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Consider the following matrix:

$R = \begin{bmatrix} 1 & 2 & 4 & 8 \\ 1 & 3 & 9 & 27 \\ 1 & 4 & 16 & 64 \\ 1 & 5 & 25 & 125 \end{bmatrix}$

The absolute value of the product of Eigen values of $R$ is _______
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I did the worst. Square the elements of column2 and subtracted it from column3.
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smart
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Hi All,

Can we apply linear transformation and covert a matrix to lower/upper or diagonal matrix while finding Eigen values of a matrix?

As per my understanding applying linear transformations or converting a matrix to lower/upper/diagonal matrix affects the Eigen values of matrix. Please correct me if i am wrong.

Thanks

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4 Answers

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68 votes
Best answer

Important properties of Eigen values:

  1. Sum of all eigen values $=$ Sum of leading diagonal(principle diagonal) elements $=$ Trace of the matrix.
  2. Product of all Eigen values $= \det(A)= \mid A \mid$
  3. Any square diagonal(lower triangular or upper triangular) matrix eigen values are leading diagonal (principle diagonal)elements itself.

Example:$A=\begin{bmatrix} 1& 0& 0\\ 0&1 &0 \\ 0& 0& 1\end{bmatrix}$

Diagonal matrix

Eigenvalues are $1,1,1$

$B=\begin{bmatrix} 1& 9& 6\\ 0&1 &12 \\ 0& 0& 1\end{bmatrix}$

Upper triangular matrix

Eigenvalues are $1,1,1$

$C=\begin{bmatrix} 1& 0& 0\\ 8&1 &0 \\ 2& 3& 1\end{bmatrix}$

Lower triangular matrix

Eigenvalues are $1,1,1$


Now coming to the actual question

$R=\begin{bmatrix} 1 &2 &4 &8 \\ 1 &3 &9 &27 \\ 1 &4 &16 &64 \\ 1 &5 &25 &125 \end{bmatrix}$

$\mid  R \mid =\begin{vmatrix} 1 &2 &4 &8 \\ 1 &3 &9 &27 \\ 1 &4 &16 &64 \\ 1 &5 &25 &125 \end{vmatrix}$

Perform 

  • $R4\rightarrow R_{4}-R_{3}$
  • $R3\rightarrow R_{3}-R_{2}$
  • $R2\rightarrow R_{2}-R_{1}$

$\implies \mid R \mid =\begin{vmatrix} 1 &2 &4 &8 \\ 0 &1 &5 &19 \\ 0 &1 &7 &37 \\ 0 &1 &9 &61 \end{vmatrix}$

Perform

  • $R4\rightarrow R_{4}-R_{3}$
  • $R3\rightarrow R_{3}-R_{2}$

$\implies \mid R \mid =\begin{vmatrix} 1 &2 &4 &8 \\ 0 &1 &5 &19 \\ 0 &0 &2 &18 \\ 0 &0 &2 &24 \end{vmatrix}$

Perform

  • $R4\rightarrow R_{4}-R_{3}$

$\implies \mid R \mid =\begin{vmatrix} 1 &2 &4 &8 \\ 0 &1 &5 &19 \\ 0 &0 &2 &18 \\ 0 &0 &0 &6 \end{vmatrix}$

 The absolute value of product of  Eigen values $= \det(A)= \text{Product of diagonal elements }  =12.$

edited by

4 Comments

@____ 

In the GATE 2014 question, they asked for the product of the non-zero eigenvalues, while in this question, they asked for the product of all eigenvalues.

In the GATE 2014 solution given, first, the eigenvalues are determined by \( |A- \lambda I |  = 0 \), and then the product of non-zero eigenvalues is calculated. Since \( |A- \lambda I | \) is a determinant, you can always perform row operations to find the determinant.

Now, regarding the given question, they are asking for the product of ALL eigenvalues. It can be found by taking the determinant of the matrix, and since you are finding the determinant, you can always perform row operations to find the determinant.

The key idea here is that whenever dealing with determinants, elementary row operations can be applied. And in both questions, we are finding the determinant. The only difference is that here we are finding the determinant of \( A \) i.e. \( |A| \), and there we are finding the determinant of \( A - \lambda I \) i.e. \( |A - \lambda I| \).

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@Sachin Mittal 1

Suppose in this question, instead of finding the determinant, if it is said to find the product of the largest 2 Eigen values, then before applying |A-XI| , can we do row operations?

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@____ 

If you want to find eigen values then you can not perform row operations on matrix A, you can perform on $A - \lambda I $ because you will find determinant of $A - \lambda I $.

If you perform row operations on A and then find the eigen values of A then it will be wrong.

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26 votes
26 votes

The given matrix is called square Vandermonde Matrix and it has the form :

$\begin{bmatrix} 1 & \alpha_1 & \alpha_1^2 & .... & \alpha_1^{n-1} \\ 1 & \alpha_2 & \alpha_2^2 & .... & \alpha_2^{n-1} \\ \vdots & \vdots & \vdots &\ddots & \vdots \\ 1 & \alpha_n & \alpha_n^2 & .... & \alpha_n^{n-1} \\  \end{bmatrix}$

and its determinant is given by : $\prod_{1 \leq i < j \leq n} (\alpha_j - \alpha_i)$. Proof is given here using induction.

So, here, answer is : $(3-2)*(4-2)*(5-2)*(4-3)*(5-3)*(5-4) = 12$.

$\Rightarrow$ For a general $4\times 4$ matrix of the form :

$\begin{bmatrix} 1 &a &a^2 &a^3 \\ 1 &b &b^2 &b^3 \\ 1 &c &c^2 &c^3 \\ 1 &d &d^2 &d^3 \end{bmatrix}$,

Determinant is given by : $(b-a)*(c-a)*(d-a)*(c-b)*(d-b)*(d-c)$ and similarly, for this type of matrix, we can find determinant of any order.

4 Comments

Good Information.
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edited by

@ankitgupta.1729

Can we write  like this, for easy remembering

$\Delta = (a-b)\:(a-c)\:(a-d)\:(b-c)\:(b-d)\:(c-d)$

I think it is valid for even order like $2\times 2, 4 \times 4,\dots$

$\Delta = -(a-b)\:(a-c)\:(a-d)\:(b-c)\:(b-d)\:(c-d)$

It is valid for odd order like $1\times 1, 3 \times 3,\dots$

Please correct me if i'm wrong?

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@Lakshman Patel RJIT

What if the size of the matrix is $5 \times 5$ ?

If no. of multiplicative terms are even then sign will be +ve outside in above formula and -ve if no. of multiplicative terms are odd according to your formula.

Here, total no. of multiplicative terms are total no. of pairs of the above nature which is $ (n-1) + (n-2) + (n-3) +...+1 = \frac{n*(n-1)}{2}$

So, if we write formula as you have written, we should have to add $(-1)^\frac{n*(n-1)}{2}$ outside in the above formula to take care of positive or negative sign.

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@ankitgupta.1729

Thank you brother

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5 votes
5 votes

Answer 12

Product of eigen values= |A| (determinant of matrix)

4 votes
4 votes

Product of eigenvalues = determinant of the matrix.

Given matrix:

R2 = R2 - R1; R3 = R3 - R1; R4 = R4 - R1

Expand via first column.

R2 = R2 - 2R1; R3 = R3 - 3R1

Expand via first column

=12

4 Comments

yes i thought the same
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@val_pro20 yes the eigenvalues will change. But in the question they have asked product of eigenvalue which is equal to det(A). So, by converting the matrix into upper triangular matrix we can easy find the product of eigenvalues. 

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edited by

Thanks for the comment @tskarthik .That was very helpful.

My understanding:

I saw earlier in questions that doing row transformations will change the eigen values. So we

 calculate eigen value without doing any row operations on matrix. But since we are being asked

for product of eigen values which is equal to the determinant of matrix, and doing row(or column)

operations do not change the determinant of a matrix. So we can safely and rightly do these

operations.

But when we are asked specifically for eigen values we cannot do these operations on matrix

before because it will change the eigen values.

-------------------------------------------------------------------------------------------------------------------------------------------------------------

PS: I misunderstood the question. I thought I have to give product of absolute values of eigen values

so I was trying to find specific eigen values. But in the question asked for absolute value of the

product of eigen values.

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