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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 _______
asked in Linear Algebra by Veteran (386k points)
edited by | 2k views
+2
under the pressure, i multiply C2 and C3 and subtracted from C3 ==> C3 is a Zero vector !

Therefore Det will be 0

What a mistake !

3 Answers

+4 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$=Det(A)=|A|$

$(3)$ 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 come to the actual question

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

$\Rightarrow$Product of all Eigen values$=Det(A)=|A|$

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

Perform some operation 

            $R4\rightarrow R_{4}-R_{3}$

            $R3\rightarrow R_{3}-R_{2}$

           $R2\rightarrow R_{2}-R_{1}$

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

Perform some operation 

            $R4\rightarrow R_{4}-R_{3}$

            $R3\rightarrow R_{3}-R_{2}$

           $R2\rightarrow R_{2}-R_{1}$

$|R|=\begin{vmatrix} 1 &2 &4 &8 \\ 0 &1 &1 &11 \\ 0 &0 &2 &18 \\ 0 &0 &2 &24 \end{vmatrix}$

Perform some operation 

            $R4\rightarrow R_{4}-R_{3}$

$|R|=\begin{vmatrix} 1 &2 &4 &8 \\ 0 &1 &1 &11 \\ 0 &0 &2 &18 \\ 0 &0 &0 &6 \end{vmatrix}$

 The absolute value of product of  Eigen values$=Det(A)=|A|=12$

answered by Boss (29.4k points)
edited by
+4 votes

Answer 12

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

answered by Active (2.5k points)
+1 vote
product of eigen values is the determinant

12 ans
answered by Junior (527 points)
Answer:

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