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Show that $$ A :=\begin{pmatrix} 1 & 0 & 0 \\ -2 & 1 & 2 \\ -2 & 0 & 3 \end{pmatrix}$$ is diagonalizable.

What I did:

First, I determined the characteristic polynomial $$\chi_A(X) = \det(X \cdot E_3-A)=(X-3)(X-1)(X-1)=X^3-5X^2+7X-3,$$

so the eigenvalues are $3$ and $1$.

I then determined the eigenspaces of each eigenvalue:

$$X=3: \left(\begin{array}{@{}ccc|c@{}} 2 & 0 & 0 & 0 \\ -2 & 2 & 2 & 0 \\ -2 & 0 & 0 & 0 \\ \end{array}\right) \leadsto \left(\begin{array}{@{}ccc|c@{}} 2 & 0 & 0 & 0 \\ 0 & 2 & 2 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array}\right),$$ so $x_1=0$, $x_2=-x_3$, $x_3=x_3$ and thus $V_3(C) = \left< \begin{pmatrix} 0\\-1\\1 \end{pmatrix} \right>$.

Analogous:

$$X=1: \left(\begin{array}{@{}ccc|c@{}} 0 & 0 & 0 & 0 \\ -2 & 0 & 2 & 0 \\ -2 & 0 & -2 & 0 \\ \end{array}\right) \leadsto \left(\begin{array}{@{}ccc|c@{}} -2 & 0 & 2 & 0 \\ 0 & 0 & -4 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array}\right),$$ so $x_1=x_3=0$, $x_2=x_2$ and thus $V_1(A) = \left< \begin{pmatrix} 0\\1\\0 \end{pmatrix} \right>$.

It now follows that $\dim(V_3(C)) + \dim(V_1(A)) = 1+1=2 \lt 3 = \dim(A)$ and because of the $\lt$, A shouldn't be diagonalizable, but it is.

So where's the mistake? Thanks in advance!

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  • $\begingroup$ The diagonal entries in both of your matrices seem wrong. Did you do $A-\lambda I$? $\endgroup$
    – Berci
    Jan 29, 2020 at 18:46
  • $\begingroup$ The matrix $A$ you gave only has eigenvalue 1... $\endgroup$ Jan 29, 2020 at 18:47
  • $\begingroup$ My mistake. The third component of the last row vector of $A$ is $3$, not $1$. $\endgroup$
    – marymk
    Jan 29, 2020 at 18:49
  • $\begingroup$ You've already received answers, don't fix typos now. That must be done before answers. Write down a new question, check it carefully...and leave this one as it is and thank the answerers $\endgroup$
    – DonAntonio
    Jan 29, 2020 at 18:54

2 Answers 2

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When $X=1$ the matrix becomes $$XI-A=I-A=\begin{bmatrix}0 & 0 & 0 \\ 2 & 0 & -2 \\ 2 & 0 & -2\end{bmatrix}$$ which has rank 1.

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Your error lies in the computation of the eigenspace corresponding to the eigenvalue $1$, which is $2$-dimensional; it is spanned by $(1,0,1)$ and by $(0,1,0)$.

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  • $\begingroup$ And I have edited my answer. $\endgroup$ Jan 29, 2020 at 18:55
  • $\begingroup$ I see but I don't understand why it's wrong. The Gaussian elimination was done properly and the results $x_1=x_3=0$, $x_2=x_2$ are right, so how come $V_1(A)=\left< \begin{pmatrix} 1\\0\\1 \end{pmatrix}, \begin{pmatrix} 0\\1\\0 \end{pmatrix} \right>$? $\endgroup$
    – marymk
    Jan 29, 2020 at 18:58
  • $\begingroup$ Note that$$A-\operatorname{Id}=\begin{bmatrix}0 & 0 & 0 \\ -2 & 0 & 2 \\ -2 & 0 & 2\end{bmatrix}.$$You had a $-2$, which should be a $2$. $\endgroup$ Jan 29, 2020 at 19:00

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