Have you checked out the rules? Pretty cool stuff there!
In particular, we cannot provide answers to your questions. We can (and will, to the best of our ability) help you find it. So, you have to show us some effort from your side.
If you know nothing about eigenvectors, Google is a good place to start.
hey sourabh, i did try to solve the question.
i got eigen values as 0, 0 , 0
and after using [A-lambdaI]X=0
i am getting 0X1+0X2+aX3=0
which makes eigen vector as [0 0 0 ]
whereas the answer is [0 0 a]
[0, 0, 0] cannot an eigenvector. But [0 0 a] is also not correct. The definition of "eigenvalue" is that [itex]\lambda[/itex] is an eigenvalue of A if and only if there exist a non-zero vector, v, such that [itex]Av= \lambda v[/itex]. Yes, the only eigenvalue is 0. Applying the given matrix to a vector [x, y, z] gives [az, 0, 0] and that must be equal 0[x, y, z]= [0, 0, 0] so we must have z= 0. We have NO information about x or y so they can be anything.
sorry hallsofevy, the correct answer is [a 0 0]
but after solving the eqn [A-lamdaI][X]=0 or AX=lamdaX
i got az+0+0 = 0
so how to proceed further and arrive at the correct answer?
What do you mean "the answer is ..."? Do you understand what an eigenvector is? If a given vector is an eigenvector so is any multiple of it. You cannot just say "the eigenvector" is any specific vector. I said before, "we must have z= 0. We have NO information about x or y so they can be anything."
That includes your [a, 0, 0], taking x= a, y= 0. It also includes [1, 0, 0], [0, 0 1], and any linear combination x[1, 0, 0]+ y[0, 1, 0]= [x, y, 0].
For any x, y,
[tex]\begin{bmatrix}0 & 0 & a \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}x \\ y \\ 0\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}= 0\begin{bmatrix}x \\ y\\ 0\end{bmatrix}[/tex]