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For the vector A: EDIT: I had originally multiplied the matrix by -1. Apologies. $$ \begin{bmatrix} 4 & 2 & 2 \\ 2 & 4 & 2 \\ 2 & 2 & 4 \\ \end{bmatrix} $$

I got $\lambda I_n - A$ as:

\begin{bmatrix} (\lambda - 4) & -2 & -2\\ -2 & (\lambda - 4) & -2\\ -2 & -2 & (\lambda - 4)\\ \end{bmatrix}

(Apologies for the bad formatting, it's a 3x3 (diagonal? or orthogonal?) matrix.

With corresponding Eigenvalues of $\lambda = 0, \lambda = 6$

For $\lambda = 0;$

I got an Identity Matrix as follows,

$$ \left[ \begin{array}{ccc|c} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \end{array} \right]$$

where the right hand column of 0's is the 0 vector you want when you multiply the left-hand side by \begin{bmatrix}v_1\\v_2\\v_3\\\end{bmatrix}

Is $v_1 = v_2 = v_3 = 0$ a valid Eigenvector/Eigenspace?

EDIT: I got the eigenvalues as follows:

Using Sarrus' Rule with the $\lambda I_n - A$ matrix:

$X = (\lambda -4)(\lambda -4)(\lambda -4) + (-2)^3 + (-2)^3$ (along the \ diagonal)

$Y = (-2)(-2)(\lambda -4) + (\lambda -4)(-2)(-2) + (-2)(\lambda -4)(-2)$ (along the / diagonal from the right)

Let $Z = X - Y$, thus giving me the polynomial.

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  • $\begingroup$ That's not $\lambda I_n + A,$ what you calculated was $\lambda I + A$. None of the eigenvalues of $A$ is equal to $0$. $\endgroup$ – 5xum Aug 25 '15 at 13:20
  • $\begingroup$ Also, how did you decide what the eigenvalues are? For instance, $0$ is not an eigenvalue of your matrix. $\endgroup$ – Omnomnomnom Aug 25 '15 at 13:22
  • $\begingroup$ @5xum I actually wrote down the wrong matrix! There were no negative numbers in the original, could you take a look at this one please? $\endgroup$ – Gabi Aug 25 '15 at 13:22
  • $\begingroup$ @Omnomnomnom I calculated the $det(\lambda I_n - A)$ and using Sarrus' rule got the polynomial $\lambda^3 - 12\lambda^2 + 36\lambda = 0$ and got $\lambda = 6$ or $0$. $\endgroup$ – Gabi Aug 25 '15 at 13:25
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    $\begingroup$ @Gabi Sarrus' Rule is fine, but you made a mistake on the sign. You should have ended up with $-16 - 16 = -32$ as a constant term rather than $-16 + 16 = 0$ $\endgroup$ – Omnomnomnom Aug 25 '15 at 13:34
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As stated in the comments: you've calculated the determinant incorrectly. When you fix your mistake, you should end up with the characteristic polynomial $$ \lambda^3 - 12 \lambda^2 + 36\lambda - 32 = (\lambda - 8)(\lambda - 2)^2 $$

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