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

@S_Sandeep

I messed up with the concept.

i will update my answer for now and will come with a good conclusion.

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@S_Sandeep bhai i  am confused 
suppose we have 10 by 10 matrix and its nullity is 8 that means we will have one of the eigen value zero and corresponding to eigen value zero we will have 8 linear independent eigen vector solution

there are 8 linearly independent eigenvectors corresponding to 0.
how can we definetly state this 

can you tell reasoning behind it i am confused

actually we have this is annotated notes means vice versa is also true (below image )

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for a real symmetric matrix, AM=GM and all E. Vectors are orthogonal .. so when rank < n means the nullity (n-k) is corresponding to Eigen Value Zero and the AM of that zero E. Value is K.

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

NO . This is not the Practical way of solving these kind of Problems. Try this Technique : 

https://gateoverflow.in/2013/gate-cse-2014-set-2-question-47?show=516020#a516020

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@Arjun the moment we got to know that one of the values is 2 the question is over there itself because we know that am and gm of lambla 0 is 3 from rank of matrix -----> at max 2 more eigen values will exist and 2 cant be repeated cuz sum is 5 other values has to be 3 

am i correct sir

 

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No, if we start with row go on row only, if started with column then go with columns

Answer 2

 

$A$ is Given  a real-symmetric matrix , and given $Rank(A)$ is $2$ , $Nullity$  here is $n- Rank(A) =3$ that means corresponding to eigen value $0$ there are exactly $3$ LI eigen vectors.

 Possible characteristic eq can be $λ^3 (λ-λ1)(λ-λ2)$

Thus $ λ1+λ2+λ3+λ4+λ5=5$ 

$⇒ $ $ λ1+λ2=5 .................(1)$

Now  Refer  the sachin sir's comment to know how  

$$
\begin{array}{r}
\operatorname{trace}\left(A^{\top} A\right)=\left\|a_{1}\right\|^{2}+\left\|a_{2}\right\|^{2}+\left\|a_{3}^{2}\right\|
\end{array}
$$

For symmetric matrix  $𝐴^⊤𝐴=𝐴^2$

Eigen values of $𝐴=0,0,0,𝜆1,𝜆2$

Eigen values of $𝐴^2=0,0,0,𝜆1^2,𝜆2^2$


$trace(𝐴^T𝐴)=trace(𝐴^2)=13$

$0+0+0+𝜆1^2+𝜆2^2=13$

$⇒𝜆1^2+𝜆2^2=13.............(2) $

Now Do squres Both side Of equ $1$

$\lambda1^2 + \lambda2^2+ 2* \lambda1*\lambda2 = 25 $

After Putting the Values of  $\lambda1^2 + \lambda2^2$ we'll get   $\lambda1*\lambda2$ Value which is Asked by Question which Will Come to be $6$.

Comments

@Prashant_Dubey

Best answer

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 Possible characteristic eq can be $\lambda^3(\lambda-\lambda_1)(\lambda-\lambda_2)=0$

I was confused as to why only this characteristic equation was considered and not $\lambda^4(\lambda-\lambda_1)=0$

This is because $\lambda=0$ has Algebraic multiplicity of $4$, meaning it could have 1 LI eigen vector, 2 LI eigen vector, 3 LI eigen vector or 4 LI eigen vector. 

When it will have $4$ LI eigen vector then it would mean nullity of $A$ is $4$ since we are talking about $\lambda=0$ this would imply $Rank(A)=1$ but that is shouldn't be possible as we have already concluded that $Rank(A)=2$

Therefore, only one characteristic is possible here. 

Thank you for your attention. 

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But you don't  need to do A^2 because trace of A^2 is the sum of squares of all the elements of matrix A. Just calculate the squared sum of each element of A. No need to calculate A^2.

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

Here I want to add a point,

As we can see the sum of eigen values is 5 (trace of square matrix), and from the characteristic equation we know that 3 eigen values are 0 (rank = 2, nullity = 3) so we have only 2 non-zero eigen values whose sum is 5.

Those 2 eigen values can be (1,4) or (2,3). Now we can check if 1 or 2 is our eigen value.

If 1 is our eigen value then 4 must be other eigen value and

if 2 is our eigen value then 3 must be the other eigen value

Since after checking, we got that 1 cannot be our eigen value since after putting 1 as eigen value in the characteristic equation the det is changed (original det is zero) and not zero .

Hence 2 must be our eigen value and the other eigen value must be 3.

 

For this question I guess we can think this way also.

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yes we can. solve by eliminiating..

 

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As we can see the sum of eigen values is 5 (trace of square matrix), and from the characteristic equation we know that 3 eigen values are 0 (rank = 2, nullity = 3) so we have only 2 non-zero eigen values whose sum is 5.

How can we say these? 

Answer 4

 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 5

Tryout this approach

Comments

Waah!!

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nice answer @Aditya_Ranjan

 

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This approach is good but i am very unsure that this will click in the exam hall 👾

Answer 6

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