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I have a matrix, $$A = \begin{bmatrix} 3-c & -c & 1 \\ c-1 & 2+c & -1 \\ c+1 & c & 3 \end{bmatrix}$$ and I am asked for what values of $c \in \mathbb{R}$ will make this matrix have eigenvectors that form a basis for $\mathbb{R}^3$. It seems quite messy to solve the characteristic polynomial, however. Is there another easier way to look at this?

If the eigenvectors $v_1,v_2, v_3$ form a basis, then for any $x \in \mathbb{R}^3$, we can write $x = c_1v_1 + c_2v_2 + c_3v_3$, then perhaps multiply the equation by A? Then $Ax = c_1Av_1 + c_2Av_2 + c_3Av_3$ = $(c_1\lambda_1)v_1 + (c_2\lambda_2)v_2 + (c_3\lambda_3)v_3$. I'm not exactly sure where to go from here, or whether I'm going in the right direction. Perhaps something having to do with the nullspace of $A$??

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  • $\begingroup$ Hint: The CP is given by: $$| A - \lambda I| = -\lambda ^3+8 \lambda ^2-20 \lambda +16 = -(\lambda -4) (\lambda -2)^2$$ $\endgroup$
    – Moo
    Apr 16, 2018 at 17:01
  • $\begingroup$ Yes, I just finished computing this:) $\endgroup$ Apr 16, 2018 at 17:04

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Hint: The characteristic polynomial is given by $$ \chi_t(A)=t^3 - 8t^2 + 20t - 16=(t - 2)^2(t - 4), $$ which is easy to compute using the rule of Sarrus. In particular, it does not depend on $c$.

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