# Diagonalizability of a matrix

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!

• The diagonal entries in both of your matrices seem wrong. Did you do $A-\lambda I$? Jan 29, 2020 at 18:46
• The matrix $A$ you gave only has eigenvalue 1... Jan 29, 2020 at 18:47
• My mistake. The third component of the last row vector of $A$ is $3$, not $1$. Jan 29, 2020 at 18:49
• 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 Jan 29, 2020 at 18:54

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.
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 $$(0,1,0)$$.
• 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 $V_1(A)=\left< \begin{pmatrix} 1\\0\\1 \end{pmatrix}, \begin{pmatrix} 0\\1\\0 \end{pmatrix} \right>$? Jan 29, 2020 at 18:58
• 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 $2$. Jan 29, 2020 at 19:00