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

$\begin{pmatrix} 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 1 & 0 & 0 & 0 & 1 \end{pmatrix}$
| 12.8k views
0
Is there some short method of solving this...?
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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?
+10
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

+1

@Arjun sir we know the sum of eigen value = sum of diagonal element.

can we solve this question by using intuition becoz here the sum of eigen value is 5 . 2+3 or 3+2 so we can product them.. 2*3=6

0 is not allowed cleardy said in question.

correct me if i am wrong.

+3

1+4 is also equal to 5 ! so we can't crack this way

0
How to do this if you have just 30 seconds in the exam:-

Rank = 2, clearly. (Only two linearly independent rows) This means non-zero eigenvalues are 2.

Sum of eigenvalues = trace of the matrix. = 5.

So, it all boils down to: x + y = 5.
What could be x and y? 1,4 and 2,3.

So, the answer is either 4 or 6. Choose one.

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)

$2(x_1+x_5) = \lambda(x_1+x_5)$
$\implies \lambda = 2.$

Similarly,
$x_2+x_3+x_4 = \lambda x_2$
$x_2+x_3+x_4 = \lambda x_3$
$x_2+x_3+x_4 = \lambda x_4$

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

So, product of non-zero eigenvalues of the matrix $=2 \times 3 = 6.$

selected by
+7
best method @subhashini
+12

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.

+5

Chhotu This is really most imp point .

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

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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?
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This method is less time consuming
+1
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 )
+7
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?
+10
@Ayush, It should be like,

$\begin{bmatrix} 1& 0& 0& 0& 1\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 1& 0& 0& 0& 1\\ \end{bmatrix} \begin{bmatrix} x1\\ x2\\ x3\\ x4\\ x5 \end{bmatrix} + \begin{bmatrix} 0& 0& 0& 0& 0\\ 0& 1& 1& 1& 0\\ 0& 1& 1& 1& 0\\ 0& 1& 1& 1& 0\\ 0& 0& 0& 0& 0\\ \end{bmatrix} \begin{bmatrix} x1\\ x2\\ x3\\ x4\\ x5 \end{bmatrix} =$ $2\begin{bmatrix}x1 \\0 \\ 0 \\ 0 \\ x5 \end{bmatrix} + 3\begin{bmatrix}0 \\x2 \\ x3 \\ x4 \\ 0 \end{bmatrix}$
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Okay, got it. Thanks Shubhgupta :)
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I am also having Ayush's doubt but unable to understand your explanation. Can you please elaborate on it further.

0
How is the rank of matrix is 2?

Thank You
+9

@guptashekhar

Following  are the only two linearly independent rows in the given matrix.

$\begin{bmatrix} 1& 0&0 & 0 &1 \\ 0& 1& 1& 1& 0 \end{bmatrix}$

and rank is equal to the number of linearly independent rows.

So the rank of the given matrix is 2.

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Thanks Annu
0

@ Is it like for same matrix two different vectors are streched by 2 & 3 units respectively.

my doubt is then before streching what was the actual vector ???

or, $\begin{bmatrix} x1\\ x2\\ x3\\ x4\\ x5\\ \end{bmatrix}$ is the vector formed by two basis vector   2$\begin{bmatrix} x1\\ 0\\ 0\\ 0\\ x5\\ \end{bmatrix}$ + 3$\begin{bmatrix} 0\\ x2\\ x3\\ x4\\ 0\\ \end{bmatrix}$

I'm confused after seeing your expression.please clear my doubt.

0
but elementary transformation changes determinant of the matrix  am i right
+1
Can anyone plss explain how this statement is valid "since rank of matrix is 2 so we can have maximum 2 non zero eigen values"
+1
It is due to RANK-NULLITY Theorem, which states that the rank is n minus the dimension of the eigen space corresponding to 0 eigen value.
+4
You can just find the 'λ' = 2 then you don't need to do all those calculations to find the next  'λ' . As there are only 2 non-zero values and one is = 2 then another one  = Trace of the matrix (sum of principle diagonal) - 1st λ

= (1+1+1+1+1) - 2 = 3
+1

Why can't we use row transformation here : like in this(https://gateoverflow.in/302804/gate2019-44)we have used row transformation

+2

Most voted ans in GO? @Arjun

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 the 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 $0.$

$\begin{pmatrix} 1-X&0&0&0&1\\ 0&1-X&1&1&0\\ 0&1&1-X&1&0\\ 0&1&1&1-X&0\\ 1&0&0&0&1-X \end{pmatrix}$

$R_1 ← R_1 + R_5, R_4 ← R_4 - R_3$

$\implies \begin{pmatrix} 2-X&0&0&0&2-X\\ 0&1-X&1&1&0\\ 0&1&1-X&1&0\\ 0&0&X&-X&0\\ 1&0&0&0&1-X \end{pmatrix}$

Taking $X$ out from $R_4, 2-X$ from $R_1,$ (so, $X = 2$ is one eigen value)

$\implies \begin{pmatrix} 1&0&0&0&1\\ 0&1-X&1&1&0\\ 0&1&1-X&1&0\\ 0&0&1&-1&0\\ 1&0&0&0&1-X \end{pmatrix}$

$R_2 ← R_2 - R_3, R_5 ← R_5 - R_1$

$\implies \begin{pmatrix} 1&0&0&0&1\\ 0&-X&X&0&0\\ 0&1&1-X&1&0\\ 0&0&1&-1&0\\ 0&0&0&0&-X \end{pmatrix}$

$C_3 ← C_3 + C_4$

$\implies \begin{pmatrix} 1&0&0&0&1\\ 0&-X&X&0&0\\ 0&1&2-X&1&0\\ 0&0&0&-1&0\\ 0&0&0&0&-X \end{pmatrix}$

Taking $X$ out from $R_2$

$\implies\begin{pmatrix} 1&0&0&0&1\\ 0&-1&1&0&0\\ 0&1&2-X&1&0\\ 0&0&0&-1&0\\ 0&0&0&0&-X \end{pmatrix}$

$R_3 ← R_3 + R_2$

$\implies\begin{pmatrix} 1&0&0&0&1\\ 0&-1&1&0&0\\ 0&0&3-X&1&0\\ 0&0&0&-1&0\\ 0&0&0&0&-X \end{pmatrix}$

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

So, $(3-X) (-X) = 0 \implies X = 3$ or $X = 0.$ So, $X = 3$ is another eigen value and product of non-zero eigen values $= 2 \times 3 = 6.$

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

by Veteran
edited by
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is it possible to have x= 2 , 2 , 1 and product as 4.
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No. That is not possible. How you got that?
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i can see how u r  getting 2 , bt how u get 3 nt clear.
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Sorry. There was some mistake in the derivation. Have corrected it now.
+2
@arjun sir, can we use column and row transformation interchangeably during elimination ?
+5
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......
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@Arjun Sir Can we apply row and coloumn operations interchangeably?
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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

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.

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@Bikram sir I have a general doubt that

is itright way to eliminate/reduce matric before calculating eigen values of matrix ?
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@ 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)  ??
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Why we not change the matrix into upper triangular and then take the diagonal elements as Eigen values
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@Arjun sir, from where did we get 2?
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Rank of matrix= numbers of non zero eigen values
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can we apply row and column transformation simultaneous to find determinant of matrix?
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After getting one of the eigenvalues as 2,

trace of matrix  = 5 hence other eigenvalue should be 3, hence product = 6

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.

+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 = $n*k$.

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 : $3*2=6$

by Boss
0

Since, determinant is 0, atleast one of the eigen values is 0. I assume say λ5=0

Can you please explain this ??

0
Got it late...Determinant is product of eigen values .

If Determinant is zero ==> Product of eigen values is zero

==> Atleast one eigen value should be zero ..right ??
0
yes
0
Check accepted answer guys.This one is complicated.Ignore it.
0

how have u taken summation of all eigen values are 5 at the beginning??

It is not correct I think.

+1

@srestha-Trace of the matrix.

0

yes correct.

One more thing , how multiplication of five eigen values are $0?$

Do u got it by determinant of $5\times 5$ matrix??

+1
Yes,

product of all eigen values = determinant of the matrix.
+1 vote

Simply we can do it like this

$\begin{bmatrix} (1-\lambda) & 0 & 0 & 0 &1 \\ 0& (1-\lambda ) & 1 & 1 & 0\\ 0& 1 & (1-\lambda) & 1 &0 \\ 0& 1 & 1 &(1-\lambda ) &0 \\ 1 & 0 & 0 & 0 & (1-\lambda ) \end{bmatrix}=0$

$=>\left ( 1-\lambda \right )\begin{bmatrix} (1-\lambda ) & 1 & 1 & 0\\ 1 & (1-\lambda) & 1 &0 \\ 1 & 1 &(1-\lambda ) &0 \\ 0 & 0 & 0 & (1-\lambda ) \end{bmatrix}$$+1.\begin{bmatrix} (1-\lambda ) & 1 & 1 & 0\\ 1 & (1-\lambda) & 1 &0 \\ 1 & 1 &(1-\lambda ) &0 \\ 0 & 0 & 0 & (1-\lambda ) \end{bmatrix}=0$

$=>\left ( 1-\lambda \right )\left ( 1-\lambda \right )\begin{bmatrix} (1-\lambda ) & 1 & 1 \\ 1 & (1-\lambda) & 1 \\ 1 & 1 &(1-\lambda ) \end{bmatrix}+\left ( -1 \right )\begin{bmatrix} (1-\lambda ) & 1 & 1 \\ 1 & (1-\lambda) & 1 \\ 1 & 1 &(1-\lambda ) \end{bmatrix}=0$

$=>\left ( 1-\lambda \right )^{2}\left [ \left ( 1-\lambda \right )\left \{ -2\lambda +\lambda ^{2} \right \}+\lambda +\lambda \right ]-\left [ \left ( 1-\lambda \right )\left \{ -2\lambda +\lambda ^{2} \right \}+\lambda +\lambda \right ]=0$

$=>\lambda ^{3}\left [ 3-\lambda \right ].\left [ \lambda -2 \right ]=0$

So, $=>\lambda =0,2,3$

Ans $6.$

Another procedure here https://gateoverflow.in/216642/matrix

by Veteran
Rank is less then n so det=o

Trace=5

So non-zero eigen values r 2 and 3
by Junior
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.
0

@daksirp

@A_i_\$_h

Initially I also did the same way but applying elementary operations change eigen values , that's why above approach fails here . So don't apply reduction , solve it directly as mentioned in above comments by bikram sir.

https://socratic.org/questions/do-elementary-row-operations-change-eigenvalues

+2
i also did the same thing it means elementary operation changes the eigen value of the matrix so this approach fails.
Convert the matrix into lower triangular matrix by row transformations , then we know that principal diagonal elements of this matrix will be Eigen values,  out of these 5 EIgen values three will be 0 and Two eigenvalues  will be 1 .so product of non zero eigenvalues values =1X1 =1 ans.
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.
by Loyal
+4
Wat hav u done ??? very tough to understand ...
+2
:D this is clearly a copy paste