x^{T}Ax=begin{bmatrix} 0.85 &0.52 end{bmatrix}_{1times 2}times begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}times begin{bmatrix} 0.85 \0.52 end{bmatrix}_{2times 1}$

$x^{T}Ax=begin{bmatrix} 3.6 end{bmatrix}_{1times 1}$

and similarly do for $lambda=frac{5-sqrt{5}}{2}$

 $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=k begin{bmatrix}(frac{1-sqrt{5}}{2}) \1 end{bmatrix}$

Now, find the unit vectors of $A:$

$hat{u}=frac{vec{X}}{|vec{X}|}$

$|vec{X}|=sqrt{(frac{1-sqrt{5}}{2})^{2}+1^{2}}=1.2$

$hat{u}=frac{begin{bmatrix}(frac{1-sqrt{5}}{2}) \1 end{bmatrix}}{1.2}$

$hat{u}=begin{bmatrix} -0.52\ 0.83 end{bmatrix}$

here in given question  $hat{u}=x=begin{bmatrix} -0.52\0.83 end{bmatrix}$ and $x^{T}=begin{bmatrix} -0.52 &0.83 end{bmatrix}$

Now $x^{T}Ax=begin{bmatrix} -0.52 &0.83 end{bmatrix}_{1times 2}times begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}times begin{bmatrix} -0.52 \0.83 end{bmatrix}_{2times 1}$

$x^{T}Ax=begin{bmatrix} 1.33 end{bmatrix}_{1times 1}$

So,I got maximum value $x^{T}Ax=begin{bmatrix} 3.6 end{bmatrix}_{1times 1}=begin{bmatrix} frac{5+sqrt{5}}{2} end{bmatrix}$

and minimum value $x^{T}Ax=begin{bmatrix} 1.33 end{bmatrix}_{1times 1}=begin{bmatrix} frac{5-sqrt{5}}{2} end{bmatrix}$

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answered yesterday

Lakshman Patel

11

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up vote
0
down vote

Here $A=begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}$

For finding the eigen value:

Chracteristic equation:$|A-lambda I|=0$ where $I$ is called identity matrix.

                                     $Rightarrow|A-lambda I|=begin{vmatrix} 3 &1 \ 1 & 2 end{vmatrix}-lambdabegin{vmatrix}  1&0 \ 0 & 1 end{vmatrix}=0$

                                     $Rightarrow begin{vmatrix} 3 &1 \ 1 & 2 end{vmatrix}-begin{vmatrix}  lambda&0 \ 0 & lambda end{vmatrix}=0$

                                     $Rightarrow begin{vmatrix} 3-lambda &1 \ 1 & 2-lambda end{vmatrix}=0$

                                     $Rightarrow(3-lambda)(2-lambda)-1=0$

                                      $Rightarrow 6-3lambda-2lambda+lambda^{2}-1=0$

                                       $Rightarrow lambda^{2}-5lambda+5=0$

Now$,lambda =frac{-(-5)pm sqrt{25-20}}{2}$

 $lambda =frac{5pm sqrt{5}}{2}$

 So$,lambda_{1} =frac{5+sqrt{5}}{2}$ and $lambda_{2} =frac{5-sqrt{5}}{2}$

Let Eigen vector $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}_{2times 1}$

Now find the Eigen Vectors:

           $AX=lambda X$

       $Rightarrow AX-lambda X=begin{bmatrix} 0 end{bmatrix}$

       $Rightarrow(A-lambda I )X=begin{bmatrix} 0 end{bmatrix}$

  $Rightarrow begin{bmatrix} 3-lambda &1 \ 1 & 2-lambda end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$------------$>(1)$

           Put $lambda=frac{5+sqrt{5}}{2}$

  $Rightarrow begin{bmatrix} 3-(frac{5+sqrt{5}}{2}) &1 \ 1 & 2-(frac{5+sqrt{5}}{2}) end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

  $Rightarrow begin{bmatrix}frac{6-5-sqrt{5}}{2} &1 \ 1 &frac{4-5-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

  $Rightarrow begin{bmatrix}frac{1-sqrt{5}}{2} &1 \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

Perform some row operation

$R_{1}rightarrowfrac{1+sqrt{5}}{2}R_{1}$

$Rightarrow begin{bmatrix}(frac{1-sqrt{5}}{2}).(frac{1+sqrt{5}}{2})&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

$Rightarrow begin{bmatrix}frac{1-5}{4}&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

$Rightarrow begin{bmatrix}-1&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

Again do some operation $R_{2}rightarrow R_{2}+R_{1}$

$Rightarrow begin{bmatrix}-1&frac{1+sqrt{5}}{2} \ 0 &0 end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

Here $Rank (A)=1$ and $number$  $of$  $unknowns$ $ =2$

clearly see $r(A)<UK(1<2),$So $UK-r(A)=2-1=1$ value assign to the one unknowns.[This is case of infinite solutions]

So$,x_{2}=k$ and $-x_{1}+(frac{1+sqrt{5}}{2})x_{2}=0$

                                  $ -x_{1}=-(frac{1+sqrt{5}}{2})k$

                                   $ x_{1}=(frac{1+sqrt{5}}{2})k$

Therefor $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix}(frac{1+sqrt{5}}{2})k \k end{bmatrix}$

               $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=k begin{bmatrix}(frac{1+sqrt{5}}{2}) \1 end{bmatrix}$

Now, find the unit vectors of $A:$

$hat{u}=frac{vec{X}}{|vec{X}|}$

$|vec{X}|=sqrt{(frac{1+sqrt{5}}{2})^{2}+1^{2}}=1.9$

$hat{u}=frac{begin{bmatrix}(frac{1+sqrt{5}}{2}) \1 end{bmatrix}}{1.9}$

$hat{u}=begin{bmatrix} 0.85\ 0.52 end{bmatrix}$

here in given question  $hat{u}=x=begin{bmatrix} 0.85 \0.52 end{bmatrix}$ and $x^{T}=begin{bmatrix} 0.85 &0.52 end{bmatrix}$

Now $x^{T}Ax=begin{bmatrix} 0.85 &0.52 end{bmatrix}_{1times 2}times begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}times begin{bmatrix} 0.85 \0.52 end{bmatrix}_{2times 1}$

$x^{T}Ax=begin{bmatrix} 3.6 end{bmatrix}_{1times 1}$

and similarly do for $lambda=frac{5-sqrt{5}}{2}$

 $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=k begin{bmatrix}(frac{1-sqrt{5}}{2}) \1 end{bmatrix}$

Now, find the unit vectors of $A:$

$hat{u}=frac{vec{X}}{|vec{X}|}$

$|vec{X}|=sqrt{(frac{1-sqrt{5}}{2})^{2}+1^{2}}=1.2$

$hat{u}=frac{begin{bmatrix}(frac{1-sqrt{5}}{2}) \1 end{bmatrix}}{1.2}$

$hat{u}=begin{bmatrix} -0.52\ 0.83 end{bmatrix}$

here in given question  $hat{u}=x=begin{bmatrix} -0.52\0.83 end{bmatrix}$ and $x^{T}=begin{bmatrix} -0.52 &0.83 end{bmatrix}$

Now $x^{T}Ax=begin{bmatrix} -0.52 &0.83 end{bmatrix}_{1times 2}times begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}times begin{bmatrix} -0.52 \0.83 end{bmatrix}_{2times 1}$

$x^{T}Ax=begin{bmatrix} 1.33 end{bmatrix}_{1times 1}$

So,I got maximum value $x^{T}Ax=begin{bmatrix} 3.6 end{bmatrix}_{1times 1}=begin{bmatrix} frac{5+sqrt{5}}{2} end{bmatrix}$

and minimum value $x^{T}Ax=begin{bmatrix} 1.33 end{bmatrix}_{1times 1}=begin{bmatrix} frac{5-sqrt{5}}{2} end{bmatrix}$

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answered yesterday

Lakshman Patel

11

add a comment | 

up vote
0
down vote

up vote
0
down vote

Here $A=begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}$

For finding the eigen value:

Chracteristic equation:$|A-lambda I|=0$ where $I$ is called identity matrix.

                                     $Rightarrow|A-lambda I|=begin{vmatrix} 3 &1 \ 1 & 2 end{vmatrix}-lambdabegin{vmatrix}  1&0 \ 0 & 1 end{vmatrix}=0$

                                     $Rightarrow begin{vmatrix} 3 &1 \ 1 & 2 end{vmatrix}-begin{vmatrix}  lambda&0 \ 0 & lambda end{vmatrix}=0$

                                     $Rightarrow begin{vmatrix} 3-lambda &1 \ 1 & 2-lambda end{vmatrix}=0$

                                     $Rightarrow(3-lambda)(2-lambda)-1=0$

                                      $Rightarrow 6-3lambda-2lambda+lambda^{2}-1=0$

                                       $Rightarrow lambda^{2}-5lambda+5=0$

Now$,lambda =frac{-(-5)pm sqrt{25-20}}{2}$

 $lambda =frac{5pm sqrt{5}}{2}$

 So$,lambda_{1} =frac{5+sqrt{5}}{2}$ and $lambda_{2} =frac{5-sqrt{5}}{2}$

Let Eigen vector $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}_{2times 1}$

Now find the Eigen Vectors:

           $AX=lambda X$

       $Rightarrow AX-lambda X=begin{bmatrix} 0 end{bmatrix}$

       $Rightarrow(A-lambda I )X=begin{bmatrix} 0 end{bmatrix}$

  $Rightarrow begin{bmatrix} 3-lambda &1 \ 1 & 2-lambda end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$------------$>(1)$

           Put $lambda=frac{5+sqrt{5}}{2}$

  $Rightarrow begin{bmatrix} 3-(frac{5+sqrt{5}}{2}) &1 \ 1 & 2-(frac{5+sqrt{5}}{2}) end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

  $Rightarrow begin{bmatrix}frac{6-5-sqrt{5}}{2} &1 \ 1 &frac{4-5-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

  $Rightarrow begin{bmatrix}frac{1-sqrt{5}}{2} &1 \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

Perform some row operation

$R_{1}rightarrowfrac{1+sqrt{5}}{2}R_{1}$

$Rightarrow begin{bmatrix}(frac{1-sqrt{5}}{2}).(frac{1+sqrt{5}}{2})&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

$Rightarrow begin{bmatrix}frac{1-5}{4}&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

$Rightarrow begin{bmatrix}-1&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

Again do some operation $R_{2}rightarrow R_{2}+R_{1}$

$Rightarrow begin{bmatrix}-1&frac{1+sqrt{5}}{2} \ 0 &0 end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

Here $Rank (A)=1$ and $number$  $of$  $unknowns$ $ =2$

clearly see $r(A)<UK(1<2),$So $UK-r(A)=2-1=1$ value assign to the one unknowns.[This is case of infinite solutions]

So$,x_{2}=k$ and $-x_{1}+(frac{1+sqrt{5}}{2})x_{2}=0$

                                  $ -x_{1}=-(frac{1+sqrt{5}}{2})k$

                                   $ x_{1}=(frac{1+sqrt{5}}{2})k$

Therefor $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix}(frac{1+sqrt{5}}{2})k \k end{bmatrix}$

               $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=k begin{bmatrix}(frac{1+sqrt{5}}{2}) \1 end{bmatrix}$

Now, find the unit vectors of $A:$

$hat{u}=frac{vec{X}}{|vec{X}|}$

$|vec{X}|=sqrt{(frac{1+sqrt{5}}{2})^{2}+1^{2}}=1.9$

$hat{u}=frac{begin{bmatrix}(frac{1+sqrt{5}}{2}) \1 end{bmatrix}}{1.9}$

$hat{u}=begin{bmatrix} 0.85\ 0.52 end{bmatrix}$

here in given question  $hat{u}=x=begin{bmatrix} 0.85 \0.52 end{bmatrix}$ and $x^{T}=begin{bmatrix} 0.85 &0.52 end{bmatrix}$

Now $x^{T}Ax=begin{bmatrix} 0.85 &0.52 end{bmatrix}_{1times 2}times begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}times begin{bmatrix} 0.85 \0.52 end{bmatrix}_{2times 1}$

$x^{T}Ax=begin{bmatrix} 3.6 end{bmatrix}_{1times 1}$

and similarly do for $lambda=frac{5-sqrt{5}}{2}$

 $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=k begin{bmatrix}(frac{1-sqrt{5}}{2}) \1 end{bmatrix}$

Now, find the unit vectors of $A:$

$hat{u}=frac{vec{X}}{|vec{X}|}$

$|vec{X}|=sqrt{(frac{1-sqrt{5}}{2})^{2}+1^{2}}=1.2$

$hat{u}=frac{begin{bmatrix}(frac{1-sqrt{5}}{2}) \1 end{bmatrix}}{1.2}$

$hat{u}=begin{bmatrix} -0.52\ 0.83 end{bmatrix}$

here in given question  $hat{u}=x=begin{bmatrix} -0.52\0.83 end{bmatrix}$ and $x^{T}=begin{bmatrix} -0.52 &0.83 end{bmatrix}$

Now $x^{T}Ax=begin{bmatrix} -0.52 &0.83 end{bmatrix}_{1times 2}times begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}times begin{bmatrix} -0.52 \0.83 end{bmatrix}_{2times 1}$

$x^{T}Ax=begin{bmatrix} 1.33 end{bmatrix}_{1times 1}$

So,I got maximum value $x^{T}Ax=begin{bmatrix} 3.6 end{bmatrix}_{1times 1}=begin{bmatrix} frac{5+sqrt{5}}{2} end{bmatrix}$

and minimum value $x^{T}Ax=begin{bmatrix} 1.33 end{bmatrix}_{1times 1}=begin{bmatrix} frac{5-sqrt{5}}{2} end{bmatrix}$

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answered yesterday

Lakshman Patel

11

Here $A=begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}$

For finding the eigen value:

Chracteristic equation:$|A-lambda I|=0$ where $I$ is called identity matrix.

                                     $Rightarrow|A-lambda I|=begin{vmatrix} 3 &1 \ 1 & 2 end{vmatrix}-lambdabegin{vmatrix}  1&0 \ 0 & 1 end{vmatrix}=0$

                                     $Rightarrow begin{vmatrix} 3 &1 \ 1 & 2 end{vmatrix}-begin{vmatrix}  lambda&0 \ 0 & lambda end{vmatrix}=0$

                                     $Rightarrow begin{vmatrix} 3-lambda &1 \ 1 & 2-lambda end{vmatrix}=0$

                                     $Rightarrow(3-lambda)(2-lambda)-1=0$

                                      $Rightarrow 6-3lambda-2lambda+lambda^{2}-1=0$

                                       $Rightarrow lambda^{2}-5lambda+5=0$

Now$,lambda =frac{-(-5)pm sqrt{25-20}}{2}$

 $lambda =frac{5pm sqrt{5}}{2}$

 So$,lambda_{1} =frac{5+sqrt{5}}{2}$ and $lambda_{2} =frac{5-sqrt{5}}{2}$

Let Eigen vector $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}_{2times 1}$

Now find the Eigen Vectors:

           $AX=lambda X$

       $Rightarrow AX-lambda X=begin{bmatrix} 0 end{bmatrix}$

       $Rightarrow(A-lambda I )X=begin{bmatrix} 0 end{bmatrix}$

  $Rightarrow begin{bmatrix} 3-lambda &1 \ 1 & 2-lambda end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$------------$>(1)$

           Put $lambda=frac{5+sqrt{5}}{2}$

  $Rightarrow begin{bmatrix} 3-(frac{5+sqrt{5}}{2}) &1 \ 1 & 2-(frac{5+sqrt{5}}{2}) end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

  $Rightarrow begin{bmatrix}frac{6-5-sqrt{5}}{2} &1 \ 1 &frac{4-5-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

  $Rightarrow begin{bmatrix}frac{1-sqrt{5}}{2} &1 \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

Perform some row operation

$R_{1}rightarrowfrac{1+sqrt{5}}{2}R_{1}$

$Rightarrow begin{bmatrix}(frac{1-sqrt{5}}{2}).(frac{1+sqrt{5}}{2})&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

$Rightarrow begin{bmatrix}frac{1-5}{4}&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

$Rightarrow begin{bmatrix}-1&frac{1+sqrt{5}}{2} \ 1 &frac{-1-sqrt{5}}{2} end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

Again do some operation $R_{2}rightarrow R_{2}+R_{1}$

$Rightarrow begin{bmatrix}-1&frac{1+sqrt{5}}{2} \ 0 &0 end{bmatrix}begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix} 0\0 end{bmatrix}$

Here $Rank (A)=1$ and $number$  $of$  $unknowns$ $ =2$

clearly see $r(A)<UK(1<2),$So $UK-r(A)=2-1=1$ value assign to the one unknowns.[This is case of infinite solutions]

So$,x_{2}=k$ and $-x_{1}+(frac{1+sqrt{5}}{2})x_{2}=0$

                                  $ -x_{1}=-(frac{1+sqrt{5}}{2})k$

                                   $ x_{1}=(frac{1+sqrt{5}}{2})k$

Therefor $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=begin{bmatrix}(frac{1+sqrt{5}}{2})k \k end{bmatrix}$

               $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=k begin{bmatrix}(frac{1+sqrt{5}}{2}) \1 end{bmatrix}$

Now, find the unit vectors of $A:$

$hat{u}=frac{vec{X}}{|vec{X}|}$

$|vec{X}|=sqrt{(frac{1+sqrt{5}}{2})^{2}+1^{2}}=1.9$

$hat{u}=frac{begin{bmatrix}(frac{1+sqrt{5}}{2}) \1 end{bmatrix}}{1.9}$

$hat{u}=begin{bmatrix} 0.85\ 0.52 end{bmatrix}$

here in given question  $hat{u}=x=begin{bmatrix} 0.85 \0.52 end{bmatrix}$ and $x^{T}=begin{bmatrix} 0.85 &0.52 end{bmatrix}$

Now $x^{T}Ax=begin{bmatrix} 0.85 &0.52 end{bmatrix}_{1times 2}times begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}times begin{bmatrix} 0.85 \0.52 end{bmatrix}_{2times 1}$

$x^{T}Ax=begin{bmatrix} 3.6 end{bmatrix}_{1times 1}$

and similarly do for $lambda=frac{5-sqrt{5}}{2}$

 $X=begin{bmatrix} x_{1}\x_{2} end{bmatrix}=k begin{bmatrix}(frac{1-sqrt{5}}{2}) \1 end{bmatrix}$

Now, find the unit vectors of $A:$

$hat{u}=frac{vec{X}}{|vec{X}|}$

$|vec{X}|=sqrt{(frac{1-sqrt{5}}{2})^{2}+1^{2}}=1.2$

$hat{u}=frac{begin{bmatrix}(frac{1-sqrt{5}}{2}) \1 end{bmatrix}}{1.2}$

$hat{u}=begin{bmatrix} -0.52\ 0.83 end{bmatrix}$

here in given question  $hat{u}=x=begin{bmatrix} -0.52\0.83 end{bmatrix}$ and $x^{T}=begin{bmatrix} -0.52 &0.83 end{bmatrix}$

Now $x^{T}Ax=begin{bmatrix} -0.52 &0.83 end{bmatrix}_{1times 2}times begin{bmatrix} 3 &1 \ 1 & 2 end{bmatrix}_{2times 2}times begin{bmatrix} -0.52 \0.83 end{bmatrix}_{2times 1}$

$x^{T}Ax=begin{bmatrix} 1.33 end{bmatrix}_{1times 1}$

So,I got maximum value $x^{T}Ax=begin{bmatrix} 3.6 end{bmatrix}_{1times 1}=begin{bmatrix} frac{5+sqrt{5}}{2} end{bmatrix}$

and minimum value $x^{T}Ax=begin{bmatrix} 1.33 end{bmatrix}_{1times 1}=begin{bmatrix} frac{5-sqrt{5}}{2} end{bmatrix}$

share|cite|improve this answer

answered yesterday

Lakshman Patel

11

share|cite|improve this answer

share|cite|improve this answer

answered yesterday

Lakshman Patel

11

answered yesterday

Lakshman Patel

11

answered yesterday

Lakshman Patel

11

11

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