GATE CSE 1993 | Question: 01.1

Question

The eigen vector $(s)$ of the matrix $$\begin{bmatrix} 0 &0 &\alpha\\ 0 &0 &0\\ 0 &0 &0 \end{bmatrix},\alpha \neq 0$$ is (are)

• A. $(0,0,\alpha)$
• B. $(\alpha,0,0)$
• C. $(0,0,1)$
• D. $(0,\alpha,0)$

Comments

what about (alpha,alpha,0) vectors? will they be eigen vectors too.I think they are.

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Bind blowing explanation satyajeet sir

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Yes they will be eigen vector@ gow

Answer 1

 

 

solution

Answer 2

clearly matrix is upper triangular matrix and for upper triangular ,lower triangular matrix ,diagonal matrix and scaler matrix ,its diagonal element is eigen values.so here eigen value=0,0,0.

now we have to find eigen vector of this matrix (A-λ)X=0  where λ=0 so we get A matrix  [ 0 0 α] has rank=1  means we get two free variable (n-r) 

x1= β x2=γ x3=0 then eigen vector  (x1,x2,x3)=(β,γ,0)= β(1,0,0)+γ(0,1,0) so  B and  D  are  right answer .

 

Answer 3

 

https://youtu.be/VqP2tREMvt0

Check @32:57

After solving the characteristic equation $\det(A – \lambda I_n)=0$ we get $\lambda=0,0,0$. Using this $\lambda=0$ in $A-\lambda I_n$ we have:

 

Here $C_1$ and $C_2$ are any real. We may even choose them as $C_1=C_2=\alpha$ to get options $B$ and $D$.