GATE CSE 2014 Set 2 | Question: 47

Question

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

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

Comments

Yes, It is not true.

Above link is correct

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

Here in this, it mentioned the number of distinct eigenvalues is atmost rank of A.

And rank = 2

Even the eigen values are 2,2,3. Then the distict are 2.

Why can’t this be true?

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@Abhilash Behera this is right. 

But here the given matrix is symmetric so number of non-zero eigen values will be equal to rank of the matrix.

Answer 1

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

Comments

LIke applying r1<-r1+r5 and then taking (2-x) gives x =2 as one of the eigen values. 

We can also do r2 <- r2 + r3 + r4 ;

This makes the row 2 as [ 0 , 3-x , 3-x,3-x , 0 ] , then we take 3-x out and other eigen value be x = 3

And 3*2 = 6 

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@Sachin Mittal 1

In this question we are first doing A-XI then doing row transformation, but in question 6.3.22 we are doing row transformation first.

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@____

Please check this comment:

https://gateoverflow.in/302804/gate-cse-2019-question-44?show=424213#c424213

Answer 2

 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.

Comments

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 = $n*k$.

Answer 3

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 : $3*2=6$

 

Comments

@Ayush Upadhyaya

yes correct.

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

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

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Yes,

product of all eigen values = determinant of the matrix.

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Loved the way u approached the question

Answer 4

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

Comments

You interchanged two columns in the second step while finding the minor of 1. I think we are not allowed to do it.

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Expanding to Find the Determinant

Here are the steps to go through to find the determinant.

1. Pick any row or column in the matrix. It does not matter which row or which column you use, the answer will be the same for any row. There are some rows or columns that are easier than others, but we'll get to that later.
2. Multiply every element in that row or column by its cofactor and add. The result is the determinant.

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

it dosent matter which row you pick while calculating determinant, you will get same answer

we will take this to our advantage and take rows which have more zeros