Fast way to tell if this matrix is diagonalizable?

Question

I have to say if this matrix is diagonalizable on the real numbers:

$$A=\begin{pmatrix}0 & 0 & 1 \\ 0 & 1 & 2 \\ 1 & 2 & 0 \\ \end{pmatrix} \ \ \ \ \ \ A\in \mathbb{R}^{3x3} $$

So i calculated the characteristic polynomial of $A$ as:

$$P(\lambda)=\det(A-\lambda I)=\begin{vmatrix}-\lambda & 0 & 1 \\ 0 & 1-\lambda & 2 \\ 1 & 2 & -\lambda \\ \end{vmatrix}=-\lambda^3+\lambda^2+5\lambda-1$$

Now to go on i needed to solve the equation $-\lambda^3+\lambda^2+5\lambda-1=0$ for $\lambda$ to get the eigenvalues of $A$, which i wasn't able to do by hand, so plugged the equation into WolframAlpha to see the intermediate steps, but they are insane! (See the equation on WolframAlpha)

I think i must be missing something because i should be able to solve this exercise in 3 minutes, since it comes from a linear algebra test made of 10 questions that has to be done in 30 minutes.

Can anyone help?

Answer 1

Every symmetric matrix is diagonalizable.

Alternatively it suffices to show that the characteristic polynomial of $A$ is of the form $p_{A}(\lambda) = -(\lambda - r_1)(\lambda - r_2)(\lambda - r_3)$ where $r_i$ are distinct.

In our case $p_{A}(\lambda)=-\lambda^3+\lambda^2+5\lambda-1$. Now, $p_{A}(0)=-1,p_{A}(1)=4$. By the Intermediate Value Theorem $p_A$ has at least one root in each of the intervals $(-\infty,0), (0,1),(1,\infty) $, and since $p_A$ has degree $3$, $p_A$ has distinct roots.

Answer 2

If you really just need a yes/no answer, I suppose you have that.

ADDED: this ($P^T H P = D$) can also be done with just integers, this time.

$$\left( \begin{array}{rrr} 2 & -1 & 1 \\ 1 & - 1 & 0 \\ 1 & - 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 0 & 0 & 1 \\ 0 & 1 & 2 \\ 1 & 2 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 2 & 1 & 1 \\ -1 & - 1 & - 1 \\ 1 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \\ \end{array} \right) $$

For symmetric real matrices, there is another definition of "diagonalize" that does not use eigenvalues; given symmetric $H,$ we solve $P^T H P = D,$ with $D$ diagonal, and $\det P = \pm 1.$ This is often called repeated completing the square. Maybe I should add that the original and the diagonal matrix here are called "congruent" (or in some cases "equivalent") rather than "similar."

$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 0 & 1 & 0 \\ 0 & - 2 & 1 \\ 1 & - \frac{ 1 }{ 2 } & \frac{ 1 }{ 4 } \\ \end{array} \right) \left( \begin{array}{rrr} 0 & 0 & 1 \\ 0 & 1 & 2 \\ 1 & 2 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 0 & 0 & 1 \\ 1 & - 2 & - \frac{ 1 }{ 2 } \\ 0 & 1 & \frac{ 1 }{ 4 } \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 4 & 0 \\ 0 & 0 & \frac{ 1 }{ 4 } \\ \end{array} \right) $$

$$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrr} 0 & 0 & 1 \\ 0 & 1 & 2 \\ 1 & 2 & 0 \\ \end{array} \right) $$

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$$ E_{1} = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 1 & 0 & 2 \\ 0 & 0 & 1 \\ 2 & 1 & 0 \\ \end{array} \right) $$

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$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & - 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & - 2 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 0 & 1 & 2 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & - 4 \\ \end{array} \right) $$

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$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 0 & 0 & 1 \\ 1 & - 2 & 0 \\ 0 & 1 & 0 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 0 & 1 & 2 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 4 & 1 \\ 0 & 1 & 0 \\ \end{array} \right) $$

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$$ E_{4} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & \frac{ 1 }{ 4 } \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{4} = \left( \begin{array}{rrr} 0 & 0 & 1 \\ 1 & - 2 & - \frac{ 1 }{ 2 } \\ 0 & 1 & \frac{ 1 }{ 4 } \\ \end{array} \right) , \; \; \; Q_{4} = \left( \begin{array}{rrr} 0 & 1 & 2 \\ - \frac{ 1 }{ 4 } & 0 & 1 \\ 1 & 0 & 0 \\ \end{array} \right) , \; \; \; D_{4} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 4 & 0 \\ 0 & 0 & \frac{ 1 }{ 4 } \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 0 & 1 & 0 \\ 0 & - 2 & 1 \\ 1 & - \frac{ 1 }{ 2 } & \frac{ 1 }{ 4 } \\ \end{array} \right) \left( \begin{array}{rrr} 0 & 0 & 1 \\ 0 & 1 & 2 \\ 1 & 2 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 0 & 0 & 1 \\ 1 & - 2 & - \frac{ 1 }{ 2 } \\ 0 & 1 & \frac{ 1 }{ 4 } \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 4 & 0 \\ 0 & 0 & \frac{ 1 }{ 4 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 0 & - \frac{ 1 }{ 4 } & 1 \\ 1 & 0 & 0 \\ 2 & 1 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 4 & 0 \\ 0 & 0 & \frac{ 1 }{ 4 } \\ \end{array} \right) \left( \begin{array}{rrr} 0 & 1 & 2 \\ - \frac{ 1 }{ 4 } & 0 & 1 \\ 1 & 0 & 0 \\ \end{array} \right) = \left( \begin{array}{rrr} 0 & 0 & 1 \\ 0 & 1 & 2 \\ 1 & 2 & 0 \\ \end{array} \right) $$