If the matrix, $$A = \begin{pmatrix}-3&0&0 \\\ -4&-7&8\\-4&-4&5 \end{pmatrix}$$

  • Find the characteristic polynomial of A and, hence, find all the eigenvalues of A.

  • Parametrizing (as a subset of $\mathbb{R}^3)$ the eigenspace of A that corresponds to the eigenvalue $\lambda = -3$. Is it possible to diagonalize A?

So far I have taken the determinant of $(A-\lambda I)$ and got the characteristic polynomial $-\lambda^{2} -5\lambda^{2} -3\lambda +9$. Then calculated the eigenvalues to be $\lambda = 1, -3$. After this, I am unsure how to proceed.

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    $\begingroup$ What did you do? $\endgroup$ – Bernard May 7 at 14:28
  • $\begingroup$ There’s an easier way for this matrix. Hint: At least two eigenvalues are easily found by inspection, and you can always get the last one “for free” by considering the trace of the matrix. $\endgroup$ – amd May 8 at 0:02
  • $\begingroup$ @Bernard I am not familiar with the trace of the matrix, how does that work? $\endgroup$ – Dylan Grimes Larkin May 8 at 7:04
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    $\begingroup$ The trace of a square matrix is the sum of its diagonal coefficients. It happens that it is also the sum of its eigenvalues (counted with multiplicity). You can see more details on Wikipedia. $\endgroup$ – Bernard May 8 at 9:38

First, you should not expand the determinant of $A-\lambda I$, but try to factor it: \begin{align}\begin{vmatrix} -3-\lambda &&0 \\ -4&-7-\lambda &8 \\ -4&-4&5-\lambda\end{vmatrix}&=\begin{vmatrix} -3-\lambda &&0\\-4&-7-\lambda &8 \\ 0&3+\lambda&-3-\lambda\end{vmatrix}=(3+\lambda)\begin{vmatrix} -3-\lambda &&0\\-4&-7-\lambda &8 \\ 0&1&-1\end{vmatrix}\\[1ex] &=(3+\lambda)\Bigl[(-3-\lambda)(\lambda+7-8)\Bigr]=-(\lambda+3)^2(\lambda-1) \end{align}

Second, to determine if the matrix is diagonalisable, you have to check whether the geometric multiplicity is equal to the algebraic multiplicity of each of the eigenvalues.

For $1$, which is a simple eigenvalue, there is no problem. For the double eigenvalue $-3$, this means you have to check whether the corresponding eigenspace $\ker(A+3I)$ has dimension $2$.


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