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1. Obtain the eigen values of the matrix

$$A=\begin {bmatrix} 1 & 2 & 34 & 49 \\ 0 & 2 & 43 & 94 \\ 0 & 0 & -2 & 104 \\ 0 & 0 & 0 & -1 \end{bmatrix}$$

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$5(a)$ the eigen value for upper triangular/lower triangular/diagonal matrices are the diagonal elements of the matrix
answered by Active (2.4k points)
edited
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we know for triangular/diagonal the determeinant is product of principal diagonal elements so in A-(lambda)I every value of lambda as any of of the principal diagonal elements will give result as 0. so all the diagonal principal diagonal elements are eigen values.
+4

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 $1,1,1$

$A=\begin{bmatrix} 1& 2 &34 &49\\ 0& 2&43 &94\\ 0& 0 & -2&104\\ 0& 0& 0&-1 \end{bmatrix}$

$|A-\lambda I |=0$

$\begin{vmatrix} 1-\lambda& 2 &34 &49\\ 0& 2-\lambda&43 &94\\ 0& 0 & -2-\lambda&104\\ 0& 0& 0&-1-\lambda \end{vmatrix}=0$

$( 1-\lambda)( 2-\lambda)( -2-\lambda)(-1-\lambda)=0$

$\lambda=1\ ,\ -1\ ,\ 2 \ , \ -2$
answered by Boss (40.8k points)
+1 vote
Eigen value $lambda$=1,-1,2,-2 A-$lambda$I=0 just solve it and for the 2nd question i think building a truth table is a naive way of answering it if any1 has better solution then plz reply
answered by Boss (14.4k points)
if the matrix is either upper triangular or lower triangular matrix then the principal diagonal element are called eigen value.
1,2,-1,-2 for this question ...
answered by Active (4.1k points)

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