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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. $\frac{(5 - √5)}{2}$
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3 votes

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.

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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.

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unit vector in 2D space = draw circle with center as (0,0) with radius 1 unit

all points lying on the circle are unit vectors.

unit vector in 3D space = draw hollow sphere with center as (0,0,0) with radius 1 unit

all points lying on the surface area of sphere are unit vectors.

so in any case distance from (0,0) to that vector is 1.

now we can define anyyy vector which is arbitrary , as an unit vector.

say, (7,4) in 2D space,

( 7/sqrt[7^2+4^2]  , 4/sqrt[7^2+4^2] ) is the unit vector corresponding to (7,4)

for recheck find distance from (0,0) ; it will be 1

now given A a matrix, let (p,q) is its E. vector

now the corresponding unit E. vector is (r,s) column vect. say.

r=p/sqrt(p^2+q^2)

s=q/sqrt(p^2+q^2)

so, xT.A.x= xT.L.x (as Ax=Lx) =L.xT.x =L.(r,s).TR(r,s) =L.(r^2+s^2) =L.1= L

so maximum e.val. is the answer.
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It is given that $A$ is $2\times2$ matrix.

$=>$ At most two Eigen Values are possible

$=>$ Two Eigen Vectors are possible corresponding to each Eigen Value

$=>$ For one Eigen Vector there is specific only one Unit Eigen Vector possible (therefore total two Unit Eigen Vectors possible)

 

Let’s say the Eigen Vectors in this case are $x_1$ and $x_2$.

We need to find $max(x_1^{T}Ax_1, x_2^{T}Ax_2)$ as maximum is taken over all $x$ that are Unit Eigen Vectors, and only Two unit Eigen Vectors are possible in this case.

(We can try to find $x_1$ and $x_2$ in this case, and take the maximum out of it. But let’s try one easier method.)

$=>$ $x^{T}Ax$

[$\because$ Cayley-Hamilton Theorem suggests every square matrix is its own characterstic equation $=>$ Meaning A can be replaced by $\lambda$]

$=>$ $x^{T}\lambda x$

$=>$ $\lambda x^{T}x$

[$\because$ We are only searching in Unit Eigen Vectors, $x^{T}x = 1$]

$=>$ $\lambda$

This point derives that the question is only asking for maximum Eigen Value.

 

$=>$ $\begin{bmatrix}
3-\lambda & 1\\
1 & 2-\lambda
\end{bmatrix} $

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

$=>$ $\lambda^{2} – 5 \lambda + 5 = 0$

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

And $max(\frac{5\pm\sqrt{5}}{2}) = \frac{5+\sqrt{5}}{2}$

 

$\therefore$ Answer is B.
Answer:

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