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The eigenvalues of my matrix are $x_1= 1$

and

$x_2=3$

I get an eigenvector

$V = t~[ 4~~~~~~ 3 ~~~~~1 ]^T $ but how can I diagonalize the matrix if I have the same column repeated twice. Should I just use different values for t since they are all eigenvectors of the same matrix?

matrix $\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 8 & 6 &3 \\ \end{bmatrix}$

Edit: I realized I had the eigenvalues wrong , now fixed

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  • $\begingroup$ Please show us the matrix. $\endgroup$ – GFauxPas Apr 17 '15 at 17:28
  • $\begingroup$ Please check your definition of eigenvalues. The eigenvalues of your matrix are $1$ and $3$. $\endgroup$ – TZakrevskiy Apr 17 '15 at 17:36
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I will assume that you are capable of finding the eigenvector for the simple root, and only focus on the repeated eigenvalue $x_1=1$. Let $\mathbf{v}=\begin{bmatrix} a \\ b \\ c \\ \end{bmatrix}$ and consider $(A-I)\mathbf{v}=\mathbf{0}$. This yields $\begin{bmatrix} 0 \\ 0 \\ 8a+6b+2c \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix}$ in which case $8a+6b+2c=0$. Let $b$ and $c$ be free variables, giving us $\mathbf{v}=\begin{bmatrix} -\frac{3}{4}b-\frac{1}{4}c \\ b \\ c \\ \end{bmatrix}=b\begin{bmatrix} -\frac{3}{4} \\ 1 \\ 0 \\ \end{bmatrix}+c\begin{bmatrix} -\frac{1}{4} \\ 0 \\ 1 \\ \end{bmatrix}$.
I hope this helps.

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  • $\begingroup$ yes it does! thank you! $\endgroup$ – Alida Apr 17 '15 at 18:32

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