GATE CSE 2015 Set 3 | Question: 15

Question

In the given matrix $\begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & 0 \\ 1 & 2 & 1 \end{bmatrix}$ , one of the eigenvalues is $1.$ The eigenvectors corresponding to the eigenvalue $1$ are

• A. $\left\{a\left(4,2,1\right) \mid a \neq 0, a \in \mathbb{R}\right\}$
• B. $\left\{a\left(-4,2,1\right) \mid a \neq 0, a \in \mathbb{R}\right\}$
• C. $\left\{a\left(\sqrt{2},0,1\right) \mid a \neq 0, a \in \mathbb{R}\right\}$
• D. $\left\{a\left(- \sqrt{2},0,1\right) \mid a \neq 0, a \in \mathbb{R}\right\}$

Comments

Answer: B. 

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see this

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To find an eigenvector corresponding to the eigenvalue 1

Step 1: Plug in the eigenvector to get $A- \lambda I$

$\begin{bmatrix} 0 & -1&2 \\ 0&0 &0 \\ 1&2 & 0 \end{bmatrix}$

 

Step 2: Use $(A- \lambda I)x=0$

$\begin{bmatrix} 0 & -1&2 \\ 0&0 &0 \\ 1&2 & 0 \end{bmatrix}$ $\begin{bmatrix} x\\ y\\ z\\ \end{bmatrix}=0$

 

So,

$-y+2z=0$

and

$x+2y=0$ (Only Option B satisfies this)

 

Now, make as many linearly dependent vectors out of it. Option B has one such.

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{a(−4,2,1)∣a≠0,a∈R}

@Lakshman Patel RJIT   , @JashanArora

Can anyone pls explain

What does  a(-4,2,1) and a!=0 means in the option B ?

 

P.S:

I understand the solution, but confused about what  a(-4,2,1) and a!=0 actually means

Answer 1

check that perspective 

Answer 2

SImply putting value of $\lambda$ in Equation, we get

$\lambda$=1

A-$\lambda$I=0

A-I=0 (putting $\lambda$=1)

Answer 3

We can use the following property :

Sum of eigen values = trace of matrix

Say a, b, 1 are the 3 eigen values

=> a+b+1 = 3

=> a+b = 2

Among the following options only OPTION B satisfies this and the other options can be eliminated

Answer 4

Eigen value is that value by which eigen vectors gets scaled up/down upon linear transformation done by the given matrix. 

In the given problem, eigen value = 1.  This implies that the vectors on linear transformation (matrix multiplication) do not change at all. We can simply find matrix-vector product for each option and check for which option is the product same as the original vector. 

Ans: B