made easy ME previou year question

Question

which on of the following is an eigenvector of the matrix

[5  0  0  0

 0  5  5  0

 0  0  2  1

 0  0  3  1]

a)  [1 -2 0 0 ]^t

b) [0 0 1 0]^t

c) [1 0 0 -2]^t

d) [1 -1 2 1]^t

 

Comments

is it A?

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it is but i need the logic behind it

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We know that AX=⋋X where A is the matrix given, X is the eigen vector and ⋋ is the eigen value.

Our A is a upper triangular matrix and the diagonal elements are the eigen values i.e. 5,5,2,3.

In such questions try matching with the options because it takes lesser time.

Starting with Option A,

AX=⋋X and we put X=(1 -2 0 0)^T as it is the eigen vector.

$\begin{bmatrix} 5 & 0& 0& 0\\ 0 & 5& 5& 0\\ 0 & 0& 2& 1\\ 0 & 0& 3& 1 \end{bmatrix} \begin{bmatrix} 1\\ -2\\ 0\\ 0 \end{bmatrix} = \begin{bmatrix} 5\\ -10\\ 0\\ 0 \end{bmatrix} => \begin{bmatrix} 5 & 0& 0& 0\\ 0 & 5& 5& 0\\ 0 & 0& 2& 1\\ 0 & 0& 3& 1 \end{bmatrix} \begin{bmatrix} 1\\ -2\\ 0\\ 0 \end{bmatrix} = 5\begin{bmatrix} 1\\ -2\\ 0\\ 0 \end{bmatrix}$

This is of the form AX=⋋X where ⋋=5 (one of the eigen values).

If you put any other options and calculate you won't be able to take out anything common.

For eg: Take option D where X=(1 -1 2 1)^T

$\begin{bmatrix} 5 & 0& 0& 0\\ 0 & 5& 5& 0\\ 0 & 0& 2& 1\\ 0 & 0& 3& 1 \end{bmatrix} \begin{bmatrix} 1\\ -1\\ 2\\ 1 \end{bmatrix} = \begin{bmatrix} 5\\ 5\\ 5\\ 7 \end{bmatrix}$

From RHS you cannot take anything common such that it gets in the form of ⋋(1 -1 2 1)^T

 

 

 

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i think so if this is a triangular matrix than 5,5,2,1 should be the eigen values.

Answer 1

$(A-\lambda I)X=0$

$\text{take first option}$

$\begin{bmatrix} 5-\lambda &0 &0 &0 \\ 0&5-\lambda & 5 &0 \\ 0&0 &2-\lambda &1 \\ 0& 0&3 &1-\lambda \end{bmatrix}\begin{bmatrix} 1\\ -2\\ 0\\ 0 \end{bmatrix}$ =$\begin{bmatrix} 0\\ 0\\ 0\\ 0 \end{bmatrix}$

consider the first row -- $5-\lambda+0=0$

$\equiv$ $\lambda=5$ it is eigen value for this eigen vector [1 -2  0 0]^t

$Ax=\lambda x$

$\begin{bmatrix} 5 &0 &0 &0 \\ 0&5 & 5 &0 \\ 0&0 &2&1 \\ 0& 0&3 &1\end{bmatrix}\begin{bmatrix} 1\\ -2\\ 0\\ 0 \end{bmatrix}$=$\begin{bmatrix} 5\\ -10\\ 0\\ 0 \end{bmatrix}$

$\begin{bmatrix} 5 &0 &0 &0 \\ 0&5 & 5 &0 \\ 0&0 &2&1 \\ 0& 0&3 &1\end{bmatrix}\begin{bmatrix} 1\\ -2\\ 0\\ 0 \end{bmatrix}$=$5$$\begin{bmatrix} 1\\ -2\\ 0\\ 0 \end{bmatrix}$

we can see $\lambda =5$

so first is correct option.