GATE IT 2007 | Question: 2

Question

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

• A. $5$
• B. $\frac{(5 + √5)}{2}$
• C. $3$
• D. $\frac{(5 - √5)}{2}$

Comments

x^T * x  = 1 in any cases so A will remain by this we find Lambda value right then why we are taking the concept of unit vector can any one explain

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@pravallika_07

Why $x^Tx=1$ will always be true if you are not considering $x$ as a unit vector. 

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Observation :

The expression $R(A, x) = \frac{x^T Ax}{x^T x}$ is known as the Rayleigh Quotient. A  theorem in linear algebra states that for a symmetric matrix $A$:

• The maximum value of the Rayleigh Quotient is equal to the largest eigenvalue    ($\lambda_{max}$) of the matrix.

• This maximum occurs precisely when $x$ is the corresponding eigenvector.

 

If $x$ is a unit eigenvector of matrix $A$ corresponding to the eigenvalue $\lambda$, the definition of an eigenvector gives us:

$$Ax = \lambda x$$

To relate this to the quadratic form, we multiply both sides by $x^T$ on the left:

$$x^T Ax = x^T (\lambda x)$$

Since $\lambda$ is a scalar, we can move it outside the product:

$$x^T Ax = \lambda (x^T x)$$

Because $x$ is a unit vector, its length is 1, meaning the dot product $x^T x = \|x\|^2 = 1$. This simplifies our expression beautifully to:

$$x^T Ax = \lambda$$

Finally we are able to know we have to find lambda.

So maximum lambda = (5+ sqrt(5))/2

Answer 1

Let $x = [ x_1 , x_2 ]$ be a unit eigen vector

Given  $\sqrt {x_{12} + x_{22} } = 1$

i.e., $x_{12} + x_{22} = 1$    $\because$ $x$ is a unit Eigen vector

$Ax = \lambda x,$ where $\lambda$ is eigen value

$x^TAx = x^T\lambda x=  \lambda x^Tx  = \lambda [x_1,x_2]^T [x_1,x_2] = \lambda [ x_{12} + x_{22} ] = \lambda (1) = \lambda.$

The maximum value of $\lambda= \frac{5 + \sqrt 5}{2}.$

Hence, maximum value of $\text{x}^{T} Ax$ is $\frac{5 + \sqrt 5}{2}.$

Option B

Comments

because L is a scalar.

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Note:- In above answer 

             x12 + x22    means    x1^2 + x2^2

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Whether multiplication of an eigen vector and its transpose is a scalar?

Why we say x(T):*x = ||x||, where ||x|| is magnitude of vector x, which is scalar.

Answer 2

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

For finding the eigen-value:

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

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

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

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

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

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

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

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

 $\lambda =\frac{5\pm \sqrt{5}}{2}$

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

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

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
$R_{1}\rightarrow\frac{1+\sqrt{5}}{2}R_{1}$

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

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

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

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 $A:$

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

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

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

$\hat{u}=\begin{bmatrix} 0.85\\ 0.52 \end{bmatrix}$  $[$For unit vector $|\hat{u}|=\sqrt{(0.85)^{2}+(0.52)^{2}}=1$ Always$.]$

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

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

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

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

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

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

A unit vector is a vector of length 1, sometimes also called a direction vector.The unit vector $\hat{v}$ having the same direction as a given (non-zero) vector $\vec{v}$ have.

Unit vector is defined as $\hat{v}=\frac{\vec{v}}{|\vec{v}|}$ and $|\hat{v}|=1$  $always.$

Comments

http://mathworld.wolfram.com/Eigenspace.html

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would it right to say that the maximum value will depend on the maximum eigen value is it correct

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@Shaik Masthan Sir,    (X)T . X = (x1)2  + (x2)2 = 1    Always? or X should be a unit vector.

Answer 3

I cannot provide a standard reference for this, but there is a theorem which states that:

The maximum value of $X^{T} A X$ , where the maximum is taken over all X that are the unit eigen vectors of A is the maximum eigen value of A.

The question was of 1 mark so I'm sure they wanted us to use this theorem directly, because solving otherwise would take a lot of time.

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

From here, $λ^{2} - 5λ + 5 = 0$ (Characteristic equation)

So the eigen values of A are 

$\left (\frac{5+√5 }{2} , \frac{5-√5}{2} \right )$

where the maximum eigen value is $\frac{5+√5 }{2}$ . This is the required answer.

Comments

you are correct

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The statement you've provided is known as the Rayleigh quotient theorem.

Answer 4

∵ x is a unit eigen vector

∴ | x | ^ 2 = x.x^T = 1

∴ X^TAX = A = eigen values

And max of eigen values will the ans

Comments

why X^TAX = A? How have u derived this?

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x'Ax=x'λx=λx'x=λ(1)=λ

We can do this because λ is scalar and can be shown in any order when multiplying..

Answer 5

two eigen values $\lambda = \frac{5+\sqrt{5}}{2} ,\frac{5-\sqrt{5}}{2}$

using both one by one calculate value of $x^TAx$

using $\lambda = \frac{5+\sqrt{5}}{2}$ the value of $x^TAx = \begin{bmatrix} 18.131 & 21.231 \\ 21.231 & 34.331 \end{bmatrix}$

using $\lambda = \frac{5-\sqrt{5}}{2}$ the value of $x^TAx = \begin{bmatrix} 14.300 & 6.200\\ -0.700 & 1.200 \end{bmatrix}$

Hence, for $\lambda = \frac{5+\sqrt{5}}{2}$ the value of $x^TAx$ is maximum.

Comments

I am doing same as @ amarVashishth

Is this correct approach @Bikram sir 

https://www.youtube.com/watch?v=3msR2mna0_s but it takes lot of time any other method?

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@ set2018 

Yes, it is correct approach .

No other method i  know.

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The value of x^tAx is Wrong in your solution because  x^tAx is a number i.e 1- by-1 matrix not 2-by-2 matrix.

Answer 6

ans is B.

here, |A-(L*I)| = 0

(3-L)(2--L)-1 = 0

6-3L-2L+L²+1=0

L²-5L+5=0

now find root, L= (5 + root(25-4(1)(5))) / 2(1) = 5+root(5) / 2, which is max value

another root with negative sign which will not be max value.

Comments

But isn't that just the maximum eigen value? How do we find maximum of xTAx?

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This is based on theorem ,

$max\ value\ of\ x^TAx\ =\ max\ eigen\ value\ of\ A $, if $x=$ unit eigen vector of A