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

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

8 Answers

77 votes
77 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 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

edited by
13 votes
13 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.

10 votes
10 votes

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 : $3*2=6$

 

3 votes
3 votes

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

Answer:

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