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For matrix  $p=\begin{bmatrix} 3 &-2 &2 \\ 0 &-2 &1 \\ 0& 0 & 1 \end{bmatrix}$if one of the eigen values is equal to – 2, then which of the following is an eigen vector?

(a) [ 3 −2 1 ]

(b) [ −3 2 −1 ]

(c) [ 1 −2 3 ]

 (d) [ 5 2 0 ]

 

asked in Linear Algebra by Junior (663 points)
edited by | 181 views
+1
Eigen vector associated to LAMDA = -2 is

[2 ,5,0]
0

In any Triangular Matrix(Upper Triangular or Lower triangular Matrix), Eigenvalues are Leading Diagonal Elements itself.

In this example$:\lambda_{1}=3,\lambda_{2}=-2,\lambda_{3}=1$

0

@srestha Ma'am

please edit option also

1 Answer

+4 votes
Best answer

Give that the matrix $A=\begin{bmatrix} 3 &-2 &2 \\ 0 &-2 &1 \\ 0&0 &1 \end{bmatrix}$

Let suppose the Eigen Vector $X=\begin{bmatrix} x\\ y\\z \end{bmatrix}$

We know that $AX=\lambda X$

$AX-\lambda X=\begin{bmatrix} 0 \end{bmatrix}$

$(A-\lambda I )X=\begin{bmatrix} 0 \end{bmatrix}$---->(1) where $I=\begin{bmatrix} 1 &0 &0 \\ 0&1 &0 \\ 0& 0 &1 \end{bmatrix}$Identity matrix.

$(A-\lambda I )=\begin{bmatrix} 3 &-2 &2 \\ 0&-2 &1 \\ 0& 0 &1 \end{bmatrix}-\lambda\begin{bmatrix} 1 &0 &0 \\ 0&1 &0 \\ 0& 0 &1 \end{bmatrix}$

Given that $\lambda=-2$

$(A-\lambda I )=\begin{bmatrix} 3 &-2 &2 \\ 0&-2 &1 \\ 0& 0 &1 \end{bmatrix}-(-2)\begin{bmatrix} 1 &0 &0 \\ 0&1 &0 \\ 0& 0 &1 \end{bmatrix}$

$(A-\lambda I )=\begin{bmatrix} 3 &-2 &2 \\ 0&-2 &1 \\ 0& 0 &1 \end{bmatrix}+2\begin{bmatrix} 1 &0 &0 \\ 0&1 &0 \\ 0& 0 &1 \end{bmatrix}$

$(A-\lambda I )=\begin{bmatrix} 3 &-2 &2 \\ 0&-2 &1 \\ 0& 0 &1 \end{bmatrix}+\begin{bmatrix} 2 &0 &0 \\ 0&2 &0 \\ 0& 0 &2 \end{bmatrix}$

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

Now,from the equation$(1),$ 

$(A-\lambda I )X=\begin{bmatrix} 0 \end{bmatrix}$

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

Now,we can apply some operation on matrix

$R_{2}\rightarrow3R_{2}$

$\begin{bmatrix} 5 &-2 &2 \\ 0&0 &3 \\ 0& 0 &3 \end{bmatrix}\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} 0\\0 \\0 \end{bmatrix}$

Again apply some operation on matrix

$R_{3}\rightarrow R_{3}-R_{2}$

$\begin{bmatrix} 5 &-2 &2 \\ 0&0 &3 \\ 0& 0 &0 \end{bmatrix}\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} 0\\0 \\0 \end{bmatrix}$

In $R_{3}$,we create Nullity.So Rank of this matrix is $2$ [Rank:Maximum number of independent Rows/Colums]

$ r=2,$ and number of unknowns ( Variables $=3$ )

Here clearly see, $r<n$

So,there are Infinitely many solution are possible,and $n-r=3-2=1 $variable give some value and other will be evaluate accordingly.

$\begin{bmatrix} 5x-2y+2z\\0x+0y+3z \\0x+0y+0z \end{bmatrix}=\begin{bmatrix} 0\\0 \\0 \end{bmatrix}$

Let say $y=k$

$z=0$ ---->(2)

$5x-2k+2z=0$

$5x-2k+0=0$

$5x=2k$

$x=\frac{2k}{5}$

Finally we got the Eigen Vector$,X=\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} \frac{2k}{5}\\ k \\ 0 \end{bmatrix}$

Put the Value of $k=5$

and we get$,X=\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} \frac{2*5}{5}\\ 5 \\ 0 \end{bmatrix}$

$X=\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} 2 \\ 5 \\ 0 \end{bmatrix}$

Actually,we get Infinitely many number of solutions.So,we can get Infinte number of Eigen Vector,but for $k=5$,we got one Eigen Vector.

answered by Boss (36.3k points)
selected by
0
$\begin{bmatrix} 1 & -2 & 2 \\ 0 & -4 & 1 \\ 0& 0& -1 \end{bmatrix}$

rank of this matrix  = 3

n- r = 0 --> No of linearly independent solution  = 0

therefore X =  Y = Z = 0

right ???
0
lambda = -2, but you substituted as +2 in (A−λI).

this is leads to z =0, and 5x-2y =0

5x-2y=0 ===> one eqn but two variables,

therefore, it leads to many solutions..

ex:- x=2,y=5,z=0

and

x=14, y=35, z=0 .... Etc
0

Shaik Masthan 

My approach was is (A - $\lambda$ I) X  

A X =  $\lambda$ X

A is given

$\lambda$  = -2

just put the eigen vectors from the options and check LHS  = RHS

but none of the options are matches

I think that options are wrong

dunno I have doubt in it !!

0
yes those options are wrong... Otherwise i will include the values in example which are matched in the options.

your approach is correct.
0

@Shaik Masthan

what is the correct answer?

0
wrt me

$\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$
0
i think  $[2 ,5,0]$ the answer.
0
@lakshman

already i gave that example in my comment, right?
0
what is the wrong in my answer

i'm not understood your approach

can you please tell me why infinite many solutions are?
0
@Laksman

when u r doing vector operation, u shouldnot do row or column addition or subtraction on it
0
why ma'am?
this is not a vector operation.
this is simple transformation, rank doesn't effect.
0
this step not correct

$\begin{bmatrix} 5 &-2 &2 \\ 0&0 &3 \\ 0& 0 &0 \end{bmatrix}\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} 0\\0 \\0 \end{bmatrix}$

because here u also doing multiplication
0
I think this is right because I want to find the value of $x,y,z?$
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How you take k= 5
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You can take any value of the $k$ and find the answer.

Because in this question an infinite number of Eigen Vector are possible.

I take $K=5$ and get the answer.
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If infinite number of solution are not possible then what k value can be take???
0
Read the highlighted line in the answer.

Rank < Variable.

So this is the case of the Infinite number of solutions, and we assign the $(n-r)$ value to the any variable.

If Infinite number other f solution are not possible then there is no need of $K$

We can easily solve the equation if Rank=Variable(Unique solution)
+1
@Srestha

I m not getting !!! why r u saying that step is not correct. ??? can u explain
0
@akash

actually these steps following no rule

but we can ignore it for this question

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