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26 votes
26 votes
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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4 Answers

Best answer
68 votes
68 votes

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 $= \det(A)= \text{Product of diagonal elements }  =12.$

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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 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

Answer:

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