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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 $A?$

  1. $5$
  2. $\frac{(5 + √5)}{2}$
  3. $3$
  4. $\frac{(5 - √5)}{2}$
in Linear Algebra edited by
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We have give matrix $A_{2 x 2}$  which means it can have max 2 EigenVectors say $x_{1} ,x_{2}$  and their   eigenValue $\lambda_{1},\lambda_{2} $ respectively

We know if $x$ is EigenVector of A then :

$Ax =\lambda x$ …..eq1

 if we solve $x^{T}Ax  =  \lambda x^{T}x$ (as $Ax = \lambda x $ from eq1) .….eq2
 $ \lambda x^{T}x = \lambda||x||^{2}$

(Here  $ x^{T}x $ is a dot product  and  $ x^{T}x = ||x||^{2} =$ sum of sqaure of elements of vector $ x$
And in the question it is given at we have to consider only unit eigen vectors among all eigen vectors i.e. $||x|| = 1$
where ||x|| is also known as $Distance$ )

$ \lambda x^{T}x = \lambda||x||^{2} = \lambda $ (as $||x|| =1$ ,given)
After simplification we just have  to  find $max(\lambda_{1},\lambda_{2})$

$|A – \lambda I| = 0$

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

$\lambda^{2} – 5\lambda + 5 = 0$ (on solving this quadratic equation)

$\lambda = \frac{5\pm \sqrt{5}}{2} $

max of $\lambda = \frac{5 + \sqrt{5}}{2}$

Option B is correct

@Sachin Mittal 1

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amazing
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very good explanation
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8 Answers

78 votes
78 votes
Best answer

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

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4 Comments

edited by
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.
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39 votes
39 votes

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 $|\hat{v}|=1$  $always.$

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4 Comments

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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.

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23 votes
23 votes

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.

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2 Comments

you are correct
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The statement you've provided is known as the Rayleigh quotient theorem.
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5 votes
5 votes
∵ 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

2 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..
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