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The product of the non-zero eigenvalues of the matrix is ____

$\begin{pmatrix} 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 1 & 0 & 0 & 0 & 1 \end{pmatrix}$
asked in Linear Algebra by Veteran (106k points) | 6.8k views
0
Is there some short method of solving this...?
0
Can solve this by using the characteristic equation (A-λI)X = 0 which gives AX=λX. where λ is eigen value and X eigen vector. but i suspect there is a shorter method for solving this. Anybody?
+4
I too suspected a short cut but I think that they have given this just for confusion that students keep thinking about shortcut.. I don't think that it had a shortcut. And the reality is that the time we spent while searching shortcut is more than if we would have solved it with straight method.
0

it will be helpful to understand why ''since rank of matrix is 2, we can have maximum 2 non-zero eigen values."

https://math.stackexchange.com/questions/146927/relation-between-rank-and-number-of-non-zero-eigenvalues-of-a-matrix

6 Answers

+114 votes
Best answer
Since rank of matrix is $2,$ we can have maximum $2$ non-zero eigen values.

$$\begin{bmatrix}1 &0&0&0&1\\0 &1&1&1&0\\0 &1&1&1&0\\0 &1&1&1&0\\1 &0&0&0&1\\\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix} = \lambda \begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix}$$

$$\because AX = \lambda X$$

$x_1 + x_5 = \lambda x_1$ (from first row)
$x_1 + x_5 = \lambda x_5$ (from last row)
Adding these two,
$2(x_1+x_5) = \lambda(x_1+x_5)$
$\implies \lambda = 2.$

Similarly,
$x_2+x_3+x_4 = \lambda x_2$
$x_2+x_3+x_4 = \lambda x_3$
$x_2+x_3+x_4 = \lambda x_4$
Adding these three,
$3(x_2+x_3+x_4) = \lambda(x_2+x_3+x_4)$
$\implies \lambda = 3.$

So, product of non-zero eigenvalues of the matrix $=2 \times 3 = 6.$
answered by (273 points)
selected by
+4
best method @subhashini
+7

JFI 

Elementary row operations change eigenvalues of given matrix (https://socratic.org/questions/do-elementary-row-operations-change-eigenvalues)

Sometime we can not use row and column transformation together. 

https://www.quora.com/Can-we-use-both-row-and-column-transformation-in-same-question-of-matrices

Notice -> Row or column transformation is basically a kind of matrix multiplication in original matrix to get another matrix. 

+3

Chhotu This is really most imp point .

"Elementry row transformation does not affect the ORDER and RANK  of the matrix ".

0
very good approach

@Subhashini
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But in exam won't you get confused if there are more eigen values other than 0,3,2?
0
This method is less time consuming
0
I think this is easy,  but we will need one more point here,  

No.  Of non zero eigen values <= Rank of the matrix

So the only non zero eigen values can be 2 and 3 ( As the rank of the matrix is 2 )
+1
I have a doubt here.

You did $2(x_1+x_5)=\lambda(x_1+x_5)$.................(1)

and

$3(x_2+x_3+x_4)=\lambda(x_2+x_3+x_4)$.................(2)

and from 1 your $\lambda=2$ and from 2 your $\lambda=3$

But my query is that the eigen value problem is $AX=\lambda X$ ........(3)

for the components $x_1$ and $x_5$ your $\lambda$ takes value 2 and for components $x_2,x_3,x_4$ your $\lambda$ takes value 3 two values for one vector at a same time?

I mean by (3)

$\lambda \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\\ x_5 \end{bmatrix}$

how $\lambda$ is different for different components of vector X?It should be same right for all components of X?
0
@Ayush, It should be like,

 

$\begin{bmatrix} 1& 0& 0& 0& 1\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 1& 0& 0& 0& 1\\ \end{bmatrix} \begin{bmatrix} x1\\ x2\\ x3\\ x4\\ x5 \end{bmatrix} + \begin{bmatrix} 0& 0& 0& 0& 0\\ 0& 1& 1& 1& 0\\ 0& 1& 1& 1& 0\\ 0& 1& 1& 1& 0\\ 0& 0& 0& 0& 0\\ \end{bmatrix} \begin{bmatrix} x1\\ x2\\ x3\\ x4\\ x5 \end{bmatrix} =$ $2\begin{bmatrix}x1 \\0 \\ 0 \\ 0 \\ x5 \end{bmatrix} + 3\begin{bmatrix}0 \\x2 \\ x3 \\ x4 \\ 0 \end{bmatrix}$
0
Okay, got it. Thanks Shubhgupta :)
0

@Shubhgupta

I am also having Ayush's doubt but unable to understand your explanation. Can you please elaborate on it further. 

+46 votes

We can see that the rank of the given matrix is 2 (since 3 rows are same, and other 2 rows are also same). Sum of eigen values = sum of diagonals. So, we have two eigen values which sum to 5. This information can be used to get answer in between the following solution. 

Let Eigen value be X. Now, equating the determinant of the following to 0 gives us the values for X. To find X in the following matrix, we can equate the determinant to 0. For finding the determinant we can use row and column additions and make the matrix a triangular one. Then determinant will just be the product of the diagonals which should equate to 0. 

1-X 0 0 0 1
0 1-X 1 1 0
0 1 1-X 1 0
0 1 1 1-X 0
1 0 0 0 1-X

R1 ← R1 + R5, R4 ← R4 - R3

2-X 0 0 0 2-X
0 1-X 1 1 0
0 1 1-X 1 0
0 0 X -X 0
1 0 0 0 1-X

Taking X out from R4, 2-X from R1, (so, X = 2 is one eigen value)

1 0 0 0 1
0 1-X 1 1 0
0 1 1-X 1 0
0 0 1 -1 0
1 0 0 0 1-X

R2 ← R2 - R3, R5 ← R5 - R1

1 0 0 0 1
0 -X X 0 0
0 1 1-X 1 0
0 0 1 -1 0
0 0 0 0 -X


C3 ←  C3 + C4

1 0 0 0 1
0 -X X 0 0
0 1 2-X 1 0
0 0 0 -1 0
0 0 0 0 -X

Taking X out from R2

1 0 0 0 1
0 -1 1 0 0
0 1 2-X 1 0
0 0 0 -1 0
0 0 0 0 -X

R3 ← R3 + R2

1 0 0 0 1
0 -1 1 0 0
0 0 3-X 1 0
0 0 0 -1 0
0 0 0 0 -X

Now, we got a triangular matrix and determinant of a triangular matrix is product of the diagonal. 

So (3-X) (-X) = 0 => X = 3 or X = 0. So, X = 3 is another eigen value and product of non-zero eigen values = 2 * 3 = 6. 

https://people.richland.edu/james/lecture/m116/matrices/determinant.html

answered by Veteran (363k points)
edited by
0
is it possible to have x= 2 , 2 , 1 and product as 4.
0
No. That is not possible. How you got that?
0
i can see how u r  getting 2 , bt how u get 3 nt clear.
0
Sorry. There was some mistake in the derivation. Have corrected it now.
+1
@arjun sir, can we use column and row transformation interchangeably during elimination ?
+4
yes, for finding determinant we can do both.
+1
A matrix doesn't change by row or colum transformations, so if matrix is changed to upper triangular matrix by applying few row transformations. then its easy.

$R_5 \leftarrow R_5-R_1$

$R_3 \leftarrow R_3-R_2$

So on......
0
@Arjun Sir Can we apply row and coloumn operations interchangeably?
0
As product of eigen value of any matrix is determinant of the matrix and here determinant of this matrix is 0. then why not answer is 0?
+3
See, they are asking product of two non-zero Eigen value. Product of two non-zero real number can't be zero!!
0
Can we apply direct elimination method to reduce the matrix and after this we have to find eigenvalues of the given matrix [email protected] Sir @Bikram Sir

(1 0 0 0 1

0 1 1  1 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0)

is this correct but according to this trace of matrix =2 ?
0

set2018 

No, we should do that in straight forward way.. in this question no such elimination method will work to reduce the matrix.

you must see above comment by @subhashini and  also see best answer for it.

0
@Bikram sir I have a general doubt that

is itright way to eliminate/reduce matric before calculating eigen values of matrix ?
0
@ set2018

No, it is not right way to reduce matrix .
In fact in this matrix you don't need to reduce it , solve it straight way means solve it without reducing .
0
does rank of matrix = no. of eigen values  , as in 3rd line of solution how came to know that there will be two eigen values .    Also from where did u got eigen value as 2 ( 3 and 0 are fine which are got from the equation)  ??
0
Why we not change the matrix into upper triangular and then take the diagonal elements as Eigen values
0
@Arjun sir, from where did we get 2?
+3 votes

 We can use the concept of "Block Matrices" to solve this problem quicker. So, its kinda of a shortcut...I have attached my solution using block matrices below.

answered by (207 points)
+1
you can reduce this by applying det. property----->

if

 $\begin{vmatrix} k &k &k \\ k &k &k \\ k &k &k \end{vmatrix}_{n*n}$

then no. of $(n-1)$ eigen values are $0$.

and $n^{th}$ eigen value is = $n*k$.
+1 vote

Here is what my analysis and I was not satisfied with any analysis so I posted this let me know if it is correct.

By seeing the matrix given, the sum of eigenvalues=5 and product(the determinant of the matrix)=0.

Let our five eigen values be $\lambda_1,\lambda_2,\lambda_3,\lambda_4,\lambda_5$

$\lambda_1+\lambda_2+\lambda_3+\lambda_4+\lambda_5=5$

and

$\lambda_1.\lambda_2.\lambda_3.\lambda_4.\lambda_5=0$

Since, determinant is 0, atleast one of the eigen values is 0. I assume say $\lambda_5=0$

Now, I can rewrite $\lambda_1+\lambda_2+\lambda_3+\lambda_4=5$

one of the possible eigen values I can think of is 1,1,1,2 but is 1 really an eigen value? If yes, then for the matrix $A-1.I$, you must be able to find a non-zero vector X such that $|A-I|.X=0$, means the determinant of $|A-I|$ (where A is our matrix given in the question) must be 0.

$|A-I|=$$\begin{bmatrix} 0 & 0 & 0 & 0& 1\\ 0& 0 &1 & 1& 0\\ 0 & 1 & 0&1 &0 \\ 0 &1 &1 & 0 &0\\ 1 & 0 &0 &0 & 0 \end{bmatrix}$

In this, all the rows and columns are independent and hence for this determinant is not zero.Hence, 1 is not our eigen value.

is 2 an eigen value?

$|A-2I|=$$\begin{vmatrix} -1& 0 & 0&0 &1 \\ 0& -1 & 1 &1 &0 \\ 0 & 1& -1& 1 &0 \\ 0 &1 & 1 & -1& 0\\ 1 & 0 &0 &0 & -1 \end{vmatrix}$

To this, vector X=$\begin{bmatrix} 1\\ 0\\ 0\\ 0\\ 1 \end{bmatrix}$ is such that $|A-2I|.X=O$

Hence, 2 is one of the eigen value.

Now, say $\lambda_1=2$ and we assumed $\lambda_5=0$

Now, $\lambda_2+\lambda_3+\lambda_4=3$

The only way this is possible is( 1 is not an eigenvalue), if one of the eigenvalues is 3 and others are zero.

so let us assume $\lambda_2=3$ and $\lambda_3=\lambda_4=0$ and 0 is an eigenvalue for the given matrix so this combination validates our equation

$\lambda_1+\lambda_2+\lambda_3+\lambda_4+\lambda_5=5$ and

$\lambda_1.\lambda_2.\lambda_3.\lambda_4.\lambda_5=0$

Any other eigen values possible?, no because atleast 1 of them has to be 0 and 1 is not an eigen-value.!!

Hence. answer : $3*2=6$

 

answered by Boss (16.1k points)
0 votes
Rank is less then n so det=o

Trace=5

So non-zero eigen values r 2 and 3
answered by Junior (609 points)
0
rank = ?
0
@Gaterank1

Since Determinant of Given Matrix is 0 AND Rank of Given Matrix is 2

and Trace = 5 then How can u say that eigen values are  must be 2 and 3

it may be 1 and 4

becz in both case we got Trace=5
0
@arjun sir

i apply transformations on row  

apo i get this one

1 0 0 0 1

0 1 1 1 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

now its an upper triangular matrix

for which eigen values are the diagonal elements

the only non zero diagonal element is 1

i am no where near the right answer :(

what went wrong
0
same doubt as @aish, please any1 comment, what went wrong in this approach.
–1 vote
The characteristic equation is : | A - zI | = 0 , where I is an identity matrix of order 5. i.e. determinant of the below shown matrix to be 0. 1-z 0 0 0 1 0 1-z 1 1 0 0 1 1-z 1 0 0 1 1 1-z 0 1 0 0 0 1-z Now solve this equation to find values of z. Steps to solve : 1) Expand the matrix by 1st row. (1-z) [ ( 1-z , 1 , 1, 0 ) (1 , 1-z, 1, 0) (1, 1, 1-z, 0) (0, 0, 0, 1-z ) ] + 1. [ ( 0, 1-z, 1, 1 ) ( 0, 1, 1-z, 1) ( 0, 1, 1, 1-z )( 1, 0, 0, 0) ] Note: ( matrix is represented in brackets, row wise ) 2) Expand both of the above 4x4 matrices along the last row. (1-z)(1-z) [ (1-z, 1, 1) (1, 1-z,1) (1 , 1, 1-z ) ] + 1.(-1) [ (1-z, 1, 1) (1, 1-z,1) (1 , 1, 1-z ) ] 3) Apply row transformations to simplify above matrices ( both the matrices are same). C1 <- C1 + C2 + C3 R2 <- R2- R1 R3 <- R3 - R1 result is : (1-z)(1-z) [ ( 3-z, 1, 1) (0, -z, 0) (0, 0, -z ) ] - 1. [ ( 3-z, 1, 1) (0, -z, 0) (0, 0, -z ) ] 4) Solve the matrix by expanding 1st column. result is : (1-z)(1-z)(z)(z) - (3-z)(z)(z) = 0 solve further to get : z^3 ( 3-z ) ( z-2 ) = 0 hence z = 0 , 0 , 0 , 3 , 2 Therefore product of non zero eigenvales is 6.
answered by Loyal (8.6k points)
+2
Wat hav u done ??? very tough to understand ...
0
:D this is clearly a copy paste
Answer:

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