GATE CSE 2025 | Set 1 | Question: 31

Question

Let $A$ be a $2 \times 2$ matrix as given.
$$ A=\left[\begin{array}{rr} 1 & 1 \\ 1 & -1 \end{array}\right]$$

What are the eigenvalues of the matrix $A^{13}$ ?

• A. $1,-1$
• B. $2 \sqrt{2},-2 \sqrt{2}$
• C. $4 \sqrt{2},-4 \sqrt{2}$
• D. $64 \sqrt{2},-64 \sqrt{2}$

Comments

For 2 marks this is very easy question

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Let's not go too deep, lets just two statements from linear algebra:

1. Product of eigen values of A = Determinant of A.
2. $|A^k| = |A|^k$, note that the $k$ has taken outside.

 

Here for given matrix the $|A| = (1 \times -1) - (1 \times 1) = -2$  then $|A^{13}| = |A|^{13} = (-2)^{13} = -2^{13}$

 

Take product of eigen values mentioned in each option, by the time you came across option D:

For option D:- $(64 \times \sqrt{2}) \times (-64 \times \sqrt{2}) = - (2^6 \times 2^6 \times 2) = -2^{13}$

 

So the answer is option D.

 

Checking the trace is not an useful option here, Since for every option the sum of eigen values is zero.

 

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If the eigenvalue of a matrix $A$ is $\lambda$, then the eigenvalue for the matrix $A^k$ is $\lambda^k$.

Furthermore, the eigenvector remains exactly the same!

if someone is interested in proof:

the 

eigenvalue and eigenvector for a matrix $A$:

$$Av = \lambda v$$

Now, let's see what happens if we multiply both sides on the left by matrix $A$ again:

$$A(Av) = A(\lambda v)$$

$$A^2 v = \lambda (Av)$$

Since we know that $Av = \lambda v$, we can substitute that back into the right side:

$$A^2 v = \lambda (\lambda v)$$

$$A^2 v = \lambda^2 v$$

This shows that for $A^2$, the eigenvalue is $\lambda^2$. If you repeat this process by multiplying by $A$ a total of $k$ times, you get the general rule:

$$A^k v = \lambda^k v$$

Note that the vector on the left side ($v$) is the exact same vector on the right side ($v$). It was never swapped or modified; it was only scaled by a larger multiplier ($\lambda^2$). This pattern continues for any power $k$.

Answer 1

Option - d

Answer 2

    Given matrix     $A=\left[\begin{array}{rr} 1 & 1 \\ 1 & -1 \end{array}\right]$

Characteristic Equation is $ |A  -\lambda I| = 0$

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

$\implies (1-\lambda)(-1-\lambda)-1=0$

$\implies -1-\lambda+\lambda+\lambda^{2}-1=0$

$\implies \lambda^{2}-2=0$

$\implies \lambda=+\sqrt 2 $ and $-\sqrt 2$

According to properties of Eigen values,

eigen values of $A^{13}=$ (eigen value of A)$^{13}$

$=(\sqrt 2)^{13}$ and $(-\sqrt 2)^{13}$

$=(\sqrt 2)^{13}$  $= 2^{\frac{13}{2}} = 2^6 \sqrt 2 = 64 \sqrt 2$

$=(- \sqrt 2)^{13} = (- \sqrt 2)^{13} = - 64 \sqrt 2$

Hence, Ans is option (D).

Credit : Narayan Kunal 

Answer 3

Let x , y are the eigen values

then, x+y=0 , xy = det A = -2

x=root2  ,  y= - root2

if eigen value of A is x , then Eigen value of A^13 is x^13

so x^13 = 64root2 , y^13= -64root2

Answer 4

Easy Question it's better to remember 2x2 shortcut to Directly get the Characteristics Equation 

Answer 5

|A| = -2  ,   |A|^13  =  -2  ^13

det(A) = product of eigen values 
so option (D) 

Answer 6

using property of eigenvalues i.e., if the eigenvalue of matrix A is K, then the eigenvalue of A^13 will be K^13.  Therefore, option D is correct answer