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+2 votes

The number of linearly independent Eigen vectors of ais  


a) 0

b) 1 

C) 2

d) infinite

asked in Linear Algebra by Veteran (16.9k points) | 766 views

2 Answers

+3 votes
Best answer

eigen values are $\lambda1=2,\Lambda2 = 2$

this eigen value must satisfy the characteristic equation,

$\left | A- \lambda I \right | X= 0$

$\Rightarrow \begin{bmatrix} (2-2) &1\\ 0& (2-2) \end{bmatrix}*\begin{bmatrix}x1\\ x2 \end {bmatrix} = 0$

$\Rightarrow \begin{bmatrix} 0 &1\\ 0& 0 \end{bmatrix}*\begin{bmatrix}x1\\ x2 \end {bmatrix} = 0$

$\Rightarrow 0*x1 + 1*x2 = 0 $

$\Rightarrow x2 = 0 $

let  $ x1 =k$

So the eigen vector corresponding to $\lambda1 = 2 $ will be $\begin {bmatrix} k\\ 0 \end {bmatrix} $

There are possible infinite many eigen vectors but all those linearly dependent on each other because you always get some constant to satisfy $C1*X1 + C2*X2 = 0$ So only one linearly independent eigen vector possible.


answered by Boss (9.2k points)
selected by
Which is that one linearly independant eigen vector?
+8 votes

No of linearly independant (distinct) eigen vectors = No of distinct eigrn values.

Here, as it is an upper triangular matrix, eigen values are 2,2. Distinct value = 2

Thus. only 1 vector. Hence, option (B).


Btw, lot of people are confused between linearly independent eigen vectors, linearly independent solutions and linearly independent Rows or Columns. Here is a quick solution to solve such problems....

**Linearly Independent ............

1) Eigen Vectors : No. of distinct Eigen values

2) Solutions : N-R (i.e, Variables - Rank)

3) Rows or Columns : R 


There are many questions in the past for all 3 types mentioned above. Do practice it well :)

answered by Loyal (2.9k points)

@Tushar , sometimes same Eigen Value can also generate multi-dimensional Eigen Vectors.

                  And Answer to this was given as (d) infinite , but i guess that is wrong.smiley 

(D) would have been correct if they had asked for 'number of eigen vectors?'..

And even multidimensional eigen vectors sound complex. I guess, they won't give such qstn in exam ;)

for 1st Q. answer should be 2 as only 2 eigen values possible (0 and n)

And I was unable to solve 2nd (other than eigen values are all real because its a symmetric matrix), until I read it as a standard solution that "Eigen vectors for (real) Symmetric matrix are are pair-wise orthogonal ". That means their dot product should be 0. Thus a.b = 0. Option (D)

(P.S. - Eigen values of hermitian matrices are also real and they posses some similar properties like symmetric matrices except few. So, Above logic applies to them as well, ie., Hermitian Matrix has orthogonal vectors for distinct eigen values.)

I guess answer should be n instead of '2'.

try to solve for Eigen Value 0 - U will get n-1  constants in Eigen Vector.

What does this Vector signify ?  How many linealy Independent Eigen Vector it may give ?
Total 'n' solutions are there for that matrix. (n-1) being 0 and one being 'n'. So total 2 distinct eigen values => 2 eigen vectors
Ok, what I am saying is try to find Eigen Vector for Eigen Value 0.

Ohh.. So dumb of me... And well then the relation  n​​​x1+x2+x3+.....+xn = 0 specifies that vector is "Any non-zero vector where sum on all (n-1)components (from x2 to xn) should be equal to negation of order with first component (i.e. x1)".  

And it will still give only 1 independent vector as 'n' is constant. So , two vectors created with eigen val 0 will be proporional. So, independent is only 1. 

I think , It should be   X1 + X2 + X3 + . . . + Xn = 0

How u got nX1 ?

So, here we have one equation, n Unknowns , so it will give n-1 constants -

 Assuming, X1 = k1

X2 = k2
X3 = k3
Xn-1 = kn-1

This given  Xn = -(k1 + k2 + k3 + . . . + K n - 1 ) 
This gives how many independent vectors ?

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