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

my sol 

Answer 2

\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}
Do row operations on $(A-I\lambda) = 0$ to make it a triangular matrix and find the determinat.

Note:

1. $det(A) = \prod_{i=1}^{n}\lambda_i$

2. $trace(A) = \sum_{i = 1}^{n} \lambda_i$

Answer 3

The product of the non-zero eigenvalues of the matrix is \textbf{6}.

 Step-by-Step Explanation

Let the given \( 5 \times 5 \) matrix be \textbf{A}.

\textbf{1. Analyze the Matrix Rank}

First, observe the rows of the matrix:
* Rows 2, 3, and 4 are identical: `(0 1 1 1 0)`
* Rows 1 and 5 are identical: `(1 0 0 0 1)`

Since there are only two unique, linearly independent rows, the \textbf{rank of the matrix A is 2}.

\textbf{2. Relate Rank to Eigenvalues}

For a symmetric matrix (like this one), the rank is equal to the number of non-zero eigenvalues. Since the rank is 2, the matrix has \textbf{two non-zero eigenvalues}. The other three eigenvalues are 0.

Let the five eigenvalues be \( \lambda_1, \lambda_2, \lambda_3, \lambda_4, \lambda_5 \). We know that:
* \( \lambda_1 \neq 0 \)
* \( \lambda_2 \neq 0 \)
* \( \lambda_3 = \lambda_4 = \lambda_5 = 0 \)

\textbf{3. Use the Trace Property}

A key property of matrices is that the sum of the eigenvalues is equal to the \textbf{trace} of the matrix (the sum of the elements on the main diagonal).

* \textbf{Trace(A)} = \( 1 + 1 + 1 + 1 + 1 = 5 \)
* \textbf{Sum of eigenvalues} = \( \lambda_1 + \lambda_2 + 0 + 0 + 0 = \lambda_1 + \lambda_2 \)

Therefore, we have the equation:
$$
\lambda_1 + \lambda_2 = 5
$$

\textbf{4. Find the Eigenvalues}

We can find the non-zero eigenvalues by inspecting the structure of the matrix using the eigenvector equation \( Ax = \lambda x \).

* \textbf{Case 1:} Consider an eigenvector of the form \( x = (1, 0, 0, 0, 1)^T \).
    \( Ax = \begin{pmatrix} 1 & 0 & 0 & 0 & 1 \\ \vdots \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1+1 \\ 0 \\ 0 \\ 0 \\ 1+1 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \\ 0 \\ 0 \\ 2 \end{pmatrix} = 2x \)
    This shows that \textbf{\( \lambda_1 = 2 \)} is an eigenvalue.

* \textbf{Case 2:} Consider an eigenvector of the form \( x = (0, 1, 1, 1, 0)^T \).
    \( Ax = \begin{pmatrix} \vdots \\ 0 & 1 & 1 & 1 & 0 \\ \vdots \end{pmatrix} \begin{pmatrix} 0 \\ 1 \\ 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1+1+1 \\ 1+1+1 \\ 1+1+1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 3 \\ 3 \\ 3 \\ 0 \end{pmatrix} = 3x \)
    This shows that \textbf{\( \lambda_2 = 3 \)} is an eigenvalue.

We have found the two non-zero eigenvalues: 2 and 3. Note that their sum is \( 2+3=5 \), which correctly matches the trace.

\textbf{5. Calculate the Product}

The question asks for the product of the non-zero eigenvalues.
$$
\text{Product} = \lambda_1 \times \lambda_2 = 2 \times 3 = \bf{6}
$$

Answer 4

Ans

Comments

You cannot use column operation in row echelon form, its a wrong approach

Answer 5

people are here kill this question,

after analysis

see trace of non zero eigen values =5

but det=0,so you cant find directly

but , using row echolean form you can find the answer

6.

Answer 6

i am going to call it trick but is not since addition of first and last two row is 2 and addition of row 2,3,4 is 3 since they are reapeating more than one time thus 2 and 3 must be the eigen value of given matrix  and addition of diagonal should be 5 and these two sum are already 5 thus other 2 eigen value will be 0,0 and if u will find determinent then it will be 0 only